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3 years ago

While driving fast around a sharp right turn, you find yourself pressing against the car door. What is happening?

Physics

1 answer:

sergiy2304 [10]3 years ago

3 0

Answer:

option C

Explanation:

The correct answer is option C

When the driver takes the sharp right turn the door will exert rightward pressure on the driver.

When the driver takes the sudden right turn the tendency of the body is to be in the straight line by the vehicle moves in the circular path so, as the vehicle turns it applies a rightward force on you.

The pushing of the door to you because of the centripetal force acting on the car due to sudden sharp turn.

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A bucket weighing 5 lbs is lifted at a constant rate from the bottom of a 100 ft well by a rope which weighs 5 lbs. The bucket h

8_murik_8 [283]

Answer:

$W_{bucket} = 24934.85\,lbf\cdot ft$

Explanation:

The system is modelled after the Principle of Energy Conservation and the Work-Energy Theorem:

$W_{bucket} = U_{g,B, bucket} - U_{g,A,bucket} +U_{g, B, water}-U_{g,A,water} +U_{g, B, rope} -U_{g,A,rope}$

$W_{bucket} = (5\,lb) (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft-0\,ft) + (25\,lb)\cdot (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft) - (30\,lb)\cdot (32.174\,\frac{ft}{s^{2}})\cdot (0\,ft)+(5\,lb) (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft-50\,ft)$

$W_{bucket} = 24934.85\,lbf\cdot ft$

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4 years ago

A 26.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 11.0 m th

frozen [14]

Answer:

18 m/s

Explanation:

Force is given by mass × acceleration

F = ma

a = F/m

For the first 11.0 m, there is no friction. Hence, the constant force is applied fully on the crate. The acceleration is

a = 225/26.0 m/s²

By the equation of motion, v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance travelled.

Using the values for the first part of the motion,

v² = 0² + 2 × 8.65 × 11.0

v = 13.8 m/s

This is the initial velocity for the second part of the motion. This part has friction of coefficient 0.20.

The frictional force = 0.20 × weight = 0.20 × 26.0 × 9.8 = 50.96 N

The effective force moving the crate in the second motion = 225 - 50.96 N = 174.04 N

This gives an acceleration of 174.04/26.0 = 6.69 m/s².

Using parameters for the equation of motion, v² = u² + 2as

v² = 13.8² + 2 × 6.69 × 10 = 324.24

v = 18 m/s

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4 years ago

zhannawk [14.2K]

Average speed= total distance/total time =12km/h

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3 years ago

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kow [346]

Answer:

9013 m/s

Explanation:

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3 years ago

A horizontal line labeled B has an arrow labeled A strike it from right and above and then another arrow D emerges from the stri

patriot [66]

Answer:

c is the actual answer.

Explanation:

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3 years ago

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