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crimeas [40]
4 years ago
8

A boy whirls a stone with uniform speed in a horizontal circle of radius 2 m and at height 2 m above the ground. The string brea

ks and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of 2 m. What was the magnitude of the stone's centripetal acceleration (in m/s2) when it was in circular motion?
Physics
1 answer:
svp [43]4 years ago
8 0

Answer:

Explanation:

162.7 m/s²

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A student pulls out his or her chair in order to sit down. The student pulls the chair 0.75 m with a force of 20 N. How much wor
IRINA_888 [86]
Work = Force x displacement
Force= 20N
Displacement= 0.75m

W= 20(0.75)
W= 15 Nm
3 0
3 years ago
16. A 95kg fullback, running at 8.2m/s, collided in midair with a 128 kg defensive tackle moving in the opposite direction. Both
Daniel [21]

a) 779 kg m/s

The momentum of an object is given by:

p = mv

where

m is the mass of the object

v is its velocity

For the fullback before the collision,

m = 95 kg

v = 8.2 m/s

Therefore, his momentum was:

p=mv=(95)(8.2)=779 kg m/s

b) -779 kg m/s

After the collision, both the fullback and the tackle come to a stop: this means that their momentum after the collision is zero,

p' = 0

The initial momentum of the fullback was

p = 779 kg m/s

Therefore, his change in momentum is

\Delta p = p' -p =0-779  = -779 kg m/s

where the negative sign indicates that the direction is opposite to the initial direction of motion.

c) -779 kg m/s

Here we can apply the law of conservation of momentum. In fact, the total momentum before and after the collision must be conserved. So we can write:

p_f + p_t = p'

where

p_f is the initial momentum of the fullback

p_t is the initial momentum of the tackle

p' is the final combined momentum after the collision

We already know that

p_f = 779 kg m/s\\p' = 0

Therefore, we can find the tackle's original momentum:

p_t = p'-p_f = 0-(779) = -779 kg m/s

where the negative sign indicates that the direction is opposite to the initial direction of motion of the fullback.

e) -6.1 m/s

To find the velocity of the tackle, we can use again the equation of the momentum:

p = mv

where here we have

p=-779 kg m/s is the original momentum of the tackle

m = 128 kg is his mass

Solving the equation for v, we find the tackle's original velocity:

v=\frac{p}{m}=\frac{-779}{128}=-6.1 m/s

So, he was moving at 6.1 m/s in the direction opposite to the fullback.

4 0
3 years ago
Assume your car reaches a speed of 21.7 m/s at a steady rate for 5.05 s after the light turns green. (a) What distance have you
boyakko [2]

Answer:

The distance and average speed are 54.79 m and 10.85 m.

Explanation:

Given that,

Speed = 21.7 m/s

Time = 5.05 s

(a). We need to calculate the distance

Firstly we will find the acceleration

Using equation of motion

v = u+at

a = \dfrac{v-u}{t}

Where, v = final velocity

u = initial velocity

t = time

Put the value in the equation

a = \dfrac{21.7-0}{5.05}

a = 4.297 m/s^2

Now, using equation of motion again

For distance,

s = ut+\dfrac{1}{2}at^2

s = 0+\dfrac{1}{2}\times4.3\times(5.05)^2

s=54.79\ m

The distance is 54.79 m.

(b). We need to calculate the average speed during this time

v_{avg}=\dfrac{D}{T}

Where, D = total distance

T = time

Put the value into the formula

v_{avg}=\dfrac{54.79}{5.05}

v_{avg}=10.85\ m/s

Hence, The distance and average speed are 54.79 m and 10.85 m.

3 0
3 years ago
Which of the following measurements involve a direction
insens350 [35]

Answer:

I think it is Velocity idk but does it hurt to try

8 0
3 years ago
7.9x10^9 km is equal to?
Georgia [21]

Explanation:

7.9x10^9 km is equal to

=7900000000km

<em>Please</em><em> </em><em>mark</em><em> </em><em>it</em><em> </em><em>as</em><em> </em><em>brainliast</em>

3 0
3 years ago
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