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stira [4]
3 years ago
15

Imagine you have just witnessed a small avalanche on a mountain while skiing, and two slushy snowballs just crashed together in

a perfectly inelastic collision. They are moving as one larger snowball, as a combined mass. Before the collision, snowball A was 7 kg and had initial momentum of –14 kg · m/s; therefore, its velocity must have been ? m/s
Snowball B had initial momentum of 15 kg ∙ m/s, and a velocity of 5 m/s; therefore, its mass must have been ? kg.

Recognizing that momentum is conserved in inelastic collisions, the total momentum of the combined snowballs after the collision must be ? kg · m/s.
Physics
2 answers:
Black_prince [1.1K]3 years ago
8 0
The first is -2.

the second is 3. 

the third is 1. 
inn [45]3 years ago
6 0
We have that the momentum p is given by the formula p=mv where m is the mass and v is the velocity. Since for A p=-14kgm/s and m=7, we have that the velocity is -14/7=-2m/s. Hence its speed is 2 m/s.
For b we have that p=15kgm/s and v=3m/s. Because m=p/v, we have m=3kg.
We also have that the momentum is conserved in this system. Hence, the net sum of the momentum of the 2 snowballs equals the momentum of the single giant ball. Hence, p(total)=p(combined)=-14+15=1kgm/s (momentum is a vector; the positive sign means that it tends to the positive direction).

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A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2 , and 1.85 ms (1 ms = 10−3
ankoles [38]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 =  u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

4 0
3 years ago
You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of
solong [7]

Answer:

The velocity of the man from the frame of  reference of a stationary observer is, V₂ = 5 m/s

Explanation:

Given,

Your velocity, V₁ = 2 m/

The velocity of the person, V₂ =?

The velocity of the person relative to you, V₂₁ = 3 m/s

According to the relative velocity of two

                                V₂₁ = V₂ -V₁

∴                               V₂ =  V₂₁ + V₁

On substitution

                                 V₂ = 3 + 2

                                      = 5 m/s

Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s

8 0
3 years ago
What does the balloon of the air capacitor represent in an electrical capacitor?
bulgar [2K]

Answer:

The balloon prohibits the flow of air through the air capacitor.

Explanation:

Just like an electric capacitor has an insulator between the plates, the air capacitor has a balloon between the chambers.

6 0
3 years ago
Read 2 more answers
A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi
nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

W = F * d

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

d=h*N, for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

V_f^2-V_i^2 = 2a\Delta X

Where,

V_f = Final velocity

V_i = Initial Velocity

a = Acceleration

\Delta X = Displacement

PART A) For the particular case of work we know then that,

W = F*d

W = m*g*(h*N)

W = 50*9.8*(0.3*30)

W = 4.41kJ

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

V_f^2-V_i^2 = 2a\Delta X

Here,

V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow3 steps in one second

v_f = 0

Replacing,

V_f^2-V_i^2 = 2a\Delta X

0-0.9^2=2a(30*0.3)

Re-arrange for a,

a = -\frac{0.9^2}{2*30*0.3}

a = -45*10^{-3}m/s^2

At this point we can calculate the time, which is,

t = \frac{\Delta V}{a}

t = \frac{0-0.9}{-45*10^{-3}}

t = 20s

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0
3 years ago
Please anyone post any class 9 physics question​
aalyn [17]

Answer:

I will but can you just wait for some minutes cus I am in a hurry now.

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6 0
2 years ago
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