All categories

2 years ago

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

You might be interested in

If a student flicks a stationary 0.1 kg ball with 5N of force for 0.1 seconds. What is its final

Alisiya [41]

5m/s

Explanation:

Given parameters:

Mass of ball = 0.1kg

Force on the ball = 5N

time taken = 0.1s

Unknown:

final speed of the ball = ?

Solution:

According to newton's second law "the net force on a body is the product of its mass and acceleration".

Force = mass x acceleration equation 1

Acceleration =

V is the final velocity

U is the initial velocity

T is the time taken

U = O since it is a stationary body;

a = $\frac{V}{T}$

Input "a" into equation 1

F = m x $\frac{V}{T}$

5 = 0.1 x $\frac{V}{0.1}$

V = 5m/s

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0

3 years ago

Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t

Musya8 [376]

Answer:

$n = 1.266\times 10^{12}$

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is $\theta = 13 degree$

thickness of string 300 mm = 0.3 m

$sin\theta =\frac{2}{0.3 m}$

r =0.3 sin 13 = 0.067 m

$Fe = \frac{ kq_1 q-2}{d^2}$

$Fe = \frac{kq^2}{(2r)^2} = mg tan\theta$

$q^2 = mg tan\theta \frac{(2r)^2}{k}$

$= 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}$

$q^2 = 4.10\times 10^{-14}$

$q = 2.026 \times 10^{-7} C$

q = ne

$n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}$

$n = 1.266\times 10^{12}$

3 0

2 years ago

A boy 11.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the i

ryzh [129]

12414253

Explanation:

8 0

2 years ago

Which statement best describes an isolated system? (1 point)

bonufazy [111]

Answer:

idk

Explanation:

6 0

3 years ago

Read 2 more answers

A 700 kg racecar slowed from 30 m/s to 15 m/s. What is the change in momentum

Lerok [7]

Momentum = (mass) x (speed)

Original momentum = (700 kg) x (30 m/s) = 21,000 kg-m/s

Final momentum = (700 kg) x (15 m/s) = 10,500 kg-m/s

Change in momentum = - 10,500 kg-m/s .

8 0

3 years ago