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4 years ago

15

Two objects of different mass have equal, non-zero kinetic energies. Which object has the greater momentum?

1 answer:

Alchen [17]4 years ago

4 0

Momentum= mass*velocity thus if you keep velocity constant an increase in mass increases momentum so the heavier object (assuming constant gravity) has a higher momentum than a lighter object

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Ice caps at the north and south poles are very reflective the ice reflects light and heat rather than absorbing it. If global wa

PolarNik [594]

The way you described it, a small change will cause a bigger change.
That's Positive feedback.

5 0

3 years ago

A 6 kg block is attached to a massless rope that passes over a frictionless pulley. The pulley can be modeled as a uniformly den

scoundrel [369]

The effective mass of the system is 12 kg.

<u>Explanation:</u>

The effective mass of the system is given by the expression,

= mₓ + 1/2 m₁ + 2/5 m₂

In the above equations the subscript x denotes block B and the subscript 2 denotes the mass of the solid sphere and the subscript 1 denotes the pulley.

Expressing the above functions with the data given below. we have,

= 6 kg + 1/2 (4 kg) + 2/5 (10 kg)

= 12 kg.

with the findings above we can conclude that the effective mass of the system is 12 kg

6 0

3 years ago

Calculate the period of wave whose frequency is 50Hz

miss Akunina [59]

Explanation:

time period= 1/frequency.

= 1/50 = 0.02 second.

hope this helps you.

5 0

3 years ago

Read 2 more answers

A light ray is incident on a plane surface separating two transparent media of refractive indices of 1.65 and 1.35. If the incid

Vesna [10]

Explanation:

Given

Refractive Index $(n_1)=1.65$

$n_2=1.35$

angle of incident $=35 ^{\circ}$

Light originates in the medium of higher refractive index

Let $\theta _1$ and $\theta _2$ be the incident and refraction angles respectively thus according to snell's law

$n_1sin\theta _1=n_2sin\theta _2$

$1.65\times sin35=1.35\times sin\theta _2$

$sin\theta _2=0.701$

$\theta _2=44.507^{\circ}$

(b)Ray originates from lighter medium

$1.35\times sin35=1.65\times sin\theta _2'$

$sin\theta _2'=0.469$

$\theta _2'=27.96\approx 28^{\circ}$

8 0

4 years ago

A volleyball player bumps a ball across a net with the velocity and angle shown below. What is the maximum height of the ball?

Marrrta [24]

Answer:

D. 12.4 m

Explanation:

Given that,

The initial velocity of the ball, u = 18 m/s

The angle at which the ball is projected, θ = 60°

The maximum height of the ball is given by the formula

h = u² sin²θ/2g m

Where,

g - acceleration due to gravity. (9.8 m/s)

Substituting the values in the above equation

h = 18² · sin²60 / 2 x 9.8

= 18² x 0.75 / 2 x 9.8

= 12.4 m

Hence, the maximum height of the ball attained, h = 12.4 m

6 0

4 years ago