Answer:I think the answer is 5.43 I don’t rlly know
Explanation:
A boy throws an arrow at an original speed of 2m / s to create an angle 0 referring to the balloon at a distance of 3m from the departure point. Calculate the angle 0 and the height of the arrow. Let g = 10m / s2.
Calculate the horizontal component of the velocity. The horizontal component of the velocity is given by:
v_x = v * cos(theta)
where v is the original speed of the arrow and theta is the angle of projection.In this case, v = 2 m/s and theta is unknown. Solving for theta, we get:
theta = arccos(v_x / v)
theta = arccos(2 / 2) = 45 degrees
Calculate the vertical component of the velocity. The vertical component of the velocity is given by:
v_y = v * sin(theta)
In this case, v = 2 m/s and theta = 45 degrees. Solving for v_y, we get:
v_y = 2 * sin(45 degrees) = 1.414 m/s
Calculate the time of flight. The time of flight is given by:
t = 2 * v_y / g
In this case, v_y = 1.414 m/s and g = 10 m/s^2. Solving for t, we get:
t = 2 * 1.414 / 10 = 0.283 seconds
Calculate the height of the arrow. The height of the arrow is given by:
y = v_y * t - 0.5 * g * t^2
In this case, v_y = 1.414 m/s, t = 0.283 seconds, and g = 10 m/s^2. Solving for y, we get:
y = 1.414 * 0.283 - 0.5 * 10 * 0.283^2 = 0.303 meters
Therefore, the angle of projection is 45 degrees and the height of the arrow is 0.303 meters.
Water flows at a speed of 13 m/s through a pipe that has a diameter of 1.2 m. What is the
diameter of the smaller end of the pipe that the water comes out with a speed of 30 m/s?
The diameter of the smaller end of the pipe is approximately 0.78 meters.
To determine the diameter of the smaller end of the pipe, we can use the principle of conservation of mass. According to this principle, the mass flow rate of water should remain constant throughout the pipe.
The mass flow rate is given by the equation:
Mass flow rate = density of water * cross-sectional area * velocity
Since the density of the water remains constant, we can write:
Cross-sectional area1 * velocity1 = Cross-sectional area2 * velocity2
Given that the velocity1 is 13 m/s, the diameter1 is 1.2 m, and the velocity2 is 30 m/s, we can solve for the diameter2 using the equation:
(pi * (diameter1/2)^2) * velocity1 = (pi * (diameter2/2)^2) * velocity2
Simplifying the equation:
(1.2/2)^2 * 13 = (diameter2/2)^2 * 30
Calculating the equation:
(0.6)^2 * 13 = (diameter2/2)^2 * 30
0.36 * 13 = (diameter2/2)^2 * 30
4.68 = (diameter2/2)^2 * 30
Dividing both sides by 30:
0.156 = (diameter2/2)^2
Taking the square root of both sides:
0.39 = diameter2/2
Multiplying both sides by 2:
0.78 = diameter2
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A -9.00 μC point charge and +7.00 μC point charge are placed along the x-axis at x = 0.000 m and x = 0.4 m, respectively. Where must a third charge be placed along the x-axis (relative to the origin) so that it does not experience any net electric force due to the other two charges? The position on the x-axis is in meters and use three significant digits.
Answer:
x₃ = 0.397 m
Explanation:
For this exercise we use the vector sum of the forces, where the force is electric given by the Coulomb equation
F = \(k \frac{q_1q_2}{r^2}\)
We also use that charges of the same sign repel and charges of a different sign attract.
In this case the force between load 1 and 3 has one direction and the force between 2 and 3 has the opposite direction, in the exercise they ask that the force on load 3 be zero
∑ F = F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
we write the expressions of the forces, the distances are
r₁₃ = (x₁-x₃) ²
r₂₃ = (x₂ - x₃) ²
\(k \frac{q_1 \ q_3}{(x_1-x_3)^2} = k \frac{q_2 \ q_3}{(x_2-x_3)^2 }\)
x₁ = 0 m
\(\frac{q_1}{x_3^2 } = \frac{q_2 }{(x_2 - x_3)^2 }\)
q₁ (x₂ - x₃) ² = q₂ x₃²
\(\sqrt{\frac{q_1}{q_2} }\) (x₂ -x₃) = x₃
we substitute the values
\(\sqrt\frac{9}{7} }\) (0.4 -x₃) = x₃
x₃ (1 + \(\sqrt{9/7}\)) = \(\sqrt{9/7}\) 0.4
x₃ (1.13389) = 0.453557
x₃ = 0.453557 / 1.13389
x₃ = 0.397 m
Three objects are brought close to each other, two at a time. When objects A and B are brought together, they repel. When objects B and C are brought together, they also repel. Which of the following are true?
a. additional experiments must be performed to determine the signs of hte cahrges.
b. objects A and C possess charges of the samesign.
c. one object is neutral.
d. objects A and C possess charges of the oppostite sign.
e. all three objects possess charges of the same sign.
Answer:
option (e) = all three objects possess charges of the same sign.
Explanation:
To understand this particular problem or question, one has to take each point from the question/problem above with care;
=> "Three objects are brought close to each other, two at a time. "
Deduction: Only two are brought together at a time.
=> "When objects A and B are brought together, they repel."
DEDUCTIONS: A and B do not attract which is because A and B POSSESS THE SAME SIGN.
=> "When objects B and C are brought together, they also repel."
DEDUCTION: A and C POSSESS THE SAME SIGN.
CONCLUSION: Thus, If A and B do not attract and B and C also repel, then, it is possible that A, B and C are of the same sign. The law in which this follows is known as the law of transitivity.
How would you write the definition of heat in your own words
answer:Heat is the form of energy that is transferred between systems or objects with different temperatures (flowing from the high-temperature system to the low-temperature system). Also referred to as heat energy or thermal energy. Heat is typically measured in Btu, calories or joules.
explained:
sorry for wrong
wavelength = 5.4 cm and frequency = 9.4 Hz
What is the speed
A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 55.5 m above ground level, and the ball is fired with initial horizontal speed . The projectile lands at a distance D = 140 m from the cliff. Assume that the cannon is fired at time t = 0 and that the cannonball hits the ground at time . a. What is the value of ? b. What is the y position of the cannonball at the time c. Find the initial speed of the projectile.
a) The value of t u = 140/t`b.
b) The y position of the cannonball at the time t is 55.5 mc.
c) The initial speed of the projectile is 52.4 m/s.
Given that a cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 55.5 m above ground level, and the ball is fired with initial horizontal speed u. The projectile lands at a distance D = 140 m from the cliff. Assume that the cannon is fired at time t = 0 and that the cannonball hits the ground at time t.Now,We have to find the value of t, y position of the cannonball at the time t and the initial speed of the projectile.
a. To find the value of t:Here, we have to use the formula of distance
i.e.,S = ut + (1/2)gt², Where S = 140 m, u = u and g = 9.8 m/s².Hence,140 = u×t ………..(1)We know that, time taken by the cannonball to hit the ground can be calculated as,`(2H)/g`
Since the height of the cannon from the ground is 55.5m, the total height of the cannonball from the ground is
(2H) = 2 × 55.5
= 111 m`2H/g
= 111/9.8`
= 11.32653 s
From equation (1),u×t = 140u = 140/t
Therefore, `u = 140/t`b.
b)To find the y position of the cannonball at the time t:
Here, we have to use the formula of height i.e.,y = u×t – (1/2)gt²,
Where, y = height of the cannonball at time t, u = 140/t, t = time taken by the cannonball to hit the ground and g = 9.8 m/s².
We have already calculated the time taken by the cannonball to hit the ground in the previous step.`
y = 140 - (1/2) × 9.8 × t²`
On substituting the value of t as `t = 11.32653`,
we get,y = 140 - (1/2) × 9.8 × (11.32653)²= 55.5 mc.
c) To find the initial speed of the projectile:
To calculate the initial speed of the projectile, we need to use the formula of range of projectile
.i.e.,R = u²sin2θ/g
Where R = 140 m, g = 9.8 m/s², θ = 0° (horizontal)
u² = R × g/sin2θ
= 140 × 9.8/sin0°
= 2744m²/s²u
= \(\sqrt(2744m^2/s^2)\)
= 52.4 m/s
Hence, the initial speed of the projectile is 52.4 m/s.
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Acceleration vs. Mass
16
The graph shows the acceleration of the cart as the
mass of the cart varies.
As the mass of the cart increases, the acceleration of
the cart increases
The type of relationship that exists between the mass of
the cart and the acceleration of the cart is
relationship.
Answer:
decreases, inverse
Explanation:
if you are on edge, i'm helping you, go run
To solve this we must be knowing each and every concept related to direct relationship and inverse relationships. Therefore, there is direct relationship that exists between the mass of the cart and the acceleration of the cart.
What is direct relationship?In a direct connection, one variable's growth is directly proportionate to another's. Using the ball from the previous section as an example, a ball will bounce higher the higher it is dropped.
The way inverse relationships function varies. The quantity of y drops when x is raised. For instance, your travel time will be reduced if you go more swiftly to your destination. There is direct relationship that exists between the mass of the cart and the acceleration of the cart.
Therefore, there is direct relationship that exists between the mass of the cart and the acceleration of the cart.
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You apply a very small force, say 0.001 newtons, to a very large truck, with a mass of 2000 kilograms. What can you say for sure about what will happen to the truck?
The truck will accelerate (move from rest to motion) as long as that tiny force is larger than any force of friction that opposes it.
The truck will not move because small forces cannot move large objects.
The truck will accelerate (move from rest to motion) as long as the small applied force is large enough to overcome the inertia of the truck.
An extremely small force can never accelerate such a large truck. The truck will not move under any circumstance.
Answer:
it will stay still. unless its in space.
Explanation:
Electric Field Direction only OO Voltage Values Grid 00V a +1 nc -1 nc Sensors Now, let's look at how the distance from the charge affects the magnitude of the electric field. Select Values on the menu, and then click and drag one of the yellow E-Field Sensors. You will see the magnitude of the electric field given in units of V/m (volts per meter, which is the same as newtons per coulomb). Place the E-Field Sensor 1 m away from the positive charge (1 m is two bold grid lines away if going in a horizontal or vertical direction), and look at the resulting field strength. Consider the locations to the right, left, above, and below the positive charge, all 1 m away. For these four locations, the magnitude of the electric field is. greatest to the right of the charge. greatest below the charge. greatest above the charge. greatest to the left of the charge. O O O the same. Submit Request Answer
Considering the locations to the right, left, above, and below the positive charge, all 1 mm away. For these four locations, the magnitude of the electric field is the same.
The area, space, or field around it is an electric field of an isolated charge. There are mainly two types of electric fields i.e., static and dynamic. Moving charges produced dynamic electric fields whereas static electric fields are produced by stationary charges.
Direction and magnitude do not change over time for static electric fields. The direction can be positive or negative which is determined by the charge of the source.
The electric field formula is the electric field magnitude at a certain point from the charge Q, and it hangs on two factors- the distance r from the point to the origin Q and the amount of charge at the origin Q.
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The correct question is:
Now, let's look at how the distance from the charge affects the magnitude of the electric field. Select Values on the menu, and then click and drag one of the yellow E-Field Sensors. You will see the magnitude of the electric field given in units of V/mV/m (volts per meter, which is the same as newtons per coulomb). Place the E-Field Sensor 1 mm away from the positive charge (1 mm is two bold grid lines away if going in a horizontal or vertical direction), and look at the resulting field strength.
Consider the locations to the right, left, above, and below the positive charge, all 1 mm away. For these four locations, the magnitude of the electric field is________________.
A bar has a 20 N weight at one end, as shown in Figure 7.22. You have a weight of 15 N to hang somewhere on the bar so that the bar is in equilibrium. Where would you hang the 15 N weight on each of these bars? Consider the bar to have no mass.
The weight of 15N should be hung at a distance of 4/3 units from the fulcrum, on the opposite side of the bar to a weight of 20N.
The fulcrum is attained equilibrium when the principle of moments is equal to zero. The principle of momentum is defined as the body is said to be balanced when the clockwise movement about the point is equal to the anticlockwise movement about the same point.
When a weight of 15N is hung on the bar, it produces an anticlockwise movement with the same magnitude, opposes the clockwise moment produced by the weight of 20N on the fulcrum.
To find the distance (d) at which the 15N weight should be hung, the distance from the 20N weight to the fulcrum is 1 unit.
20×1 = 15×d
d = 20/15
=4/3 m
Thus, the weight of 15N should be hung by the distance of 4/3 units from the fulcrum, on the opposite side of the bar having the weight of 20N, to make the fulcrum to be at equilibrium.
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which are the four major components of the compression refrigeration cycle in order
Answer:
Compression, Condensation, Expansion, Evaporation
Question 15 of 32
A bungee jumper jumps off a bridge and bounces up and down several times.
She finally comes to rest 30 m below the bridge from which she just jumped.
If her mass is 50 kg and the spring constant of the bungee cord is 10 N/m,
how much energy was lost due to air resistance while she was bouncing?
(Recall that g = 9.8 m/s²)
A. 7330 N
B. 9200 N
C. 10,200 N
D. 8605 N
C. 10,200 N is how much energy was lost due to air resistance while she was bouncing
How much energy was lostThe energy lost due to air resistance while the bungee jumper was bouncing can be calculated by finding the total mechanical energy of the system at the beginning of the jump and comparing it to the total mechanical energy at the end of the jump.
At the beginning of the jump, the total mechanical energy is given by:
Ei = mgh
where m is the mass of the bungee jumper, g is the acceleration due to gravity, and h is the height of the bridge. Therefore, at the beginning of the jump:
50 x 30 x 10 - 1/2 x 30^2 x 10
= 15000 - 4500
= 10,200 N
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A key challenge with renewable energy is that the energy must be transported to the place where it's needed _ or devices that store the energy needed improvement.
A key challenge with renewable energy is that the energy must be transported to the place where it's needed batteries or devices .
Option A is correct.
What is renewable energy?Renewable energy sources such as solar, wind, and hydroelectric power are often located far away from the areas where the energy is needed. This makes it necessary to transport the energy over long distances, which can be expensive and lead to losses due to transmission and distribution. In addition, renewable energy sources are often intermittent, meaning they don't produce a constant supply of energy.
To address this, energy storage devices are needed to store excess energy produced during peak times for use during times of low energy production. However, the current technology for energy storage is not yet efficient or cost-effective enough to meet the growing demand for renewable energy. Developing more effective energy storage devices is therefore an important area of research and development in the field of renewable energy.
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Question incomplete:
A key challenge with renewable energy is that the energy must be transported to the place where it's needed _ or devices that store the energy needed improvement.
A. batteries
B. refineries
C. solar panels
D. wind turbines
Answer:
batteries
Explanation:
A car travels at a speed of 60 km/h for 4 minutes and 10 seconds. Calculate the distance it has traveled.?
Hello..! :)
We first apply the data to the problem.
Data:
V = 60km/hT = 4.10 minD = ?Now, we apply a conversion.
Conversion:
60km/h • (1000m/1km) • (1h/3600s)V = 16.6m/s 4.10 • (60s/1min)T = 246sThen, we apply the formula that is.
Formula:
D = V • TFinally we develop the problem.
Developing:
D = (16.6m/s) • (246s)
D = 4083.6m
He traveled a space of 4083.6 meters.
¿Doubts? On the comments. Greetings :D
When solving vector addition problems you can use either the graphical
method or the
Answer :the resultant of two vectors can be found using either the parallelogram method or the triangle method. don't know if this was helpful ?
Explanation:
Answer:
Analytical method.
3. What is the net force experienced by a box of mass 50 kg moving down a
frictionless plane inclined at a 60 degree angle?
If you jump off a 5 m diving platform how long will take you hit the water
Answer:
t = 1.0s
Explanation:
^d = v_0*t + (1/2)*a*t²
(5m-0m) = 0m/s*t + (.5)(9.81m/s²)*(t)²
5m = (4.905m/s²)*t²
t² = 1.01937s²
t = √1.01937s²
t = 1.0s
By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect the earth's rotational effects
Answer:
\(Weight\ loss=1.6321N\)
Explanation:
From the question we are told that:
Weight \(W=85.9kg\)
Altitude \(h= 6.33 km\)
Let
Radius of Earth \(r=6380km\)
Gravity \(g=9.8m/s^2\)
Generally the equation for Gravity at altitude is mathematically given by
\(g_s=9.8(\frac{6380}{6380+6.33})^2\)
\(g_s=9.781m/s^2\)
Therefore
Weight at sea level
\(W_s=9.8*85.9\)
\(W_s=841.82N\)
Weight at 6.33 altitude
\(W_a=9.781*85.9\)
\(W_a=840.2N\)
Therefore
\(Weight loss=W_s-W_b\)
\(Weight loss=841.82-840.2\)
\(Weight loss=1.6321N\)
3. A car traveling initially at 7.0 m/s accelerates uniformly until it reaches 80.0 m/s. If the car
took 245 s to accelerate, what is its acceleration?
Answer:
1.1111.22222
Explanation:
What is the difference between classical mechanics and quantum mechanics?
Classical mechanics describes the motion of objects on a macroscopic scale, while quantum mechanics deals with the behavior of particles on a microscopic scale. Classical mechanics is deterministic, meaning that it predicts precise outcomes based on initial conditions, while quantum mechanics is probabilistic, providing probabilities of different outcomes. Classical mechanics follows the principle of causality, where every effect has a specific cause, whereas quantum mechanics introduces inherent uncertainty and wave-particle duality. Classical mechanics is well-suited for describing everyday objects, while quantum mechanics is necessary to explain the behavior of particles at the atomic and subatomic levels.
~~~Harsha~~~
why might calcium be a important in the diet of many living things
Answer:
Calcium is an essential mineral that is important in the diet of many living things as it plays several important roles in the body:
1. Bone and teeth formation: Calcium is a key component of bones and teeth, making them strong and healthy.
2. Muscle function: Calcium plays a critical role in muscle contraction and relaxation, helping muscles function properly.
3. Nerve function: Calcium is involved in the transmission of nerve impulses, which allows for proper communication between nerve cells.
4. Blood clotting: Calcium is required for blood clotting, which is important for preventing excessive bleeding after an injury.
5. Cellular signaling: Calcium is involved in many cellular signaling pathways, helping to regulate various physiological processes in the body.
Therefore, having an adequate amount of calcium in the diet is crucial for the overall health and well-being of many living organisms.
Complete the statement: As the temperature of water changes from 105°C to 101°C.
the PE increases
the PE decreases
the KE increases
the KE decreases
As the temperature of water changes from 105°C to 101°C, the KE decreases.
Explanation:
KE, or kinetic energy, is the energy associated with the movement of particles. As the temperature of water decreases, the movement of particles slows down, which means that the kinetic energy decreases. Potential energy, or PE, is the energy associated with the position or arrangement of particles. In the case of water, the potential energy does not change significantly with changes in temperature. Therefore, as the temperature of water changes from 105°C to 101°C, the KE decreases.
A space probe is sent (from the surface of the Earth) to intercept a comet when it is at a distance
of 3 × 10^11 m from the Sun. Find the minimum energy required to move a 100-kg payload to
that location. [MEarth = 6 × 10^24 kg; REarth = 6.37 × 10^6 m; MSun = 2.0 × 10^30 kg; radius of
Earth’s orbit = 1.5 × 10^11 m.]
The energy that is required by the space probe is 2.94 * 10^14 J.
What is the energy required?Let us recall that the gravitational potential energy is the energy that is possessed by a body by virtue of the position of the body in space. We know that the energy that is possessed by the space probe is the gravitational potential energy and we have to obtain the energy required.
Recall that energy required = Work done = mgh
m = mass of the object
g = acceleration due to gravity
h = distance covered
Then;
100-kg * 9.8 m/s^2 * 3 × 10^11 m
= 2.94 * 10^14 J
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50 Joules of work in 25 seconds. How much power did she use?
P(W) = E(J) / t(s)
50/25=2
PLEASE HELP! Thank you!
Chloe and Sarah are driving bumper cars. Chloe, who is traveling west at 3.9 m/s, is behind Sarah, who is traveling west at 1.6 m/s. The total mass of Chloe’s car is 163 kg, and total mass of Sarah’s car is 179 kg. Immediately after Chloe collides with Sarah, Chloe’s velocity reduces to 0.95 m/s west. What is Sarah’s velocity immediately after the collision?
A. 5.2 m/s
B. 4.0 m/s
C. 4.3 m/s
D. 4.6 m/s
Sarah’s velocity immediately after the collision is 4.3 m/s west.
option C is the correct answer.
What is Sarah’s velocity immediately after the collision?Sarah’s velocity immediately after the collision is calculated by applying the principle of conservation of linear momentum as follows;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
m₁ is the mass of Sarah's carm₂ is the mass of Chloe's caru₁ is the initial velocity of Sarahu₂ is the initial velocity of Chloev₁ is the final velocity of Sarahv₂ is the final velocity of ChloeSarah’s velocity immediately after the collision is calculated as;
179 (1.6) + 163(3.9) = 179v₁ + 163(0.9)
922.1 = 179v₁ + 146.7
179v₁ = 775.4
v₁ = 4.3 m/s
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If two speeds are in exact opposite directions, the combined speed is the difference in the speeds.
A. True
B. False
Imagine that you are observing the region of space where a cloud of gas and dust is beginning to collapse inward to form a star (the object that initially forms in this process is called a protostar). Will the atoms in the collapsing cloud move away from one another, move closer to one another, or stay at the same locations? 2) What physical interaction, or force, causes the atoms to behave as you described they would in Question 1?
Answer:
1. The atoms in the collapsing cloud would move closer to one another.
2. Gravity
Explanation:
1. This is because there is a force of attraction between the atoms of the collapsing cloud. Thus, the atoms in the collapsing cloud would move closer to one another.
2. This is because gravity is the force of attraction between masses. Now, the collapsing cloud of dust contains mass of atoms, and as the atoms come together due to gravity, they form a larger mass which develops into a star.
(a) The atoms in the collapsing cloud would move closer to one another.
(b) Due to the force of interaction between masses that is gravity
What is force of attraction?
(a) The atoms will come closer
This is because there is a force of attraction between the atoms of the collapsing cloud. Thus, the atoms in the collapsing cloud would move closer to one another.
(b) Due to the gravitational attraction
This is because gravity is the force of attraction between masses. Now, the collapsing cloud of dust contains mass of atoms, and as the atoms come together due to gravity, they form a larger mass which develops into a star.
Hence
(a) The atoms in the collapsing cloud would move closer to one another.
(b) Due to the force of interaction between masses that is gravity
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a 16 kg box sitting at the top of an icy hill begins to slide down the essentially frictionless surface. at the bottom of the hill it collides with a spring loaded guardrail with a spring constant of 384 N/m. if the hill is 7.1 m high, how much did the box compress the spring?
The compression of the spring of spring constant 384 N/m is 2.41 m.
What is compression spring?Compression springs work by resisting and pushing back against any downward or inward force that tries to squash and hold them in a compressed state.
To calculate the compression of the spring, we use the formula below.
Formula:
e = √(2mgh/k)........... Equation 1Where:
e = Compression of the springm = Mass of the boxh = Heightg = Acceleration due to gravityk = Spring constantFrom the question,
Given:
m = 16 kgh = 7.1 mg = 9.8 m/s²k = 384 N/mSubstitute these values into equation 1
e = √(2×16×7.1×9.8/384)e = 2.41 mHence, the compression of the spring is 2.41 m.
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3. A car with a mass of 1600 kg has a kinetic energy of 125 000 J. How fast is it moving?
The car is moving at approximately 12.5 meters per second.
The kinetic energy (KE) of an object can be calculated using the formula:
KE = 1/2 * m * \(v^2\)
where
KE = kinetic energy,
m =Mass of the object, and
v = velocity.
In this case, we are given the mass (m) of the car as 1600 kg and the kinetic energy (KE) as 125,000 J. To find the velocity .
Substituting the values , we have:
125,000 J = 1/2 * 1600 kg *\(v^2\)
Now, we can solve for v by rearranging the equation:
\(v^2\) = (2 * 125,000 J) / 1600 kg
\(v^2\) = 156.25 \(m^2/s^2\)
Taking the square root, we find:
v = √156.25\(m^2/s^2\)
v ≈ 12.5 m/s
Therefore, the car is moving at approximately 12.5 meters per second.
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