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KATRIN_1 [288]
3 years ago
6

What type of relationship exists between momentum and mass?

Physics
1 answer:
alina1380 [7]3 years ago
4 0

Answer:

A. Direct

Explanation:

Thats just how it be

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A circular dartboard has a radius of 2 meters and a red circle in the center. Assume you hit the target at a random point. For w
Sati [7]

Answer:

1.549 m

Explanation:

Given:

The radius of the circular board, r = 2 m

The probability of hitting the red is given as 0.6

Now, this probability of hitting the red can be conclude as

0.6 = (Area of red)/ (Total area of the board)

Total area of the board = πr² = π × 2²

let the radius of the red area be R

thus, area of red circle, = πR²

on substituting the value of the area, we have

0.6 = (πR²)/ (π × 2²)

or

R² = 2.4

or

R = 1.549 m

Thus, the radius of the red circle is 1.549 m

3 0
3 years ago
In the diagram, the wavelength is shown by :
MAXImum [283]

Answer:

i think the anwer is C

Explanation:

6 0
4 years ago
For each question, select the right answer from the choices below:
Leni [432]

Option (ii) B is the correct option. The object on the moon has greater mass.

To resolve this, utilize the formulas Force = Mass * Acceleration.

The equation can be used to find the mass given the force in Newtons, using 9.8 m/s² for the acceleration of gravity of the earth and 1.6 m/s² for the moon.

Calculating the mass on earth:

30 N = 9.8 m/s² * mass

This results in a mass of 3.0 kg for the object on Earth.

Calculating the mass of the moon:

30 N = 1.6 m/s²2 * mass

Thus, the moon's object has a mass of 19. kg.

This can be explained by the fact that the earth has a stronger gravitational pull than the moon, producing more force per kilogram of mass. As a result, the moon's mass must be bigger to produce the same amount of force at a lower acceleration from gravity (1.6 m/s² vs. 9.8 m/s²).

To know more about Mass, refer to this link :

brainly.com/question/13386792

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3 0
1 year ago
Two football players collide head-on in midair while trying to catch a thrown football. The first player is 98.5 kg and has an i
romanna [79]

Answer:

v_f=0.825m/s

Explanation:

We must use conservation of linear momentum before and after the collision, p_i=p_f

Before the collision we have:

p_i=p_1+p_2=m_1v_1+m_2v_2

where these are the masses are initial velocities of both players.

After the collision we have:

p_f=(m_1+m_2)v_f

since they clong together, acting as one body.

This means we have:

m_1v_1+m_2v_2=(m_1+m_2)v_f

Or:

v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}

Which for our values is:

v_f=\frac{(98.5kg)(6.05m/s)+(119kg)(-3.5m/s)}{(98.5kg)+(119kg)}=0.825m/s

7 0
4 years ago
An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a p
kotykmax [81]

Answer:

a)v=1.77\times 10^6\ m/s

b)v=3.872\times 10^6\ m/s

c)v=5.5\times 10^6\ m/s

d)v=6.7\times 10^6\ m/s

e)v=7.7\times 10^6\ m/s

Explanation:

Given that

d = 2 cm

V = 200 V

u=4\times 10^5\ m/s

We know that

F = E q

F = m a

E = V/d

So

m a =  q .V/d b            

a=\dfrac{q.V}{m.d}                 ---------1

The mass of electron

m=9.1\times 10^{-31}\ kg

The charge on electron

q=1.6\times 10^{-19}\ C

Now by putting the all values in equation 1

a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2

a=1.5\times 10^{15}\ m/s^2

We know that

v^2=u^2+2as

a)

s = 0.1 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}

v=\sqrt{3.16\times 10^{12}}\ m/s

v=1.77\times 10^6\ m/s

b)

s = 0.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}

v=\sqrt{1.5\times 10^{13}}\ m/s

v=3.872\times 10^6\ m/s

c)

s = 1 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1\times 10^{-2}

v=\sqrt{3.06\times 10^{13}}\ m/s

v=5.5\times 10^6\ m/s

d)

s = 1.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1.5\times 10^{-2}

v=\sqrt{4.5\times 10^{13}}\ m/s

v=6.7\times 10^6\ m/s

e)

s = 2 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 2\times 10^{-2}

v=\sqrt{6.06\times 10^{13}}\ m/s

v=7.7\times 10^6\ m/s

5 0
4 years ago
Read 2 more answers
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