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bezimeni [28]
3 years ago
13

Label and describe what is happening in this picture

Physics
1 answer:
SOVA2 [1]3 years ago
7 0
Something is reproducing.
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In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vi
irinina [24]

To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where,

m_{1,2}= Mass of each object

v_{1,2} = Initial velocity of each object

v_f= Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

m_2v_2 = (m_1+m_2)v_f

(0.42)(21) = (90+0.42)v_f

v_f = 0.0975m/s

Therefore the velocity right after catching the ball is 0.0975m/s

8 0
3 years ago
How much work is done to move 2.10 μC of charge from the negative terminal to the positive terminal of a 3.50 V battery?
rjkz [21]

Answer:

6.3E-6

Explanation:

Workdone = V Q

4 0
3 years ago
You travel 23 meters north in 16 seconds, 5 meters south in 4 seconds, and 16 meters north in 18 seconds. Calculate your total d
dolphi86 [110]

Answer:

d: 44m

Δd:34  

Explanation:

d: 23+5+16= 44m

Δd: 23-5+16= 34m

4 0
3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
Which automobile is faster a chevrolet or a ford? Also explain pls
antiseptic1488 [7]

Answer:

depends on what type of car it is

Explanation:

6 0
2 years ago
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