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3 years ago

A 900-kg giraffe runs across a field at a rate of 50 km/hr. What is the magnitude of its momentum? 18 km/hr

2 answers:

7 0

The correct answer is the one with 45000 kg*km/hr.
the formula is p = m*v
900 *50/hr giving u 45000
I know this is the correct answer because i have already turned it in and got a 100%.

balandron [24]3 years ago

7 0

Answer: The magnitude of momentum will be equal to 12,500 kg m/s.

Explanation:

Momentum is defined as the product of the mass of the object and velocity of the object.

Mathematically,

$p=mv$

where, p = momentum

m = mass of the object ( in kg) = 900 kg

v = velocity of the object ( in m/s) = 50km/hr = $\left(\frac{50\times 1000}{3600}\right)m/s$

Putting values in above equation, we get:

$p=900kg\times \left(\frac{50\times 1000}{3600}\right)m/s=12500kg.m/s$

Hence, the magnitude of momentum will be equal to 12,500 kg m/s.

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3 years ago

Read 2 more answers

A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$

v is the speed of cat, $v=\dfrac{2\pi r}{t}$

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So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

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3 years ago