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2 years ago

The velocity of a 200 kg object is changed from 5 m/s to 25 m/s in 50 seconds by an applied constant force.

1 answer:

4 0

<h2>a) Change in momentum is 4000 kg m/s in the direction of motion</h2><h2>b) Magnitude of the force is 80 N</h2>

Explanation:

Force is given by rate of change of momentum.

Mass of object = 200 kg

Initial velocity = 5 m/s

Final velocity = 25 m/s

a) Change in momentum = 200 x 25 - 200 x 5

Change in momentum = 4000 kg m/s in the direction of motion

b) Time taken = 50 s

Rate of change of momentum = Change in momentum ÷ Time

Rate of change of momentum = 4000 ÷ 50

Rate of change of momentum = 80 N

Force = 80 N

Magnitude of the force is 80 N

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Which of these steps would most likely be part of a lab procedure?

Snezhnost [94]

Answer:

These are a part of lab procedures:

1. Write a hypothesis to answer a question.

2. Write a title at the top of a completed lab report.

3. Record the time to complete a chemical reaction.

These are NOT a part of lab procedures:

1. Create a question on the cause of a chemical reaction.

8 0

2 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

2 years ago

A 5 kg ball is suspended on one cable. Calculate the tension in the cable

ludmilkaskok [199]

Answer:

49N

Explanation:

F=ma

We know the mass is 5kg, and since the ball is suspended on one cable, the acceleration is g, 9.8m/s^2

F=5kg*9.8m/s^2

= 49N

Hope this helps!