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3 years ago

A 5.4-kg ball is thrown into the air with an initial velocity of 35.2 m/s.a

Physics

1 answer:

kompoz [17]3 years ago

4 0

The answer is

Ekin = 1/2 * m * v^2
Ekin = 1/2 * 5,4 * 35,2^2
Ekin = 3345,41 j (joule)

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Can someone please help? Thank u!

Snowcat [4.5K]

Answer:

B. 7.5 m/s^2

Explanation:

To find acceleration you need to subtract the final velocity by the starting velocity then divide that by the time

a= v-v/t

a= 60-0/8

a= 60/8

a=7.5 m/s^2

7 0

3 years ago

I need help with this work

san4es73 [151]

What work??? I don’t see anything

7 0

3 years ago

Read 2 more answers

Jim is driving a 2268-kg pickup truck at 22 m/s and releases his foot from the accelerator pedal. The car eventually stops due t

shutvik [7]

Answer:

610 meters.

Explanation:

Because Jim released the accelerator, the truck started to slow down, so the friction force will eventually stop the truck.

the kinetic energy of the truck just after Jim released the pedal is:

$E_k=\frac{1}{2}*m*v^2\\E_k=\frac{1}{2}*2268*(22)^2=548856J$

The work done by the friction force is given by:

$W_f=F_s*d\\\\d=\frac{548856J}{900N}\\\\d=610m$

6 0

3 years ago

You look down an old well, cannot see the bottom, and mutter to yourself "Oh well!". In order to estimate the depth of the well,

telo118 [61]

Answer:

The best estimate of the depth of the well is 2.3 sec.

Explanation:

Given that,

Record time,

$t_{1}=2.19\ sec$

$t_{2}=2.30\ sec$

$t_{3}=2.26\ sec$

$t_{4}=2.29\ sec$

$t_{5}=2.27\ sec$

We need to find the best estimate of the depth of the well

According to record time,

We can write of the record time

$t_{1}=2.19\approx 2.2\ sec$

$t_{2}=2.30\approx 2.3\ sec$

$t_{3}=2.26\approx 2.3\ sec$

$t_{4}=2.29\approx 2.3\ sec$

$t_{5}=2.27\approx 2.3\ sec$

Here, all time is nearest 2.3 sec.

So, we can say that the best estimate of the depth of the well is 2.3 sec.

Hence, The best estimate of the depth of the well is 2.3 sec.

6 0

3 years ago

A weightlifter raises a 50kg weight to a height of 2m in 2 minutes. What was the power spent by the weightlifter?

belka [17]

Answer:

117.72kW

Explanation:

Given data

Mass m= 50kg

height x = 2m

time taken = 2 minutes= 129 seconds

let us find the work done

WD= force * distance

WD= mgx

WD= 50*9.81*2

WD= 981 Joules

Let us find the power

Power= work * time

Power= 981*120

Power= 117720

Power= 117.72 kW

Hence the power spent is 117.72kW

8 0

3 years ago