All categories

3 years ago

Calculate the potential energy of a 5 kg object sitting at the top of a 2 meter ramp.

1 answer:

5 0

The formula used to find potential energy is P.E. = M * G * H (P.E. is potential energy, M is mass, G is gravitational pull, and H is height). So the answer to your question is 5 * 9.8 * 2, which equals 98.

You might be interested in

Estimate the acceleration due to gravity at the surface of Europa (one of the moons of Jupiter) given that its mass is 4.9×1022k

Taya2010 [7]

Answer: $g=1.97 m/s^2$

Explanation:

Given

mass of Jupiter is $M=4.9\times 10^{22} kg$

Density of Jupiter is same as Earth

$density\ of\ Earth=density\ of\ Jupiter=5510 kg/m^3$

$mass=volume\times density$

considering Jupiter to be sphere of radius r

$M=\frac{4}{3}\times \pi r^3\times \rho$

$r^3=\frac{3M}{\rho \times 4\pi}$

$r^3=\frac{3\times 4.9\times 10^{22}}{5510\times 4\pi}$

$r=(\frac{3\times 4.9\times 10^{22}}{5510\times 4\pi})^{\frac{1}{3}}$

$r=1.28\times 10^6 m$

acceleration due to gravity is given by

$g=\frac{GM}{r^2}$

$g=\frac{6.67\times 10^{-11}\times 4.9\times 10^{22}}{(1.28\times 10^6)^2}$

$g=1.97 m/s^2$

3 0

4 years ago

A driver, traveling at 22.0 m/s, slows down her 2.00 x 103 kg car to stop for a red light. What work is done by the friction for

Artemon [7]

Some work will be done on friction between wheels and road but it is negligible compared to work done on friction on breaks.

W = Ek = (m*v^2)/2 = 2000*22^2/2 = 1000*22^2 = 484KJ

Because car is not changing its potential energy, there is no work to be done on while changing it which means that all goes on changing kinetic energy (energy of motion)

6 0

4 years ago

A 297 g block is connected to a light spring with spring constant 4.34 N/m, and displaced 7.45 cm from equilibrium. It is then r

vagabundo [1.1K]

Answer:

x = A sin ω t describes the displacement of the particle

v = A ω cos ω t

a = -A ω^2 sin ω t

a (max) = -A ω^2 is the max acceleration (- can be ignored here)

ω = (K/ m)^1/2 for SHM

F = - K x^2 restoring force of spring

K = 4.34 / .0745^2 = 782 N / m

ω = (782 / .297)^1/2 = 51.3 / sec

a (max) = .0745 * 782 / .297 = 196 m / s^2

4 0

2 years ago

What is the strength of the electric field in a region where the electric potential is constant?

Roman55 [17]

Answer:

Where the electric potential is constant, the strength of the electric field is zero.

Explanation:

As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e

Eₓ = - dV / dx ----------(i)

From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.

Therefore, a constant electric potential means that electric field is zero.

4 0

3 years ago

A body is at aheight of 81m and is ascending upwards with a velocity of 12m/s .a body of 2 kg weight is dropped from it.if g=10m

butalik [34]

First the velocity drops to zero in 1.2 secs. In those seconds it went upwards for 7.2 m, then it went from 87.2 to 0m. x-x0=v0*t+1/2*g*t^2 ergo t=sqrt(2x/g) that is 4.1761 s. Finally the total time required is 5.3761 s

7 0

3 years ago