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Nitella [24]
3 years ago
9

A cheetah runs at a constant velocity of 7 m/s. What is it’s acceleration in m/s/s PLEASE HELP

Physics
1 answer:
uysha [10]3 years ago
5 0

Answer:

0 m/s²

Explanation:

Acceleration is the change in velocity over change in time.  If the velocity is constant, then the acceleration is 0.

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When a heart artery is clogged and the heart muscle dies, the condition is _______________ ischemia.
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What is density?
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A flying squirrel has a mass of 0. 765 kg. He jumps off the tree and hits the ground with 125 joules of energy. Determine how hi
uranmaximum [27]

Answer:

Height, h = 16.67 m

Explanation:

We have,

Mass of a squirrel is 0.765 kg.

He jumps off the tree and hits the ground with 125 joules of energy.

It is required to find the height up on the tree the squirrel was when it jumped.

The energy possessed by the squirrel is called its gravitational potential energy. It can be given by :

E=mgh

h is height up on the tree the squirrel was when it jumped

h=\dfrac{E}{mg}\\\\h=\dfrac{125\ J}{0.765\ kg\times 9.8\ m/s^2}\\\\h=16.67\ m

So, the squirrel will go to a height of 16.67 m.

6 0
3 years ago
3
Misha Larkins [42]

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

KE = \frac{1}{2}mv^2

KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)

Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:

U_s = \frac{1}{2}kx^2

k = Spring Constant (1.14 × 10⁷ N/m)

x = compressed distance of bumper (? m)

Since energy is conserved:

E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2

We can simplify and solve for 'x'.

mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}

Plug in the givens and solve.

x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}

3 0
2 years ago
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