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EastWind [94]
3 years ago
10

3. What average net force is required to stop a 7 kg shopping cart in 2 s if it's initially

Physics
1 answer:
Mashcka [7]3 years ago
5 0

Answer:

The force required to stop the shopping cart is, F = 12.25 N

Explanation:

Given data,

The mass of the shopping cart, m = 7 kg

The initial velocity of the shopping cart, u = 3.5 m/s

The final velocity of the shopping cart, v = 0 m/s

The time period of acceleration, t = 2 s

The change in momentum of the cart,

                                       p = m(u - v)

                                          = 7 (3.5 - 0)

                                          = 24.5 kg m/s

The force is defined as the rate of change of momentum. To stop the shopping cart, the force required is given by the formula

                                            F = p / t

                                                = 24.5 / 2

                                                = 12.25 N

Hence, the force required to stop the shopping cart is, F = 12.25 N

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Makovka662 [10]

hey you look nice (pic).

According to Newton’s first law, if no force is applied to a ball, it will continue moving at the same speed and direction as it did before. When we put the ball on the grass it stays in its place, namely it stays in zero motion since no force is applied to it. However, after we kick the ball, it will continue moving in the direction we kicked it. Its speed will drop gradually, due to friction (a force applied on the ball in the opposite direction to its motion), but the direction of its motion will remain the same.

According to Newton’s second law, a force applied to an object changes that object’s acceleration – namely, the rate at which the speed of the object changes. When we kick the ball, the force we apply to it causes it to accelerate from a speed of 0 to a speed of dozens of kilometers per hour. When the ball is released from the foot, it begins to decelerate (negative acceleration) due to the force of friction that is exerted upon it (as we observed in the previous example). If we were to kick a ball in outer space, where there is no friction, it would accelerate during the kick, and then continue moving at a constant speed in the direction that we kicked at, until it hits some other object or another force is applied to it.

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3 years ago
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lubasha [3.4K]

Answer:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

Explanation:

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3 years ago
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Water flows through a 4.50-cm inside diameter pipe with a speed of 12.5 m/s. At a later position, the pipe has a 6.25-cm inside
jek_recluse [69]

Given,

The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m

The initial speed of the water, v₁=12.5 m/s

The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m

From the continuity equation,

\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.

On substituting the known values,

\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}

Thus, the flow rate of the water at the later position is 5.99 m/s

4 0
1 year ago
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Answer:

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<em>Remember</em>, the doppler effect refers to the changes in sound (frequency of sound) observed by a person who is in a position relative to the wave source.

In this example, we notice as the train comes closer to the boy, the sound becomes louder also increasing the pitch slightly, the doppler effect sets in when the train passes the boy because the boy notices a decrease in the pitch of the moving train.

We learn from the change in the observed sound of the train that the frequency of the sound is determined by the distance of the observer from the wave source.

In other words, the closer the source of the sound to the observer; the faster it travels to the observer, however, the farther it is; the lesser it is; the greater the sound heard.

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