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4 years ago

13

A car moves to the right as shown. In order for the

1 answer:

spayn [35]4 years ago

8 0

Answer:

1

Explanation:

The car's acceleration must be equals to or less than the car's velocity, cause of the change in velocity.

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PLEASE HELP QUICK ASAP HURRY This is Earth Science

SCORPION-xisa [38]

Sun this is the answer

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3 years ago

What happens to an object as its kinetic energy increases?

Reptile [31]

Its moves more, If an object moves it has a kinetic enegy

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3 years ago

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

Oksanka [162]

Answer:

vBxf = 0.08625m/s

Explanation:

This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.

$p_f=p_i$

pf: final momentum

pi: initial momentum

The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.

Then, you have:

$m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}$ (1)

mQ: the mass of the quarterback = 80kg

mB: the mass of the football = 0.43kg

(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s

(vBx)i: the horizontal velocity of football before being thrown = 0m/s

(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?

(vBx)f: the horizontal velocity of football after being thrown = 15 m/s

You replace the values of the variables in the equation (1), and you solve for (vBx)f:

$0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}$

Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.

Hence, the speed of the quarterback after he throws the ball is 0.08625m/s

6 0

4 years ago

speed is the ratio of the distance an object moves to a. the direction the object moves. b. the amount of time needed to travel

Dennis_Churaev [7]

It is the ratio of distance an object moves to the amount of time needed to travel the distance. Option b.

5 0

3 years ago

Read 2 more answers

An α-particle has a charge of +2e and a mass of 6.64 × 10-27 kg. It is accelerated from rest through a potential difference that

zzz [600]

Answer with Explanation:

We are given that

Charge on alpha particle=q=2 e= $2\times 1.6\times 10^{-19} C$

1 e= $1.6\times 10^{-19} C$

Mass of alpha particle=m= $6.64\times 10^{-27} kg$

Potential difference,V= $1.97\times 10^6 V$

Magnetic field,B=3.49 T

a.Speed of alpha particle=v= $\sqrt{\frac{2 qV}{m}}$

By using the formula

$v=\sqrt{\frac{2\times 2\times 1.6\times 10^{-19}\times 1.97\times 10^6}{6.64\times 10^{-27}}$

$v=1.38\times 10^7 m/s$

b.Magnetic force,F $=qvB=2\times 1.6\times 10^{-19}\times 1.38\times 10^7\times 3.49=1.5\times 10^{-11} N$

F= $1.5\times 10^{-11} N$

c.Radius of circular path, r= $\frac{mv^2}{F}$

$r= \frac{6.64\times 10^{-27}\times (1.38\times 10^7)^2}{1.5\times 10^{-11}}$

r=0.084 m

5 0

3 years ago