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1 year ago

If the applied force is 60 N and the lower mass is 1.0 kg, what is the tension in the string?(frictionless surface)

Physics

1 answer:

Alik [6]1 year ago

3 0

The net force on the larger mass in the horizontal direction (where we take the direction of F to be positive) is

F - T = 2Ma

where T is the magnitude of tension and a is the mass's acceleration.

The net force on the smaller mass in the vertical direction (upward is positive) is

T - Mg = Ma

If M = 1.0 kg and F = 60 N, then we end up with two equations with two unknowns,

60 N - T = (2.0 kg) a

T - (1.0 kg) (9.8 m/s²) = (1.0 kg) a

Eliminate a and solve for T :

(60 N - T) - 2 (T - (1.0 kg) (9.8 m/s²)) = (2.0 kg) a - 2 (1.0 kg) a

60 N - T - 2T + 19.6 N = (2.0 kg) a - (2.0 kg) a

40.4 N - 3T = 0

3T = 40.4 N

T = (40.4 N) / 3

T ≈ 13 N

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Answer:

The speed of the plank is 81.68 m/s

Explanation:

Given that,

Speed of bullet = 152 m/s

Speed of wood = 128 m/s

Speed of another bullet = 97 m/s

We need to calculate the speed of plank

Using conservation of momentum

$m_{1}u_{1}=(m_{1}+m_{2})v$

Where,

u = initial velocity

v = final velocity

$152m_{1}=(m_{1}+m_{2})128$ ....(I)

$m_{1}97=(m_{1}+m_{2})v$ ....(II)

From equation(I) and equation(II)

$\dfrac{152}{97}=\dfrac{128}{v}$

$= \dfrac{128\times97}{152}$

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Hence, The speed of the plank is 81.68 m/s

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3 years ago

The two sleds shown below are about to collide. If they stick together in the collision, what will happen after the collision? (

zhannawk [14.2K]

Answer:

B. The two sleds will move to the right.

Explanation:

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2 years ago

A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 8.1 m from this surface, the potenti

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Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, $V_s$ = 450 V

potential at radial distance, $V_r$ = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

$V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}$

$450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m$

Therefore, the radius of the sphere is 4.05 m

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2 years ago

Suppose you place a ball in the middle of a wagon, and then accelerate the wagon forward. Describe the motion of the ball relati

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Answer:

The motion of the ball relative to the ground is stationary

The motion of the ball relative to the wagon is backwards

Explanation:

To describe the motion of the ball relative to the ground, we note that

Assuming the ball is perfectly round and rotate freely, then we have

Force on the ball due to motion of the wagon = 0 N,

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The motion of the ball relative to the wagon

Relative to the wagon, the ball appears to be moving in the opposite direction to the wagon, that is backwards.

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3 years ago

Read 2 more answers

A 70.0 kg man jumping from a window lands in an elevated fire rescue net 11.0 m below the window. He momentarily stops when he h

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Answer:

70.15 Joule

Explanation:

mass of man, m = 70 kg

intial length, l = 11 m

extension, Δl = 1.5 m

Let K is the spring constant.

In the equilibrium position

mg = K l

70 x 9.8 = K x 11

K = 62.36 N/m

Potential energy stored, U = 0.5 x K x Δl²

U = 0.5 x 62.36 x 1.5 x 1.5

U = 70.15 Joule

6 0

2 years ago