Answers
acceleration can not be negative hence in positive
Related Questions
An OP AMP has CMRR = 100 dB and open-loop gain =250000.
Determine the common mode gain ACM
Answers
Answer:
ACM = 2.5
Explanation:
The common mode rejection ratio, is a parameter that is used to measure the ability of an Operational-Amplifier, to reject or eliminate the noises of same polarity. The common mode rejection ratio is given by the following formula:
CMMR = AV/ACM
for the value of CMMR in deciBels (dB), the formula becomes:
CMMR = 20 log(AV/ACM)
where,
CMMR = Common Mode Rejection Ratio = 100 dB
AV = Open Loop Gain = 250000
ACM = Common Mode Gain = ?
Therefore,
100 = 20 log (250000/ACM)
100/20 = log (250000/ACM)
10⁵ = 250000/ACM
ACM = 250000/10⁵
ACM = 2.5
Question 2 of 10
According to the law of conservation of energy, which statement must be
true?
OA. The total energy of a system can increase only if energy enters the
system.
OB. Energy that is transferred cannot be transformed into a different
type of energy.
OC. A system cannot take in additional matter.
OD. The total energy in a system can only decrease over time.
SUBMIT
Answers
The total energy of a system can increase only if energy enters the
system.
option A.
What is the law of conservation of energy?The law of conservation of energy states that in a chemical reaction or an isolated system, the energy of the system is neither created nor destroyed but can be converted from one form to another.
The statements that are not true about the law of conservation of energy are;
Energy that is transferred cannot be transformed into a differenttype of energy.A system cannot take in additional matter.The total energy in a system can only decrease over time.Thus, the only true statement is;
The total energy of a system can increase only if energy enters the
system.
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6. As distance increases, gravitational force *
(10 Points)
increases
decreases
Answers
Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.
*By what factor does the speed of each object change if total work -18 J is done on each?
Answers
Answer:
the final speed of object A changed by a factor of \(\frac{1}{\sqrt{3} }\) = 0.58
the final speed of object B changed by a factor of \(\sqrt{\frac{5}{3} }\) = 1.29
Explanation:
Given;
kinetic energy of object A, = 27 J
let the mass of object A = \(m_A\)
then, the mass of object B = \(m_B = \frac{m_A}{4}\)
work done on object A = -18 J
work done on object B = -18 J
let \(v_i\) be the initial speed
let \(v_f\) be the final speed
For object A;
\(K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\\)
\(v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\\)
Thus, the final speed of object A changed by a factor of \(\frac{1}{\sqrt{3} }\) = 0.58
To obtain the change in the final speed of object B, apply the following equations.
\(K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B = \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\\)
\(\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i\)
Thus, the final speed of object B changed by a factor of \(\sqrt{\frac{5}{3} }\) = 1.29
Velocity is the rate of change of position with respect to time. The velocity of objects A and B is 0.577 and 1.29 times the initial velocities.
What is Velocity?Velocity is the rate of change of position with respect to time.
Given to us
The kinetic energy of object A before work is applied, \(KE_A = 27\rm\ J\)
Mass of object A = \(m_a\)
Mass of object B = \(m_b = \dfrac{m_a}{4}\)
Work done on both the objects, w = -18 J
We know that the kinetic energy of object A is 27 j, therefore,
\(KE_A = 27 J\\\\\dfrac{1}{2}m_av_i^2 = 27\\\\m_av_i^2 = 54\\\\m_a = \dfrac{54}{v_i^2}\)
We know that the work done on object A is -18 J, therefore, the difference in the Kinetic energy of the object will be work,
\(KE_f - KE_i = -18\\\\\dfrac{1}{2}m_Av_f^2 -\dfrac{1}{2}m_Av_i^2 = 18\\\\v_f^2 -v_i^2 = \dfrac{-18 \times 2}{m_A}\\\\v_f^2 -v_i^2 = \dfrac{-36}{m_A}\)
Substitute the value of \(m_a\),
\(v_f^2 -v_i^2 = \dfrac{-36}{m_A}\\\\v_f^2 -v_i^2 = \dfrac{-36}{\dfrac{54}{v_i^2}}\\\\v_f^2 -v_i^2 = \dfrac{-36 v_i^2}{54}\\\\v_f^2 = \dfrac{1v_i^2}{3}+v_i^2\\\\v_f^2 = \dfrac{-36 v_i^2}{54}\\\\v_f = v_i\sqrt{\dfrac{1}{3}}\\\\v_f = 0.577v_i\)
Hence, the velocity of object A is 0.577 times the initial velocity.
Kinetic Energy of the object B,
\(KE_B = \dfrac{1}{2}m_B v_i^2\\\\KE_B = \dfrac{1}{2}\dfrac{m_A}{4} v_i^2\\\\KE_B =\dfrac{m_A}{8} v_i^2\\\\KE_B =\dfrac{m_A}{8} v_i^2\\\\KE_B =\dfrac{54}{8} \\\)
We know that the work done on object B is -18 J, therefore, the difference in the Kinetic energy of the object will be work,
\(KE_f - KE_i = -18\\\\\dfrac{1}{2}m_Bv_f^2 -\dfrac{1}{2}m_Bv_i^2 = 18\\\\v_f^2 -v_i^2 = \dfrac{-18 \times 2}{m_B}\\\\v_f^2 -v_i^2 = \dfrac{-36}{m_B}\\\\v_f^2 -v_i^2 = \dfrac{-36}{\dfrac{m_A}{4}}\\\\v_f^2 -v_i^2 = \dfrac{-144}{m_A}\\\\}}\\\\v_f^2 -v_i^2 = \dfrac{-144}{\dfrac{54}{v_i^2}}\)
\(v_f^2 -v_i^2 = \dfrac{-144}{\dfrac{54}{v_i^2}}\\\\v_f^2 -v_i^2 = \dfrac{-144v_i^2}{54}\\\\v_f^2 = \dfrac{-144v_i^2}{54} + v_i^2\\\\v_f^2 = \dfrac{-144v_i^2}{54} + v_i^2\\\\v_f^2 = \dfrac{-5v_i^2}{3}\\\\ | v_f |= \sqrt{\dfrac{5v_i^2}{3}}\\\\v_f = \sqrt{\dfrac{5v_i^2}{3}}\\v_f = v_i\sqrt{\dfrac{5}{3}}\\\\v_f = 1.29 v_i\)
hence, the velocity of object B is 1.29 times the initial velocity.
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A roller coaster car is released from rest as shown in the image below. If
friction is neglected, the car will oscillate back and forth across the "dip" in
the roller coaster. What is the approximate velocity of the roller coaster car
each time it reaches the bottom of the roller coaster in the image? (Recall
that g = 9.8 m/s2.)
TAS
81 m
O A. 40 m/s
B. 25 m/s
C. 30 m/s
D. 15 m/s
Answers
Answer:
40m/s
Explanation:
a=g
u=0
s=81
v²=u²+2as
v²=2(9.81)(81)
v=√1587.6=39.8446985181≈40m/s
The velocity of the roller coaster car each time it reaches the bottom is 40 ms⁻¹. The correct option is (A).
The rate at which the position of an object changes with respect to time is described by the physical quantity known as velocity. It has both magnitude and direction because it is a vector quantity.
Given:
Initial velocity, u = 0 m/s
Acceleration, a = -9.8 ms⁻²
Distance, d = 81 m
From the third equation of motion:
v² = u² - 2as
v² = 0 - 2×(-9.8)×81
v = 40 ms⁻¹
Hence, the velocity of the roller coaster car is 40 ms⁻¹. The correct option is (A).
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9. Approximately how much kinetic energy does an object have from its
motion if its mass is 150-kg and it is moving at a speed of 5.0 meters per
second?
A. O 400 J
B. O 1,900 J
C. O 3,800 J
D. O 57,000 J
Answers
1875 N*m an object have from its motion if its mass is 150-kg and it is moving at a speed of 5.0 meters per second.
The correct option is E.
What is kinetic energy?The power an object has as a consequence of motion is known as kinetic energy. A force must be applied to an object in order to accelerate it. We must put in effort in case of applying a force. After the work has been finished, energy is transferred to the item, which further moves at a new, constant speed.
Who discovered kinetic energy?The idea of kinetic energy was first proposed in 1849 by William Thompson, who subsequently rose to the position of Lord Kelvin. The amount equal to one-half the object's mass times the square of its velocity is now what we refer to as the kinetic energy of an object.
Briefing:KE = 1/2*m*v²
KE = 1/2*150kg*(5 m/s)²
KE = 75kg * 25m²/s²
KE = 1875 kg*m²/s²
KE = 1875 N*m
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The complete question is -
Approximately how much kinetic energy does an object have from its
motion if its mass is 150-kg and it is moving at a speed of 5.0 meters per
second?
A. 400 J
B. 1,900 J
C. 3,800 J
D. 57,000 J
E. 1875 J
effort distance of a lever should be increased to lift the havier load give reason
Answers
The effort distance of a lever should be increased to lift a heavier load because it provides a mechanical advantage, allowing for easier lifting of the load.
The effort distance of a lever should be increased to lift a heavier load because it allows for a mechanical advantage that compensates for the increased weight.
In a lever system, the effort distance is the distance between the point of application of the input force (effort) and the fulcrum, while the load distance is the distance between the point of application of the output force (load) and the fulcrum. The mechanical advantage of a lever is determined by the ratio of the load distance to the effort distance.
By increasing the effort distance, the mechanical advantage of the lever system is increased. This means that for the same input force (effort), a greater output force (load) can be achieved. When dealing with a heavier load, a higher mechanical advantage is required to overcome the increased resistance.
By increasing the effort distance, the lever system can effectively multiply the applied force, making it easier to lift the heavier load. This allows for the redistribution of force and facilitates the efficient use of human effort in various applications, such as in construction, engineering, and even everyday tools like scissors and pliers.
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Write the percent as a fraction or a mixed number in simplest form.
7%
Answers
The figure shows a spherical hollow inside a lead sphere of radius R = 2.3 m; the surface of the hollow passes through the center of the sphere and “touches” the right side of the sphere. The mass of the sphere before hollowing was M = 388 kg. With what gravitational force does the hollowed-out lead sphere attract a small sphere of mass m = 27 kg that lies at a distance d = 14 m from the center of the lead sphere, on the straight line connecting the centers of the spheres and of the hollow?
Answers
Answer:
F = 303,615 10⁻¹¹ N
Explanation:
Let's analyze this problem a little, problem we are asked to find the attractive force of the large sphere and a small sphere, we can find separately the attractive force between the full large sphere and the small sphere, Let's call this force F1 and on the other hand the force is between a sphere representing the gap and the small sphere, let's call this outside F2, the net bone force of the large sphere with gap is the subtraction of these two forces.
F = F₁ -F₂
Let's start by finding the force of the sphere of complete
F₁ = G M₁ m / r²
the masses of the sphere is M = 388 kg and the distance is r = d = 14m
F₁ = 6.67 10⁻¹¹ 380 27/14²
F₁ = 356.50 10⁻¹¹ N
Now let's calculate the mass of the gap if large sphere
let's use the concept of density
ρ = m / V
the volume of the gap is
V = 4/3 π r³
For the radius of the hole they tell us that it touches one side and the center of the sphere, therefore its diameter is the radius of the large sphere
d = 2.3 m
r_hole = 1.15 m
V_hole = 4/3 π r³
V_hole = 4/3 π 1.15³
V_hole = 6.37 m³
let's look at the density of the large sphere
ρ = m / V
V = 4/3 π r³
V = 4/3 π 2.3³
V = 50,965 m³
ρ = 388 / 50,965
ρ = 7.613 kg / m³
this is the density of the sphere without the gap
the mass of the gap is
m_hole = ρ V
m_hole = 7,613 6.37
m_hole = 48.49 kg
the distance from the hole to the small building
r₂ = d - r_hueco
r₂ = 14 - 1.15
r₂ = 12.85 m
the strength of this is
F₂ = 6.67 10⁻¹¹ 48.49 27 / 12.85²
F₂ = 52,885 10⁻¹¹ N
The strength of the sphere with the gap is
F = F₁ -F₂
F = (356.50 - 52,885) 10⁻¹¹
F = 303,615 10⁻¹¹ N
The surface of the hollow passes through the center of the sphere and “touches” the right side of the sphere, the hollowed-out lead sphere attracts the small sphere with a gravitational force of approximately 2.84 × \(10^{-9\) N.
We may use Newton's law of universal gravitation, which states that the force of gravity between two objects is given by: to calculate the gravitational force the hollowed-out lead spherical exerts on the smaller sphere.
\(F = (G * m_1 * m_2) / r^2\)
V = (4/3) * π * \(R^3\)
The volume of the hollowed-out portion ( \(V_{hollow\)) is then:
\(V_{hollow\) = (1/2) * (4/3) * π * R^3 = (2/3) * π * \(R^3\)
Mass of the remaining lead sphere = M - (2/3) * M = (1/3) * M
F = \((G * m_1 * m_2) / r^2\)
F = (G * [(1/3) * M] * m) / \(d^2\)
Substituting the given values:
F = (6.674 × \(d^2 N m^2/kg^2\) * [(1/3) * 388 kg] * 27 kg) / \((14 m)^2\)
F ≈ 2.84 × \(10^{-9\)N
Thus, the hollowed-out lead sphere attracts the small sphere with a gravitational force of approximately 2.84 × \(10^{-9\)N.
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Which comparison is correct?
(1 point)
O HF is a stronger acid than HCI.
O H2S is a stronger acid than HCI.
O PH3 is a stronger acid than NH3.
O NH3 is a stronger acid than HF.
Answers
Considering their pKa, we can affirm that PH₃ is a stronger acid than NH₃.
We want to compare the strength of different acids. To do so, we need to compare their pKa.
What is the pKa?The pKa is the negative base 10 logarithm of the acid dissociation constant, Ka.
The lower the pKa, the stronger the acid.
Which comparison is correct?
HF is a stronger acid than HCI. FALSE. The pKa of HF is 3.1. and the pKa of HCl is 6.0. H₂S is a stronger acid than HCI. FALSE. The pKa₁ of H₂S is 7.04. and the pKa of HCl is 6.0. PH₃ is a stronger acid than NH₃. TRUE. The pKa of PH₃ is -5.5 and the pKa of NH₃ is 9.9. NH₃ is a stronger acid than HF. FALSE. The pKa of NH₃ is 9.9 and the pKa of HF is 3.1.Considering their pKa, we can affirm that PH₃ is a stronger acid than NH₃.
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Answer: C.) PH₃ is a stronger acid than NH₃.
Electromagnetic induction occurs when
a.
b.
a conductor is moved through a magnetic
a magnet is connected to an electrical circ
electrons are rubbed from one object onto
electrons flow from one chemical to anoth
C.
d.
your hd
Answers
Answer:
A
Explanation:
occurs when a magnetic feild and an elctric conductor move relative to one another
24- What happens to a given mass of water as it is cooled from 4°C to zero?
(a) expands
(b) contracts
(c) vaporizes
(d) Neither expands, contracts, nor vaporizes.
Answers
Explanation:
A-It starts to expand. Hence ice and water of same weight has different volumes
Expansion happens to a given mass of water as it is cooled from 4°C to zero. Option A is correct.
What is temperature?Temperature directs to the hotness or coldness of a body. In straightforward terms,
it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles more the temperature.
Water loses density when it is cooled from 4°C to 0°C. It turns out that the temperature at which liquid water has the most significant density is 4 degrees Celsius.
It will increase in size when heated or cooled. Since most liquids shrink when they are chilled, it is rare for water to expand when cooled to lower temperatures.
As a mass of water is cooled from 4°C to zero, expansion occurs.
Hence, option A is correct.
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Hendry throws an object vertically upwards at a velocity of 26,5 m⋅s-1 from the
edge of a cliff of height 120m. After some time, the projectile lands on the
ground below the cliff After 1 second, Cathy fires a second object upwards FROM THE
GROUND BELOW with a velocity of 45 m⋅s-1. Calculate the time and
distance will the two objects meet.
Answers
Hendry and Cathy will each throw an object, and the time and location at which they will collide can be determined using the laws of motion. Hendry's item had an initial velocity of 26.5 m/s, whereas Cathy's object had an initial velocity of 45 m/s. Hendry's object's equation of motion is given by: s = u*t + 0.5*a*t*2, where s is the displacement, u*t* is the starting velocity, t* is the time, and a*t* is the acceleration brought on by gravity.
The acceleration caused by gravity is negative since the item is being flung upward. The item that Cathy threw has the following equation of motion: s = u * t - 0.5 * a * t2.where s is the distance travelled, u is the starting speed, t is the passage of time, and an is the acceleration brought on by gravity. The acceleration caused by gravity is negative since the item is being flung upward.
These equations allow us to determine the location and timing of the two items' collision. By figuring out the two equations for t, one may determine the moment when the objects will collide. By changing the value of t in either equation, one may determine the distance at which the objects will collide. Therefore, using the equations of motion, it is possible to determine the moment and distance at which the two objects will collide.
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A projector lens projects an image from a 6.35 cm wide LCD screen onto a
screen 3.25 m wide. If the focal length of the projector lens is 13.8 cm, the screen
must be how far from the projector
Answers
Answer:
For any given projector, the width of the image (W) relative to the throw distance (D) is know as the throw ratio D/W or distance over width. So for example, the most common projector throw ratio is 2.0. This means that for each foot of image width, the projector needs to be 2 feet away or D/W = 2/1 = 2.0.
The graph shows the decay of two different radioactive nuclei, Source A and Source B.
How long does it take Source B to complete two half-lives?
a. 10 s
b. 5 s
c. 40 s
d. 20 s
Answers
It takes source B 40 s to complete two half lives.
What is the half life?The half-life of a chemical is the amount of time it takes for half of the atoms in a sample to decay. This concept is widely used to describe how quickly a substance decays radioactively in nuclear physics and radiochemistry. Each radioactive substance has a unique half-life due to the properties of each atomic nucleus.
Looking at the graph, we can see that the half life of the source B is 20s .
Hence the time taken to complete two half lives = 2(20 s) = 40 s.
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The equation of a displacement-time curve of a particle is given by X =10t- 5t + 6 = 0. Find the instantaneous velocities at t = 2sec and t = 5sec.
Answers
\(\\ \rm\hookrightarrow v={\displaystyle{\int}}sdt\)
Now
\(\\ \rm\hookrightarrow v={\displaystyle{\int}^5_2}10t-5t+6\)
\(\\ \rm\hookrightarrow v={\displaystyle{\int}^5_2}5t+6\)
\(\boxed{\sf {\displaystyle{\int}}x^n dx=\dfrac{x^{n+1}}{n+1}}\)
\(\\ \rm\hookrightarrow v=\left[\dfrac{5t^2}{2}+6t\right]^5_2\)
\(\\ \rm\hookrightarrow v=\dfrac{5\left\{5)^2-(2)^2\right\}}{2}+6(5-2)\)
\(\\ \rm\hookrightarrow v=\dfrac{5(25-4)}{2}+6(3)\)
\(\\ \rm\hookrightarrow v=\dfrac{5(21)}{2}+18\)
\(\\ \rm\hookrightarrow v=\dfrac{105}{2}+18\)
\(\\ \rm\hookrightarrow v=52.5+18\)
\(\\ \rm\hookrightarrow v=70.5m/s\)
An object with mass m is given initial velocity to slide across a horizontal plane AB towards a semi circle BCD with radius R.
Between the object and the plane exists a kinetic friction u_k=0.5, but only between the section FB with length R. F is in the middle of A and B. The inside of the circle is smooth.
When the object reached to point C, it applies a force of 3mg on the semi circle.
The object is going to the left.
1. Write an expression for the initial velocity at point A.
Answers
The expression for the initial velocity at point A is:
0 = (velocity at point A - 0) / time
Simplifying the equation, we find:
Velocity at point A = 0
The initial velocity at point A is zero, indicating that the object starts from rest before sliding on the horizontal plane AB.
To write an expression for the initial velocity at point A, we need to analyze the forces acting on the object and apply the principles of motion.
Given:
Mass of the object, m
Radius of the semi circle, R
Coefficient of kinetic friction, μ\(_k\) = 0.5
Force applied at point C, F = 3mg
The object is initially at rest.
Let's break down the motion into two parts: the motion on the horizontal plane AB and the motion along the semi circle BCD.
1. Motion on the horizontal plane AB:
The only force acting on the object on the horizontal plane is the force of kinetic friction. The frictional force can be calculated using:
Frictional force, f = μ\(_k\)* Normal force
The normal force is equal to the weight of the object, which is mg.
Normal force, N = mg
Frictional force, f = μ\(_k\) * mg
The frictional force acts in the opposite direction to the motion, so its magnitude is negative. Thus, the net force on the object on the horizontal plane is:
Net force = -f = -μ\(_k\)* mg
Using Newton's second law, we can relate the net force to the acceleration:
Net force = mass * acceleration
-μ\(_k\) * mg = m * acceleration
The acceleration can be expressed as the rate of change of velocity:
Acceleration = (final velocity - initial velocity) / time
Since the object is initially at rest, the initial velocity is zero.
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The image below shows an onion root viewed under a microscope.
What does this image best demonstrate?
A.
Living organisms are composed of one or more cells.
B.
Cells of different organisms have a similar composition.
C.
New cells only arise from pre-existing cells.
D.
Each cell can perform all functions necessary for life.
Answers
Answer:
Explanation:
a
What is the mass of a block of lead that is 30cm by 80cm by 60cm?
Answers
To calculate the mass of the block, we need to use the formula:
mass = density x volume
The density of lead is approximately 11.34 grams per cubic centimeter.
The volume of the block is:
30cm x 80cm x 60cm = 144,000 cubic centimeters
Therefore, the mass of the block of lead is:
11.34 g/cm3 x 144,000 cm3 = 1,632,960 grams
So the mass of the block of lead is 1,632,960 grams.
The mass of the block of lead is 1632.96 kg.
To find the mass of a block of lead with dimensions 30 cm by 80 cm by 60 cm, we need to know the density of lead. The density of lead is 11.34 g/cm³. The formula for mass is
mass = density x volume.
The volume of the lead block is 30 cm x 80 cm x 60 cm = 144000 cm³.
Therefore, mass = 11.34 g/cm³ x 144000 cm³ = 1632960 g.
The mass of the block of lead is 1632.96 kg.The mass of a block of lead is determined by the formula
mass = density x volume.
The density of lead is 11.34 g/cm³. The block of lead has dimensions of 30 cm by 80 cm by 60 cm, therefore its volume can be calculated as 30 cm x 80 cm x 60 cm = 144000 cm³.The formula for the mass of an object is
mass = density x volume.
From the provided values, we can calculate the mass of the block of lead as follows:
mass = 11.34 g/cm³ x 144000 cm³ = 1632960 g.Since 1 kg = 1000 g,
we can convert the mass from grams to kilograms by dividing the answer by 1000:
mass = 1632960 g / 1000 = 1632.96 kg.
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A rotating Steel shaft (as shown in Figure 4.35) used in a spur gear reducer is subjected to a constant torque together with lateral forces that tend to bend it downward in the center producing fluctuating stresses of 60 MPa bending and steady stresses of 80 MPa torsion. These values, however, do not take into account stress concentrations caused by the stepped shaft (shoulder). The fatigue stress concentrations caused by the shoulder of the shaft are K_fb = 1.53 for bending and K_ft = 1.28 for torsion. The surfaces are in machined and the Steel has a hardness of 200 Bhn and strength values of S_u = 700 MPa and S_y = 550 MPa The modified endurance limit is 239 MPa The stepped shaft has D/d =1.2, d = 30 mm, and r = 3 mm. Find: Determine the safety factor with respect to infinite fatigue life. Schematic and Given Data:
Answers
Answer: 24
Explanation:
A skydiver jumped from a plane. As she fell, her speed kept increasing until it reached terminal velocity of 150 miles per hour at 15 seconds. At terminal velocity, an object's speed (velocity) stops increasing due to air resistance. At 40 seconds after her jump from the plane, she opened her parachute. After the parachute opened, her descent speed rapidly decreased until it reached 20 miles per hour just after 60 seconds into the dive. Which of these graphs best represents the skydiver's kinetic energy during her dive?
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The graph that best represents the skydiver's kinetic energy during her dive is a graph that starts at 0 at the beginning of the dive, increases to a peak at 15 seconds when the skydiver reaches terminal velocity, and then decreases rapidly after 40 seconds when the parachute is opened.
How do we sip water from glass using straw?can we sip water with straw on moon?
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A particle with charge q and mass m moving with speed v in the x direction enters a magnetic field of strength B pointing in the y direction. The magnitude of the acceleration of the particle as it travels one semicircle is
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Answer:
The acceleration of the particle as it travels one semicircle is
\(a= \frac{\pi R}{t^2}\)
Explanation:'
Kindly see attached a sketch of a semi-circle
Step one:
given data
velocity =v
let the time taken be t
The path PQM is the distance covered
so distance \(d= \pi R\)
we know that time= distance/velocity
t= πR/v
step two:
velocity =distance/time
\(velocity=\frac{\pi R}{t}\)
also, we know that acceleration is velocity/time
\(a= \frac{\pi R}{\frac{t}{t} }\)
\(a= \frac{\pi R}{t}*\frac{1}{t}\)
\(a= \frac{\pi R}{t^2}\)
As you look at the side of the wheel, you can see it spinning clockwise. What are the directions of the angular velocity, and angular momentum?
If the torque is applied to the rim, causing the wheel to slow down, what are the directions of the torque and the angular acceleration?
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Answer:
The direction of angular velocity and angular momentum are perpendicular to the plane of rotation. Using the right hand rule, the direction of both angular velocity and angular momentum is defined as the direction in which the thumb of your right hand points when you curl your fingers in the direction of rotation.
Explanation:
please help
why does 45° produce a max. range?
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Answer:
i do not know
Explanation:
Suppose one Sherpa uses a force of 980 N to move a load of equipment to a height of 20 meters in 25 seconds. How much power is used?
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F = 980 N
h = 20 m
t = 25 s
P=? (power)
W=F*h (work)
P=W*t
P=F*h*t
P=980*20*25 =490000 W = 490 kW = 0.49 MW
What is the mass of a block of lead that is 30cm by 80cm by 60cm?
Answers
Calculating the mass of the block requires a bit of work. The formula for the volume of a rectangular solid is V = l*w*h, where V is the volume, l is the length, w is the width, and h is the height. Using the dimensions given, we can calculate the volume of the block as 30*80*60 = 144000 cubic centimeters.
The density of lead is approximately 11.34 grams per cubic centimeter. To calculate the mass of the block, we can use the formula m = V*d, where m is the mass, V is the volume, and d is the density. Plugging in the values we get m = 144000*11.34 = 1,634,400 grams or approximately 1.63 metric tons.
So, the mass of the block of lead is approximately 1.63 metric tons.
What force acts on a projectile in the horizontal direction?
Answers
The force that acts on a projectile in the horizontal direction is Gravitational force.
A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.
Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity. Horizontal forces are not required to keep a projectile moving horizontally. Hence, The only force acting upon a projectile is gravity.
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An 5kg object is released from rest near the surface of a planet. The vertical position of the object as a function of time is shown in the graph. All frictional forces are considered to be negligible. What is the closest approximation of the weight of the object.
a) 300N
b) 30N
c) 5N
d) 150N
Answers
Answer:
The correct option is b: 30 N.
Explanation:
First, we need to find the acceleration due to gravity (a):
\( y_{f} - y_{0} = v_{o}t - \frac{1}{2}a(\Delta t)^{2} \) (1)
Where:
\(y_{f}\): is the final vertical position (obtained from the graph)
\(y_{0}\): is the initial vertical position (obtained from the graph)
v₀: is the initial speed = 0 (it is released from rest)
Δt: is the variation of time (from the graph)
From the graph, we can take the following values of height and time:
t₀ = 0 s → \(t_{f}\) = 5 s
y₀ = 300 m → \(y_{f}\) = 225 m
Now, by entering the above values into equation (1) and solving for "a" we have:
\( a = 2\frac{y_{0} - y_{f}}{(t_{f} - t_{0})^{2}} = 2\frac{300 m - 225 m}{(5 s - 0)^{2}} = 6 m/s^{2} \)
Finally, the weight of the object is:
\( W = ma = 5 kg*6 m/s^{2} = 30 N \)
Therefore, the correct option is b: 30 N.
I hope it helps you!
A circle loop of radius 2 m is positioned in a uniform magnetic field of magnitude 1.5 N/C so that the plane of the loop makes an angle of 65° with the magnetic field. Find the flux passing through the circle loop.
Answers
The flux through the circle loop of wire is determined as 7.96 N/C.m².
What is the flux passing through the circle loop?The flux through the circle loop of wire is calculated by applying the following formula.
Ф = EA cosθ
where;
B is the magnitude of the electric fieldA is the area of the circular loopθ is the direction of the loopThe area of the wire is calculated as follows;
A = πr²
A = π (2 m)²
A = 12.57 m²
The flux through the circle loop of wire is calculated as;
Ф = EA cosθ
Ф = 1.5 x 12.57 x cos (65)
Ф = 7.96 N/C.m²
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