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Y_Kistochka [10]
2 years ago
8

A 2.0-kg ball is at rest when a horizontal force of 5.0 N is applied. In the absence of friction, what is the speed of the ball

after it ha gone 10 m?
Physics
1 answer:
Rzqust [24]2 years ago
7 0

Answer:

v = 7.1 m/s

Explanation:

Work applied will change kinetic energy

Fd = ½mv²

v = √(2Fd/m) = √(2(5.0)(10) / 2.0) = √50 = 7.07... m/s

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Sunny_sXe [5.5K]

Here in crash test the two forces are acting on the dummy in two different directions

As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.

So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors

so we can say

F_{net} = \sqrt{F_1^2 + F_2^2}

here given that

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F_2 = 4500.0 N

now we will plug in all data in the above equation

F_{net} = \sqrt{4500^2 + 130^}

F_{net} = 4501.9 N

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4 0
2 years ago
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Your favorite taco stand is 800 m away, and closes in 15 minutes. How fast do you have to run to make it on
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You must run at least 53.3333333 meters a minute.

8 0
3 years ago
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A bus slows with constant acceleration from 24.0 m/s to 16.0 m/s and moves 50.0 m in the process. (a) How much further does it t
navik [9.2K]

Answer:

(a) Bus will traveled further a distance of 40 m

(b) It will take 7.5 sec to stop the bus

Explanation:

We have given initial velocity of the bus u = 24 m/sec

And final velocity v = 16 m/sec

Distance traveled in this process s = 50 m

From third equation of motion we know that v^2=u^2+2as

16^2=24^2+2\times a\times 50

a=-3.2m/sec^2

(a) Now as the bus finally stops so final velocity v = 0 m/sec

So v^2=u^2+2as

0^2=24^2-2\times 3.2\times s

s= 90 m

So further distance traveled by bus = 90-50 =40 m

(b) Now as the bus finally stops so final velocity v= 0 m/sec

Initial velocity u = 24 m/sec

Acceleration a=-3.2m/sec^2

So time t=\frac{v-u}{a}=\frac{0-24}{-3.2}=7.5sec

7 0
3 years ago
Hey, I need help can someone help me out, please?
yan [13]

Explanation:

6) newton

7) f =ma = 15*15 = 225N

8) a= 100/20 = 5ms^-2

6 0
3 years ago
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3.00 m^3 of water is at 20.0°C.
krok68 [10]

Answer:

9m^3

Explanation:

Given data

volume  v1=  3m^3

volume  v2=  ???

Temperature T1= 20.0°C.

Temperature T2= 60.0°C.

Applying the relation for temperature and volume

V1/T1= V2/T2

substitute

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3*60= V2*20

180= 20*V2

180/20= V2

V2= 9m^3

Hence the final volume is 9m^3

6 0
2 years ago
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