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3 years ago

A 68-kg skydiver has a speed of 52 m/s at an altitude of 670 m above the

Physics

1 answer:

Arlecino [84]3 years ago

4 0

Answer:

91936J

Explanation:

We know that kinetic energy= 1/2 mv^2

M= mass = 68 Kg

v= velocity= 52 m/s

KE=1/2 × 68 × (52)^2

KE= 1/2 × 68 × 2704

KE= 91936J

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A projectile fired up into the air at an angle has a range of 235 m and a flight time of 47 s.

madam [21]

<h2>Answer:5 $ms^{-1}$ ,133.6 $m$ ,51.18 $ms^{-1}$ </h2>

Explanation:

Let $v_{x}$ , $v_{y}$ be the horizontal and vertical components of velocity.

Question a:

Horizontal component of velocity is the ratio of range and time of flight.

So,horizontal component of velocity is $\frac{235}{47}=5ms^{-1}$

So, $v_{x}=5ms^{-1}$

Question b:

Time of flight= $\frac{2v_{y}}{g}$

So, $v_{y}=\frac{47\times 9.8}{2}=51.18ms^{-1}$

Maximum height is given by $\frac{v_{y}^{2}}{2g}$

So,maximum height is $\frac{51.18^{2}}{2\times 9.8}=133.6m$

Question c:

The vertical velocity is already calculated in Question b.

$v_{y}=51.18ms^{-1}$

7 0

3 years ago

The centers of two 15.0 kg spheres are separated by 3.00 m. The magnitude of the gravitational force between the two spheres is

kompoz [17]

we have to use newtons law of gravitation which is
F=GMm/r^2
G=6.67 x 10^<span>-11N kg^2/m^2
</span>M=<span>(15kg)
</span>m=15 kg
r=(3.0m)^2<span>
</span>putting values we have
<span>=(6.67 x 10^-11N kg^2/m^2)(15kg)(15kg)/(3.0m)^2 </span>
=1.67 x 10^-9N

7 0

3 years ago

two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco

Reika [66]

Answer:

$t=750s$

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

$\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt$

$x_f$ is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have: