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BaLLatris [955]
3 years ago
13

A 44 kg block lies on a horizontal frictionless surface. A horizontal force of 110 N is applied to the block. (a) Determine the

acceleration of the block. (b) Calculate the distance the block will travel is the force is applied for 8 s. (c) Calculate the speed of the block after the force has been applied for 8 s.
Physics
1 answer:
Oksana_A [137]3 years ago
8 0
44k jkjk know the answer
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A system gains 767 kJ of heat, resulting in a change in internal energy of the system equal to +151 kJ. How much work is done?
Crazy boy [7]

Answer:

The work done on the system is -616 kJ

Explanation:

Given;

Quantity of heat absorbed by the system, Q = 767 kJ

change in the internal energy of the system, ΔU = +151 kJ

Apply the first law of thermodynamics;

ΔU = W + Q

Where;

ΔU  is the change in internal energy

W is the work done

Q is the heat gained

W = ΔU  - Q

W = 151 - 767

W = -616 kJ (The negative sign indicates that the work is done on the system)

Therefore, the work done on the system is -616 kJ

6 0
3 years ago
3. You are flying 2586 miles from San Francisco to New York. An hour into the flight, you are 600
Citrus2011 [14]

Answer:

268.2 m/s (1dp)

Explanation:

600 miles = 965,606m (1mile = 1609.34m)

965606/60 = 16,0943.43 ..... (metres traveled every minute)

16,0943.43/60 = 268.2238 .....

3 0
3 years ago
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An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.5 kg mass of the pulley is concentrated on i
alexandr1967 [171]

Explanation:

63 kg ice skater finishes her performance and crossed the finish line with a speed of 10.8 m/s

8 0
3 years ago
Write down Newton's second law in terms of momentum and acceleration. Write down this law in the form of differential equation (
svetoff [14.1K]

Explanation:

According to Newton's second law of motion, the rate of change of momentum is directly proportional to the applied unbalanced force. The mathematical expression is given by:

F=\dfrac{d(mv)}{dt}

Where

F is the applied force

m is the mass of the object

v is the velocity with which it is moving

F=m\dfrac{dv}{dt}

Momentum of a particle is given by the product of mass and velocity as :

p=mv

Hence, this is the required solution.

3 0
3 years ago
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
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