A total of 750 j of work was done when a force of 125 n was exerted on a box to move it. how far was the box moved?

Ahmad is riding his bicycle. He finds that he can accelerate from rest at 0.44 m/s^2 for 5 s to reach a speed of 2.2 m/s. The to

snow_lady [41]

Answer:

1) The force Christian can exert on his bicycle before picking up the the cargo is 529.74 N

2) The force Christian can exert on his bicycle after picking up the the cargo is 647.46 N

Therefore, Christian has to exert more force on his bike after picking up the cargo

Explanation:

The given parameters are;

The mass of Christian and his bicycle = 54 kg

The mass of the cargo = 12 kg

1) The force Christian can exert on his bicycle before picking up the the cargo = Mass of Christian and his bicycle × Acceleration due to gravity

∴ The force Christian can exert on his bicycle before picking up the the cargo = 54 kg × 9.81 m/s² = 529.74 N

2) The force Christian can exert on his bicycle after picking up the the cargo = (54 + 12) kg × 9.81 m/s² = 647.46 N

Therefore, Christian has to exert more force on his bike after picking up the cargo.

7 0

2 years ago

In a clear cup there are three substances. Their densities are 3, 1, and 2. From top of the cup to the bottom, what would the or

Vesnalui [34]

Answer:

1, 2 and 3

Explanation:

The most dense substance will settle at the bottom of the cup

4 0

2 years ago

You and a highway patrolman are driving at constant speeds in opposite directions on a straight highway. The patrolman is drivin

kompoz [17]

Answer: 75 mph

Explanation:

The Relative Speed for a mobile is equal to the diference between the object and the observer:

Relative Speed (Rs) = Object's Velocity - Observer's Velocity

Thinking on those terms, we would need to have a universal observer to do any understandable measurement on daily basics. This is why we all use earth as a Static Observer for every measurement we do everyday.

Using Earth as an observer, the Velocity for the Patrolman is:

Patrolman Velocity (Vp) = 60 mph

Because the radar gun does measure the Relative Speed for the object, which is 135 mph, we need to work with the equation to find the Velocity using Earth as a reference.

Object's Relative Velocity = Object's Velocity - Patrolman's Velocity

Object's Velocity = Object's Relative Velocity + Patrolman's Velocity

We need to keep in mind, the Patrolman is going on the opposite direction. Because of this the sign for his velocity should be negative.

Object's Velocity = 135 mph + ( -60 mph)

Object's Velocity = 135 mph - 60 mph

Object's Velocity = 75 mph

3 0

2 years ago

An athlete needs to lose weight and decides to do it by “pumping iron.” (a) How many times must an 90.0 kg weight be lifted a di

Wittaler [7]

Answer:

230kg would be the best answer

Explanation:

7 0

3 years ago

Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A

boyakko [2]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

(c) The average force on the right sled is 922.625 N

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴ 22.7 × v₁' = -3.63 × 3.05

v₁' = -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law, m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴ 22.7 × v₂' = -3.63 × 3.05

v₂' = -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = $F_{average}$ × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

$F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}$

∴ $F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s} = 922.625 \ kg\cdot m/s^2 = 922.625 \ N$

The average force on the right sled applied by the cat while landing, $\mathbf{F_{average}}$ = 922.625 N

Learn more about conservation of momentum here:

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8 0

2 years ago