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Dima020 [189]
3 years ago
6

Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0 incline. The coeffi cient of kinetic friction between the bl

ock and the incline is mk 0.436. Determine (a) the work done by the force of gravity, (b) the work done by the friction force between block and incline, and (c) the work done by the normal force. (d) Qualitatively, how would the answers change if a shorter ramp at a steeper angle were used to span the same vertical height
Physics
1 answer:
UkoKoshka [18]3 years ago
5 0

Answer:

a) W_g =61.25J

b) W_k = -46.25J

c) W_N = 0

d) W_g would be the same.

   W_k would decrease.

   W_N would be the same.

Explanation:

a) On an inclined plane the force of gravity is the sine component of the weight of the block.

F_g = mg\sin(\theta) = 5(9.8)\sin(30^\circ)\\W_g = F_g x = 5(9.8)\sin(30^\circ)2.5 = 61.25J

b) The friction force is equal to the normal force times coefficient of friction.

F_k = -mg\cos(\theta)\mu_k = -5(9.8)cos(30^\circ)0.436 = -18.5 N\\W_k = -F_kx = -46.25J

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.

d) The relation between the vertical height and the distance on the ramp is

h = x\sin(\theta)

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of x\sin(\theta).

The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.

The work done by the normal force would still be zero.

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8 0
3 years ago
A man stands at the midpoint between two speakers that are broadcasting an amplified static hiss uniformly in all directions. Th
Illusion [34]

Answer:

Explanation:

The power of each of the speakers is 0.535 W. At a distance d intensity of sound can be found by the following formula

Intensity of sound =  Power / 4π d²

= .535 / 4 x 3.14 x (27.3/2)²

= 2.286 x 10⁻⁴ J m⁻² s⁻¹

Intensity of sound due to other source = 5.715 x 10⁻⁵J m⁻² s⁻¹

Total intensity = 2 x 2.286 x 10⁻⁴J m⁻² s⁻¹

= 4.57 x 10⁻⁴J m⁻² s⁻¹

b ) In this case, man is standing at distances 18.15 m and 9.15 m from the sources .

The total intensity of sound reaching him is as follows

0.535 / (4 π x18.15² ) + 0.535 /  (4 π x9.15² )

= 1.293 x 10⁻⁴ + 5.087 x 10⁻⁴

= 6.38 x 10⁻⁴J m⁻² s⁻¹

5 0
3 years ago
A single-turn current loop, carrying a current of 3.50 a, is in the shape of a right triangle with sides 50.0, 120, and 130 cm.
dimulka [17.4K]
Thank your very much
3 0
3 years ago
A dog of mass 18 kg runs at a speed of 4 m/s. What is the momentum of the
Andrej [43]

Answer:

A, 72 kg•m/s

Explanation:

p=mv

p=18x4

p=72

6 0
3 years ago
A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude
Darya [45]

Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

\omega = \sqrt{\frac{k}{m} } \\\\\omega =  \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

8 0
2 years ago
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