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3 years ago

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A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its potential energy (PE) when it is 2.00 m above the grou

nd PLEASE HELP

1 answer:

Colt1911 [192]3 years ago

7 0

The potential energy of the lemming is 1.53 J

Explanation:

The potential energy (PE) of an object is the energy possessed by the object due to its position in the Earth's gravitational field, and it is given by:

$PE=mgh$

where:

m is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

h is the height of the object relative to the ground

In this problem:

m = 0.0780 kg is the mass of the lemming

We want to find the potential energy when the height is

h = 2.00 m

Therefore, we find:

$PE=(0.0780)(9.8)(2.00)=1.53 J$

Learn more about potential energy:

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For a 166-lb person, a dose of 452 milligrams is prescribed to treat asthma.

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2 years ago

A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

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Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

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put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

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3 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

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Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

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$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

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$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

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Thus;

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$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

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$\mathbf{Q = 4 \pi R^2 \sigma }$

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3 years ago

Consider the f(x) = cos(x-C) function shown in the figure in blue color, where 0 ≤ C ≤ 2π. What is the value of parameter C for

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The value of parameter C for the function in the figure is 2.

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2 years ago

What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x

lyudmila [28]

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

(a) What is the change in potential energy of a +2.00nC test charge, $U_{electric,b} - U_{electric,a}$ as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?

(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?

(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?

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(b) KE = 2× $10^{-6}$ J

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(a) Related to electricity, Potential Energy can be calculated as:

ΔU = Eqd

where E is the electric field (in N/C);

q is the charge (in C);

d is the distance between plaques (in m);

For a at x = - 30cm and b at x = 50 cm:

E = 2× $10^{3}$ N/C

q = 2× $10^{-9}$ C

d = 50 - (-30) = 80× $10^{-2}$ = 8× $10^{-1}$ m

ΔU = $U_{electric,b} - U_{electric,a}$ = Eqd

$U_{electric,b} - U_{electric,a}$ = 2× $10^{3}$ . 2× $10^{-9}$ . 8× $10^{-1}$

ΔU = 3.2× $10^{-6}$ J

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$KE_{i} + U_{i} = KE_{f} + U_{f}$

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