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Bess [88]
3 years ago
7

An 80-kgkg quarterback jumps straight up in the air right before throwing a 0.43-kgkg football horizontally at 15 m/sm/s . How f

ast will he be moving backward just after releasing the ball?
Physics
1 answer:
vladimir1956 [14]3 years ago
5 0

Answer:

V = 0.0806 m/s

Explanation:

given data

mass quarterback = 80 kg

mass football = 0.43 kg

velocity = 15 m/s

solution

we consider here momentum conservation is in horizontal direction.

so that here no initial momentum of the quarterback

so that final momentum of the system will be 0

so we can say

M(quarterback) ×  V = m(football) × v (football)   ........................1

put here value we get

80 ×  V  = 0.43  × 15

V = 0.0806 m/s

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To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i
ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

m_1v_1+m_2v_2=(m_1+m_2)v

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

m_1(-v_1)+m_2v_2=(m1+m2)(-v2)

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s

hence, the time is t=0.42 s

4 0
3 years ago
What is the reaction force if a girl pulls on a cow?
Papessa [141]
Answer : B) The cow pulls back on the girl.

From newton’s third law we know that every action has a reaction force pushing back. So when the girl pulls on a cow, the cow is pulling back on her.
8 0
3 years ago
An object starts at rest. Its acceleration over 30 seconds is shown in the graph below:
ddd [48]

Answer:

The instantaneous speed of the object after the first five seconds is 12.5 m/s.

(C) is correct option.

Explanation:

Given that,

An object starts at rest. Its acceleration over 30 seconds.

We need to calculate the instantaneous speed of the object after the first five seconds

We know that,

Area under the acceleration -time graph gives speed.

According to figure,

speed = area\ of\ tringle

speed=\dfrac{1}{2}\times b\times h

speed =\dfrac{1}{2}\times5\times5

speed-12.5\ m/s

Hence, The instantaneous speed of the object after the first five seconds is 12.5 m/s.

6 0
3 years ago
If a pickup is placed 16.25 cm from one of the fixed ends of a 65.00-cm-long string, which of the harmonics from n=1 to n=12 wil
Lina20 [59]

Answer:

The answer to this question can be defined as follows:

Explanation:

Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.

Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.

4 0
3 years ago
[BWS.02]If the same experiment is repeated in different parts of the world by different scientists,
guajiro [1.7K]

Answer:

the results will be the same.it may be

7 0
3 years ago
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