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3 years ago

What is the value of work done on an object when a 50–newton force moves it 15 meters in the same direction as the force?

Physics

1 answer:

laila [671]3 years ago

4 0

Answer:

Work done, W = 750 joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

It moves to a distance of, d = 15 meters

We need to find the work done on an object. We know that the product of force and distance covered is called the work done. As the force and the displacement are in same direction. So,

$W=F\times d$

$W=50\ N\times 15\ m$

W = 750 joules

So, the work done on an object is 750 joules. Hence, this is the required solution.

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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

3 years ago

White light falls on a yellow filter,if:

riadik2000 [5.3K]

Answer:

all colours are absorbed except for the colour of the filter.

Explanation:

When white light passes through a coloured filter, all colours are absorbed except for the colour of the filter. For example, an orange filter transmits orange light but absorbs all the other colours. If white light is shone on an orange filter, only the orange wavelengths will be observed by the human eye.

6 0

3 years ago

An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 5.6 m/s. A child on a bridge, 71 m abo

saveliy_v [14]

Answer:

21.28 m

Explanation:

height, h = 71 m

velocity of raft, v = 5.6 m/s

let the time taken by the stone to reach to raft is t.

use second equation of motion for stone

$h = ut + \frac{1}{2}at^{2}$

u = 0 m/s, h = 71 m, g = 9.8 m/s^2

71 = 0 + 0.5 x 9.8 x t^2

t = 3.8 s

Horizontal distance traveled by the raft in time t

d = v x t = 5.6 x 3.8 = 21.28 m

3 0

3 years ago

A car driving on the highway is going 72mph After 3 hours how many miles will the car have driven ?

Sergeu [11.5K]

Answer:

216 miles

Explanation:

you times 72 by three

7 0

3 years ago

Read 2 more answers

What is the magnitude of the orbital velocity of the earth in m/s?

Lana71 [14]

Circumference C=2πr
C=2π(1.5x10^8)=9.42x10^8 

In 365 Days there are 8760hr 

V=distance/time 

V=(9.42x10^8)/8760=107534.2km/hr

8 0

3 years ago