What was Bear B's average speed in m/min? (No units required for answer. Type your answer.)

Momentum of the truck will be 70000 kg-m/s

Momentum can be defined as the impulse gained by a moving object.

Mass of the truck = m = 2000 kg

Velocity of the truck = v = 35 m/s

Momentum = M = m * v

M = 2000 * 35

M = 70000 kg-m/s

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NGC 300 is different from our solar system because of its smaller size.

NGC 300 galaxy is different from our solar system because of its size. NGC 300 is smaller than our solar system which means that it has less number of stars as compared to our solar system.

So we can conclude that NGC 300 is different from our solar system because of its smaller size.

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Answer:

B.  

Explanation:

Correlation is defined as a statistic that determine the relation between two variables.

Both the variables are important while finding the significance of correlation and no variable can alone claim the test results are significant. A test result is significant when both the variables are related to each other because it shows a significant probability that the relationship between the two variables exists.

Hence, the correct answer is "B. "

In correlation, variables are related to each other.

Option B is correct.

When two or more sets of data are linked together, they have a high correlation.

Data sets have a positive correlation when they increase together, and a negative correlation when one set increases as the other decreases.

Correlation refers to the degree of correspondence or relationship between two variables. Correlated variables tend to change together.

Therefore, option B  is correct.

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Answer:

34.5 m

Explanation:

Given data : An elephant produces 10 Hz sound wave. Assuming the speed of sound in air is 345 m/s. To find : What is the wavelength of sound ? Answer : The wavelength of the sound is 34.5 m.

e xddd  snoAnswer:n

E

xplanation:s no

Answer:

Explanation:

7.80ly=7.379370e+16m

There are 4 fundamental types of amplifiers, though they are categorized based on input and output parameters such as, current, voltage, trans conductor and trans resistance amplifier .

To provide the greatest amplified voltage gain, the input resistance must be infinite and the output resistance must be zero. Practically speaking, however, it can be deduced that the voltage amplifier must have a high input resistance and a low output resistance.

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Answer:

the minimum thickness of the coating  is 128.25 nm

Explanation:

The computation of the minimum thickness of the coating  is shown below:

In the case of non-reflective coatings, for determining the minimum film thickness needed we need to apply the following formula  

Thickness = t = Wavelength ÷ 4n

where

n represent index of refraction of the coating material.

t = 631 ÷ (4 × 1.23)

= 128.25 nm

Hence, the minimum thickness of the coating  is 128.25 nm

Answer:

Initially the PE of the object was W * h = 30 * 10 = 300 Joules

The KE of the object when it struck the ground was 1/2 M v^2

KE = 1/2 * 30/9.8 * 13^2 = 259 J

So the object lost 41 J to friction during the fall

Since Work = Force * distance

Force = 41 J / 10 m = 4.1 N  (the average force of friction)

The parent should push vertically downward at a distance of 4.66 m from the pivot point in order to hold teeter-totter level.

Force can be a push or a pull that is always with magnitude and direction.

As, Fp × x - Mg × d - mc × (L - x) = 0

Fp is force applied by the parent, x is distance from the pivot point to the point where the parent applies force, Mg is weight of the teeter-totter, mc is the weight of the child, and L is length of the teeter-totter.

Fp × x - (38 kg × 9.81 m/s²) × 2.6 m - (19 kg × 9.81 m/s²) × (5.2 m - x) = 0

Fp × x - 976.68 N × m - 185.58 N × m + 183.39 N × m = 0

Fp × x = 978.87 N × m

x = 978.87 N × m / Fp

So, x = 978.87 N × m / 210 N = 4.66 m

Therefore, the parent should push vertically downward at a distance of 4.66 m from the pivot point in order to hold the teeter-totter level.

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Answer:

From the second law of motion:

F = ma

we are given that the force applied on the block is 20N and the block accelerates at an acceleration of 4 m/s/s

So, F=  20N   and  a = 4 m/s/s

Replacing the variables in the equation:

20 = 4* m

m = 20 / 4

m = 5 kg

The total distance travelled by the object is 175 m.

option B.

The total distance travelled by the object is calculated by applying the following kinematic equation as shown below;

Total distance = area rectangle + area of triangle

Total distance = (15 s - 0 s) x (10 m/s - 0 m/s) + ¹/₂(5s)((10 m/s)

Total distance = (15s)(10 m/s) + (5s )(5 m/s)

Total distance = 150 m + 25 m

Total distance = 175 m

Thus, the total distance travelled by the object is sum of all the distance covered from 0 second to 15 seconds.

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Answer:

Depends on position

Explanation:

Rest:

A body is said to be at rest, if it does not change its position with respect to its surroundings.

Motion:

A body is said to be in motion, if it changes its position with respect to its surroundings.

The state of rest and motion is relative

The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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The angular velocity to nearest degree per second, gives:

A) 21584°/s

B) 79217°/s

C) 215848°/s

D) 551180°/s

The formula to convert frequency (f) in Hertz (Hz) to angular velocity (ω) in radians per second is:

ω = 2πf

To convert the result to degrees per second, we can multiply by 180/π, which is the conversion factor between radians and degrees.

Using this formula, we get:

A) ω = 2π × 60 = 376.99 rad/s ≈ 21584°/s

B) ω = 2π × 220 = 1382.94 rad/s ≈ 79217°/s

C) ω = 2π × 600 = 3769.91 rad/s ≈ 215848°/s

D) ω = 2π × 1.53 × 10^3 = 9621.89 rad/s ≈ 551180°/s

Rounding these results to the nearest degree per second, we get:

A) 21584°/s

B) 79217°/s

C) 215848°/s

D) 551180°/s

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Explanation:

it has no energy when considered with respect to earth ,as it has neither height (i e potential energy) nor velocity (i.e kinetic energy).

Answer:

I KNOW THE ANSWER IT WILL COST 30$

Explanation:

The buoyant force on the cannon stays the same as long as it is fully under water because the buoyant force is determined by the volume of fluid that the cannon displaces and not by the weight or mass of the cannon itself.

Archimedes' Principle states that the buoyant force on an object in a fluid is equal to the weight of the fluid that the object displaces. This means that as long as the cannon remains fully submerged in the water and does not change its volume, the amount of water it displaces and thus the buoyant force on the cannon will also remain the same.

In other words, the buoyant force is dependent on the fluid's density and the volume of the object, not its weight. So, as long as the volume of the cannon and the density of the fluid surrounding it remain constant, the buoyant force will also stay constant.

The buoyant force on the cannon will stay the same as long as it is fully under water.

The buoyant force on an object is equal to the weight of the fluid displaced by the object. The buoyant force is always directed upwards, and it opposes the force of gravity. As long as the cannon is fully under water, the amount of water displaced by the cannon will stay the same. This means that the buoyant force on the cannon will also stay the same.

The buoyant force on an object depends on the density of the fluid, the volume of the object, and the acceleration due to gravity. The density of water is constant, so the buoyant force on the cannon will only change if the volume of the cannon changes or if the acceleration due to gravity changes.

Neither of these factors change. The volume of the cannon does not change as it is being recovered from the shipwreck. The acceleration due to gravity also does not change, as it is the same on Earth's surface as it is underwater.

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a) The final pressure and temperature for the isothermal compression are \(4.5*10^5 Pa\) and 293 K, respectively, while  b) the final pressure and temperature for the adiabatic compression are\(5.58*10^5 Pa\) and 515 K, respectively.

a. Isothermal compression:

For an isothermal process, the temperature remains constant. Therefore, we can use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.

Since the process is isothermal, we can write:

\(P_1V_1 = P_2V_2\)

where P1 and V1 are the initial pressure and volume, and\(P_2\)and\(V_2\)are the final pressure and volume.

We are given that the volume is compressed to 1/3 of its original volume, so\(V_2 = (1/3)V_1\). Substituting this into the equation above gives:

\(P_2 = (V_1/V_2)P_1 = 3P_1\) = \(4.5*10^5 Pa\)

To find the final temperature, we can use the ideal gas law again:

PV = nRT

Rearranging, we get:

T = PV/(nR)

Substituting the values we know, we get:

T = (\(1.5*10^5\)Pa)(V1)/(nR)

Since the process is isothermal, the temperature remains constant, so the final temperature is the same as the initial temperature:

T2 = T1 = 293 K

b. Adiabatic compression:

For an adiabatic process, there is no heat transfer between the gas and its surroundings. Therefore, we can use the adiabatic equation:

PV^γ = constant

where γ = Cp/Cv is the ratio of specific heats.

Since the process is adiabatic and reversible, we can write:

\(P_1V_1\)^γ = \(P_2V_2\)^γ

We are given that the volume is compressed to 1/3 of its original volume, so V2 = (1/3)V1. Substituting this into the equation above gives:

\(P_2 = P_1(V_1/V_2)\)^γ = \(P_1\)\((3)^{(1.4)\) = \(5.58*10^5 Pa\)

To find the final temperature, we can use the adiabatic equation again:

\(T_2 = T_1(P_2/P_1)\)^((γ-1)/γ) = T1(5.58/1.5)^(0.4) = 515 K

Therefore, the final pressure and temperature for the isothermal compression are \(4.5*10^5 Pa\)and 293 K, respectively, while the final pressure and temperature for the adiabatic compression are \(5.58*10^5\) Pa and 515 K, respectively.

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Answer:

Explanation:

The magnitude of the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object, as shown in this formula. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function. Given a position vector →v=⟨a,b⟩, the magnitude is found by |v|=√a2+b2. The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application.

Answer:

potential until moved than kinetic

Explanation:

The potential V(x, y, z) everywhere inside the box. Formulas give V=0 at the center of this cube. Is E=0 there\((A_{n,m}e^{a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}}+B_{n,m}e^{-a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}})=\frac{16V_{0}}{nm\pi^{2}}\: \: \: n,m =odd\)

Laplace equation in cartesian co-ordinates is

\(\frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial y^2}+\frac{\partial^2 V}{\partial z^2}=0\)

Multiply both side by \(sin\left ( \frac{n'\pi x}{a} \right )sin\left ( \frac{m'\pi z}{a} \right )\)   and integrate over x and z from 0 to a

\(\int_{0}^{a}\int_{0}^{a}V_{0}sin\left ( \frac{n\pi x}{a} \right )sin\left ( \frac{m\pi z}{a} \right )dxdz=\frac{a^{2}}{4}(A_{n,m}e^{-a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}}+B_{n,m}e^{a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}})\)

\((A_{n,m}e^{-a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}}+B_{n,m}e^{a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}})=\frac{4V_{0}}{a^{2}}\int_{0}^{a}sin\left ( \frac{n\pi x}{a} \right )dx \int_{0}^{a}sin\left ( \frac{m\pi z}{a} \right )dz\)

\((A_{n,m}e^{-a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}}+B_{n,m}e^{a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}})=\frac{16V_{0}}{nm\pi^{2}}\: \: \: n,m =odd\)

Now apply the final boundary condition V(x, y=a/2, z) = V0

Solving we get

\((A_{n,m}e^{a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}}+B_{n,m}e^{-a/2\sqrt{(n^{2}+m^{2})\pi^{2}/a^{2}}})=\frac{16V_{0}}{nm\pi^{2}}\: \: \: n,m =odd\)

The Laplace equation is a partial differential equation that describes the behavior of a scalar field in space. In its simplest form, it states that the sum of the second partial derivatives of the scalar field with respect to each of the spatial dimensions is equal to zero. This means that the scalar field has no sources or sinks, and its value is determined only by the boundary conditions.

The Laplace equation has many applications in physics, engineering, and mathematics. For example, it can be used to model the behavior of electric and gravitational fields, fluid flow, and heat transfer. It is also used in solving problems involving potential functions, which arise in many areas of physics and engineering.

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Complete Question: -

You have a cubical box (sides all of length a) made of 6 metal plates which are insulated from each other. The left wall is located at y=-a/2, the right wall is at y=+a/2. Both left and right walls are held at constant potential V=V0. All four other walls are grounded. Find the potential V(x, y, z) everywhere inside the box. Do your formulas give V=0 at the center of this cube? Is E=0 there? (Should they be??)

Answer:

Energy Stored = 36000 J = 36 KJ

Explanation:

The power of a battery is given by the formula:

P = IV

where,

P = Power delivered by the battery

I = Current Supplied to the battery

V = Potential Difference between terminals of battery = 12 volt

Now, we multiply both sides by the time period (t):

Pt = VIt

where,

Pt = (Power)(Time) = Energy Stored = E = ?

It = Battery Current Rating = 50 A.min

Converting this to A.sec;

It = Battery Current Rating = (50 A.min)(60 sec/min) = 3000 A.sec

Therefore,

E = (12 volt)(3000 A.sec)

E = 36000 J = 36 KJ

Answer:

Explanation:

angle θ = 110 arcsec

= 110 /60 arcmin

= 1.8333 arcmin

= 1.8333 / 60 degree

= .030555 degree

= .030555 x π / 180 radian

= 5.33 x 10⁻⁴ radian .

linear diameter = angular diameter in radian x distance

= 5.33 x 10⁻⁴ x 25000

= 13.325 m

linear diameter of object = 13.325 m .

The mass of Car B is -6000 kg.

To solve this problem, we can apply the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

Therefore, we can write the equation for the conservation of momentum as:

(mass of Car A * velocity of Car A) + (mass of Car B * velocity of Car B) = (mass of Car A + mass of Car B) * velocity after collision

Let's substitute the given values into the equation:

(2000 kg * 10 m/s) + (mass of Car B * 0 m/s) = (2000 kg + mass of Car B) * (-5 m/s)

Simplifying the equation:

20000 kg*m/s = -5 m/s * (2000 kg + mass of Car B)

Dividing both sides by -5 m/s:

-4000 kg = 2000 kg + mass of Car B

Subtracting 2000 kg from both sides:

mass of Car B = -4000 kg - 2000 kg

mass of Car B = -6000 kg

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Correct question is;

A tiny water droplet of radius 0.010 cm descends through air from a high building .Calculate its terminal velocity . Given that η of air = 19 × 10^(-6) kg/m.s and density of water ρ = 1000kg/ms

Answer:

1.146 m/s

Explanation:

We are given;

Radius; r = 0.010 cm = 0.01 × 10^(-2) m

η = 19 × 10^(-6) kg/m.s

ρ = 1000 kg/ms

The formula for the terminal velocity is given by;

V_t = 2r²ρg/9η

g is acceleration due to gravity = 9.8 m/s²

Thus;

V_t = (2 × (0.01 × 10^(-2))² × 1000 × 9.8)/(9 × 19 × 10^(-6))

V_t = 1.146 m/s

From the question;

1) It takes 15.75 years to decrease to 1/8

2) It takes 8.36 years to decrease to 1/3

Half-life is the length of time it takes for a chemical to degrade or go through a particular process. It frequently refers to the length of time it takes for half of a radioactive substance to decay into a stable form in the context of radioactive decay.

We know that;

\(N/No = (1/2)^t/t1/2\)

No = initial amount

N = amount at time t

t = Time taken

t1/2 = half life

\(1/8 = (1/2)^t/5.252^-3 = 2^-t/5.25\)

t = 15.75 years

Again;

\(1/3 = (1/2)^t/5.25\)

ln0.33 = t/5.25ln0.5

t = ln0.33/ln0.5 * 5.25

t = 8.36 years

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