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Scrat [10]
3 years ago
13

PLEASE HELP QUICK WILL MARK BRAINLIEST

Physics
1 answer:
ZanzabumX [31]3 years ago
7 0

Answer:

Dont you already have it marked and i think its C

hope this helps :D

if im wrong i am so so sorry

Explanation:

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A 4-kilogram ball moving at 8 m/sec to the right collides with a 1-kilogram ball at rest. After the collision,
elena-14-01-66 [18.8K]
Conservation of momentum: total momentum before = total momentum after

Momentum = mass x velocity

So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
32+0=32

Therefore after the collision
4kg x 4.8m/s = 19.2
1kg x βm/s = β
19.2 + β = 32

Therefore β = 12.8 m/s
4 0
4 years ago
In a football game, a receiver is standing still, having just caught a pass. Before he can move, a tackler, running at a velocit
ivolga24 [154]

Answer:

m_{receiver}=115Kg*3.1/(1.6)-115Kg=107.8Kg    

Explanation:

The football players collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V=1.6m/s.

We need to use the conservation of momentum Law, the total momentum is the same before and after the collision, at the initial point the receiver does not have any speed

m_{tackler}*v_{tackler}=(m_{tackler}+m_{receiver})V     (1)

We solve in order to find the receiver mass:

m_{receiver}={m_{tackler}*v_{tackler}/V}-m_{tackler}    

m_{receiver}=115Kg*3.1/(1.6)-115Kg=107.8Kg    

5 0
3 years ago
A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of b
lakkis [162]

Answer:

T = 17.26 ^oC

Explanation:

At thermal equilibrium we have heat given by aluminium must be equal to the heat absorbed by the water

so we will have

Q_1 = Q_2

m_1s_1\Delta T_1 = m_2s_2\Delta T_2

so we will have

32.5(900)(45.8 - T) = 105.3(4186)(T - 15.4)

so we have

(45.8 - T) = 15.1(T - 15.4)

so we have

16.1 T = 277.87

T = 17.26 ^oC

6 0
3 years ago
The nonreflective coating on a camera lens with an index of refraction of 1.21 is designed to minimize the reflection of 570-nm
lord [1]

Answer: 117.8 nm

Explanation:

Given,

Nonreflective coating refractive index : n = 1.21

Index of refraction: n_0 = 1.52

Wave length of light = λ = 570 nm = 570\times10^{-9}\ m

\text{ Thickness}=\dfrac{\lambda}{4n}

=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}

Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

5 0
4 years ago
A swing has a period of 10 seconds. What is its frequency ?
kow [346]

Frequency = 1 / (period)

Frequency = 1 / (10 seconds) = (1/10) ( / second) = 0.1 per second = <em>0.1 Hz</em>.


4 0
3 years ago
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