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Evgen [1.6K]
2 years ago
14

What is the momentum of a bird with a mass of 0.015 kg flying 12 m/s?

Physics
1 answer:
maksim [4K]2 years ago
6 0

Answer:

momentum = mass • velocity

v= 17.5/2.5

= 7 m/s

Explanation:

I TOOK THE TEST

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In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr
NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

A)

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

t=\frac{2u sin \theta}{g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is t.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t

B)

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

h=\frac{u^2 sin^2\theta}{2g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is h.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h

C)

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

D=\frac{u^2 sin(2\theta)}{g}

where:

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D

7 0
3 years ago
Let theta denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the Bob is m, the
posledela
The tension has to hold the part of the weight in the direction of the string:

T = mg*cos(theta)

Theta=0, whole weight, theta=90, T=0, if the pendulum is horizontal, the string will be loose! Yeah
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3 years ago
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For Jane to see an image, light must enter her eyes. What specifically is entering her eyes when she sees an image?
Sergio [31]
Energy
because once the light hits her eyes energy flows through her body so the answer is A energy
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3 years ago
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An oil film with refractive index 1.48 and thickness 290 nm is floating on water and illuminated with white light at normal inci
VikaD [51]

Answer:

572.3 nm

Explanation:

n_{oil} = refractive index of the oil film = 1.48

t_{oil} = thickness of the oil film = 290 nm

\lambda = wavelength of the dominant color

m = order

Using the equation

2 n_{oil} t_{oil} = (m + 0.5) \lambda

For m = 0

2 (1.48) (290) = (0 + 0.5) \lambda

\lambda = 1716.8 nm

For m = 1

2 (1.48) (290) = (1 + 0.5) \lambda

\lambda = 572.3 nm

For m = 2

2 (1.48) (290) = (2 + 0.5) \lambda

\lambda = 343.4 nm

Hence the dominant color wavelength is 572.3 nm

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while working out a man performed 2525j of work in 19seconds . what was his power A:132.9w. B:241.5w C 47.975w. D100.5w
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A: 132.9w because 2525\19 is how much energy transferred per second which is also known as the power
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