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3 years ago

43.5 kg. Shawn rides his bike 1.3 km in

Physics

1 answer:

Alexeev081 [22]3 years ago

6 0

Answer:

K.E is 90.625 J

Explanation:

Data:

Mass = m = 43.5kg

Distance = d = 1.3km = 1300 m

Time = t = 10.4 min = 624 sec

Accelaration = a or g = 9.8

Formula for kinrtic energy

K.E = 1/2mv²

Find velocity

V = d/t = 1300/624 = 2.0833333333333 m/s

Put values in formula

K.E = 1/2 (43.5)(2.084)² = 90.624999999995

Mark brainliest if you understand it

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ndicate whether the statement is true or false. Whenever we move, we alter the rate at which we move into the future. A. True B.

Ksju [112]

Whenever we move, we alter the rate at which we move into the future. This statement is true.

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3 years ago

From a cliff 37.6 m high. At the level of the sea, a rock sticks out a horizontal distance of 12.12 m. The acceleration of gravi

Aleks [24]

Answer:

4.3 m/sec

Explanation:

Here height of cliff = y = 37.6 m

Gravitational acceleration = g = 9.8 m/sec2

vi = 0 m/s

Let's find the time which the diver will take if jumps from there!

Using formula

y = vit+1/2gt2

==> 37.6= 0 + 0.5 ×9.8× $t^{2}$

==> $t^{2}$ = $\frac{37.6}{4.9}$

==> t = 2.8 sec

In this time the diver has to cover a horizontal distance of 12.12 m

If x = 12.12 m is the horizontal distance to be covered then using

x= Vx × t

==> Vx = x/t

==> Vx= 12.12/2.8 = 4.3 m/s

8 0

3 years ago

A shopper is pushing a cart down a grocery store aisle. Starting from rest, the shopper applies a constant force to the cart for

inysia [295]

A shopping cart that starts from rest, is accelerated for 4 s, moves at constant velocity for 4 s, and is decelerated for 4s until returning to rest, has an average acceleration of 0 m/s².

A shopper is pushing a cart down a grocery store aisle. The movement of the cart is:

It starts from rest.
From t = 0 s to t = 4.0 s it is accelerated with a constant force.
From t = 4 s to t = 8.0 s it receives just enough force to balance the friction on the cart.
From t = 8 s to t = 12 s it is decelerated until it comes to rest.

All in all, at the initial time (t = 0 s), the velocity is 0 m/s (rest) and at the final time (t = 12 s) the velocity is 0 m/s as well (rest). The average acceleration in that period is:

$a = \frac{v_{12}-v__o}{t_{12}-t_0} = \frac{0m/m-0m/s}{12s-0s} = 0 m/s^{2}$