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Scorpion4ik [409]
3 years ago
10

Pete slides a crate up a ramp with constant speed at an angle of 24.3 ◦ by exerting a 289 N force parallel to the ramp. How much

work has been done against gravity when the crate is raised a vertical distance of 2.39 m? The coefficient of friction is 0.33.
Physics
1 answer:
Bumek [7]3 years ago
5 0

Answer:

W=972.73 J

Explanation:

Given that

θ = 24.3°

F= 289 N

h= 2.39 m

μ = 0.33

Given that velocity is constant is means that acceleration is zero that is why force will be balance.

Lest take mass = m

Gravity force along ramp = m g sinθ

Friction force =  μm g cosθ

F= m g sinθ  + μm g cosθ

By putting the values

F= m g (sinθ  + μcosθ)              

289 = 10 m  ( sin 24.3°+ 0.33 cos 24.3°)                 ( take g =10 m/s²)

289 = 10 m(0.71)

m = 40.7 kg

The force against gravity W

W= m g h

W= 40.7 x 10 x 2.39 J

W=972.73 J

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Answer:

The velocity of skaters after collision is 0.55 m/s and they are moving to the left

Explanation:

Consider m₁ and m₂ are the masses of two skaters and their initial velocity be v₁ and v₂ respectively.

Consider the velocity along right direction as positive while negative for left direction.

According to the problem,

Mass of first skater, m₁ = 66.0 kg

Mass of second skater, m₂ = 75.0 kg

Velocity of first skater, v₁ = + 2.00 m/s

Velocity of second skater, v₂ = -2.80 m/s

Since, the two skaters grab each other after collision. Hence, they are moving with same final velocity v.

Applying conservation of linear momentum,

Momentum before collision = Momentum after collision

m₁v₁ + m₂v₂ = (m₁+m₂)v

Substitute the suitable values in the above equation.

66 x 2 - 75 x 2.8 = (66 + 75 )v

v= \frac{-78}{141}

v = -0.55 m/s

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4 years ago
What process occurs when all of the energy from light waves is transferred to a medium?
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Absorption occurs when <span>all of the energy from light waves is transferred to a medium! </span>
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The hubbles telescopes orbit is 5.6 x10 ^5 meters above earths suface. the telescope has a mass os 1.1 x10^4 kilograms. earth ex
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(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg

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A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonc
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(a) The tension on the wire when the two charges have opposite signs is 383.5 N.

(b) The tension on the wire if both charges were negative is 3.640.25 N.

The given parameters;

  • <em>first charge, q₁ = 8.75 μC </em>
  • <em>second charge, q₂ = -6.5 μC  </em>
  • <em>electric field, E = 1.85 x 10⁸ N/C</em>
  • <em>distance between the two charges, r = 2.5 cm</em>

<em />

(a)

The attractive force between the charges is calculated as follows;

F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N

The force on the negative charge due to the electric field is calculated as follows;

F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N

The tension on the wire is the resultant of the two forces and it is calculated as follows;

T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N

(b) when the two charges are negative

The repulsive force between the two charges is calculated as follows;

F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N

The force on the first negative charge due to the electric field is calculated as follows;

F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N

The force on the second negative charge due to the electric field is calculated as follows;

F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N

The tension on the wire is the resultant of the three forces and it is calculated as follows;

T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N

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