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3 years ago

An arrow is used to show a force acting on an object. what information does the arrow give about the force

Physics

1 answer:

Vedmedyk [2.9K]3 years ago

6 0

The Force surrounds us and penetrates us. It flows through Jedis. May the Force be with you all.

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A force of 100. newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the time it takes to do

Flauer [41]

<h3>It takes 60 seconds to do the work</h3>

Solution:

Given that,

Force = 100 newtons

Distance = 15 meters

Power = 25 watts

To find: time it takes to do the work

Find the work done:

$work = force \times displacement\\\\work = 100\ newtons \times 15\ meters\\\\work = 1500\ joule$

Find the time taken

$power = \frac{work}{time}\\\\25\ watts = \frac{1500\ joule}{time}\\\\time = \frac{1500\ joule}{25\ watt}\\\\time = 60\ second$

Thus it takes 60 seconds to do the work

3 0

3 years ago

6. If a drag racer wins the final round of herrace by going an average speed of 320 m/sin 4.5 seconds, what distance did he cove

Ivan

We want to calculate the distance covered by the drag racer. Recall, the formula for calculating distance is expressed as

Distance = speed x time

From the information given,

speed = 320 m/s

time = 4.5 s

By substituting these values into the formula, we have

Distance = 320 m/s x 4.5s

s cancels out. We are left with m. Thus,

Distance = 1440m

4 0

1 year ago

The first PCS technology used a form of time division multiplexing called ____. a. Time Division Multiple Access (TDMA) b. Gl

salantis [7]

Answer:

A. Time Division multiple Access (TDMA)

Explanation:

Time-division multiple access (TDMA) is a channel access method for shared-medium networks

It allows several users to share the same frequency channel by dividing the signal into different time slots. The users transmit in rapid succession, one after the other, each using its own time slot.

3 0

3 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago

A hockey puck with a mass of 0.159 kg slides over the ice. The puck initially slides with a speed of 4.75 m/s, but it comes to a

Anton [14]

Answer:

The energy dissipated as the puck slides over the rough patch is 1.355 J

Explanation:

Given;

mass of the hockey puck, m = 0.159 kg

initial speed of the puck, u = 4.75 m/s

final speed of the puck, v = 2.35 m/s

The energy dissipated as the puck slides over the rough patch is given by;

ΔE = ¹/₂m(v² - u²)

ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)

ΔE = -1.355 J

the lost energy is 1.355 J

Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J

5 0

3 years ago