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Svetllana [295]

3 years ago

13

A 17.5-kg block is dragged over a rough, horizontal surface by a 75-N force acting at 21° above the horizontal. The block is dis

placed 5.7 m, and the coefficient of kinetic friction is 0.150.
Requried:
a. Find the work done on the block by the 75-N force.
b. Find the work done on the block by the normal force.
c. Find the work done on the block by the gravitational force.
d. What is the increase in internal energy of the block-surface system due to friction?
e. Find the total change in the block's kinetic energy.

1 answer:

pogonyaev3 years ago

5 0

Answer:

Explanation:

given

m= 17.5kg

F= 75N

d= 5.7m

∪=0.150

θ= 21°

a. W = Fcos θ × d

75cos21° ×5.7

=399.106J

b. normal force is zero. 0 Joules

cos 90°=0

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2 years ago

A 12.0-g bullet is fired horizontally into a 109-g wooden block that is initially at rest on a frictionless horizontal surface a

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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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To find the constant k, we examine the total charge Q which is:

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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

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$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

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

6 0

3 years ago