3 years ago

Why can someone please help me out

Physics

1 answer:

denis23 [38]3 years ago

4 0

D I think hope this helps

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An athlete whose mass is 97.0 kg kg is performing weight-lifting exercises. Starting from the rest position, he lifts, with cons

SIZIF [17.4K]

Answer:

470 N.

Explanation:

Using equations of motion:

S = vi*t + 1/2*(a*(t^2))

Given:

S = 0.65 m

t = 1.5 s

vi = 0 m/s

0.65 = 1/2 * (a * (1.5)^2)

a = 1.3/2.25

= 0.578 m/s^2

Force = mass * acceleration due to gravity

= 92 * 0.578

= 53.16 N

Total force = 420 + 53.16

= 473.16 N

= 470 N.

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3 years ago

If a bus is running with a speed of 72 km/hr,calculate the distance travelled by it in 5 second.Which of the following is right

ratelena [41]

Answer:

100m

Explanation:

conversation km to meter

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3 years ago

Find the ratio of the new/old periods of a pendulum if the pendulum were transported from earth to the moon, where the accelerat

vichka [17]

The period of a pendulum is given by
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length and g is the gravitational acceleration.

We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
$\frac{T_m}{T_e} = \frac{2 \pi \sqrt{ \frac{L}{g_m} } }{2 \pi \sqrt{ \frac{L}{g_e} } }= \sqrt{ \frac{g_e}{g_m} }$
where the labels m and e refer to "Moon" and "Earth".

Since the gravitational acceleration on Earth is $g_e = 9.81 m/s^2$ while on the Moon is $g_m=1.63 m/s^2$ , the ratio between the period on the Moon and on Earth is
$\frac{T_m}{T_e}= \sqrt{ \frac{g_e}{g_m} }= \sqrt{ \frac{9.81 m/s^2}{1.63 m/s^2} }=2.45$

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3 years ago

IF YOU MOVE 50 METERS IN TO SECONDS,

Yanka [14]

You can use photo math for This

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3 years ago

How a suit of armor might be a good analogy for a function of the skeleton system

RideAnS [48]

How might a suit of armor be a good analogy for a function of the skeletal system?

It's a frame for your body and protects organs and armor protects your body from injury

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3 years ago