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PIT_PIT [208]
3 years ago
11

As the ball rolls down the ramp, there are changes in kinetic and potential energy. How much kinetic energy does the

Physics
2 answers:
Sonja [21]3 years ago
8 0

The answer is C, 50J

Please mark brainliest Have a nice day :)

pogonyaev3 years ago
5 0
C 50j the ball is at when reaches point b
You might be interested in
An insulated beaker with negligible mass contains liquid water with a mass of 0.285 kg and a temperature of 75.2 ∘C How much ice
GrogVix [38]

Answer:

Explanation:

We shall apply the theory of

heat lost = heat gained .

heat lost by water = mass x specific heat x temperature diff

= .285 x 4190 x ( 75.2 - 32 ) = 51587.28 J  

heat gained by ice to attain temperature of zero

= m x 2100 x 22.8 = 47880 m

heat gained by ice in melting = latent heat x mass

= 334000m

heat gained by water at zero to become warm at 32 degree

= m x 4190 x 32 = 134080 m

Total heat gained = 515960 m

So

515960 m = 51587.28

m = .1 kg

= 100 gm

4 0
4 years ago
Why is the sun renewable but oil is non renewable?
DochEvi [55]
Because once you use oil you can not reuse it but the sun you can because it is always here and it has solar energy
4 0
3 years ago
A pitcher throws a 0.144-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just
Solnce55 [7]

(a) 12.8 kg m/s

The impulse delivered by the bat on the baseball is equal to the change in momentum of the baseball:

I=\Delta p = m(v-u)

where we have

m = 0.144 kg is the mass of the ball

v = -47 m/s is the final velocity of the ball

u = 42 m/s is the initial velocity of the ball

Substituting into the equation, we find

I=(0.144 kg)(-47 m/s-(42 m/s))=-12.8 kg m/s

And since we are interested in the magnitude only,

I=12.8 kg m/s

(b) 2.78 kN

The impulse exerted on the ball is also equal to the product between the average force and the contact time:

I=F\Delta t

where

F is the average force exerted on the ball

\Delta t=0.0046 s is the contact time

Solving the formula for F, we find

F=\frac{I}{\Delta t}=\frac{12.8 kg m/s}{0.0046 s}=2783 N = 2.78 kN

(c) The force exerted on the ball is much larger (1988 times more) than the weigth of the ball

The weight of the ball is given by

W=mg

where

m = 0.144 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration due to gravity

Solving the equation for W, we find

W=(0.144 kg)(9.8 m/s^2)=1.4 N

So as we see, the force exerted on the ball (2783 N) is almost 2000 times larger than the weight of the ball (1.4 N):

\frac{F}{W}=\frac{2783 N}{1.4 N}=1988

8 0
3 years ago
An 80 kg astronaut has gone outside his space capsule to do some repair work. Unfortunately, he forgot to lock his safety tether
Anna [14]

Answer:

the time taken by the astronaut to reach safety = 9.8 hr

Explanation:

The equation for intensity can be written as :

I = \frac{P}{A}

where :

\frac{I}{c}= \frac{F}{A}

Replacing that into the above previous equation; we have:

\frac{P}{Ac}=\frac{F}{A}

F = \frac{P}{c}

However ; the force needed to push the astronaut is as follows:

F = ma

where ;

m = mass of the  astronaut and a = its acceleration

we as well say;

\frac{P}{c} = ma

a = \frac{P}{mc}

Replacing P with 1000 W ; m with 80 kg and 3*10^{8} \  m/s for  c

Then; a = \frac{1000 \ W}{(80)(3.0*10^8)}

a = 4.2*10^{-8} \ m/s

It is also known that the battery will run for one hour and after which the battery on the laser will run out

Then to determine the change in the position after the first hour ; we have:

\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 \ h)^2

\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 *3600 s)^2

= 0.27 m

Furthermore, the final velocity of the astronaut is determined as:

v_1 = at_1

where ;

v_1 = final \ velocity

replacing t_1 = 1.0 \ h and a =  4.2*10^{-8} \ m/s; Then:

v_1 = (4.2*10^8 \ m/s * 1.0 \ h * \frac{ 3600\ s}{1.0 \ h})

v_1 =  1.51 *10^{-4} \ m/s

Also; when he drifted 5.0 m away from the capsule; the distance is far short of the 5 m but he still have 9 hours left of oxygen . In addition to that, he acceleration is also zero and the final velocity remains the same, so:

To find the final distance traveled by the astronaut ;we have:

\Delta x_2 = d - \Delta x_1

where;

\Delta x_2 = the final distance

d = total distance

So;

\Delta x_2 = 5 m - 0.27 m \\ \\ \Delta x_2 = 4.73 \ m

The time taken to reach the final distance can be calculated as:

t_2 = \frac{\Delta x_2 }{v_1}

where;

t_2 = is the  time to  reach the final distance

Replacing 4.73 for {\Delta x_2 } and  1.51*10^{-4} m/s for v_1

t_2 = \frac{4.73 \ m }{1.51*10^{-4} \ m/s}

t_2 = 31500 \ s (\frac{1.0 \ h}{3600 \ s} )

t_2 = 8.8 \ h

We knew the laser was operated for 1 hour; thus the total time taken by the astronaut to  reach the final distance is the sum of the time taken to reach the final distance and the operated time of the laser.

Hence ; the time taken by the astronaut to reach safety = 9.8 hr

8 0
3 years ago
Please answer this and I will make you the BRAINLIST!!!!!!!! ✨✨✨
sweet-ann [11.9K]

Answer:

it may be efficient at first but over time there will be a lot of wasted energy.

Explanation:

I'm pretty sure

4 0
3 years ago
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