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3 years ago

15

A car is coasting backwards downhill at a speed of 3.0 m/s when the driver gets the engine started. After 2.5

1 answer:

astra-53 [7]3 years ago

4 0

Answer:

0.6m/s²

Explanation:

Given parameters:

Initial velocity = 3m/s

Final velocity = 4.5m/s

Time taken = 2.4s

Unknown:

Average acceleration = ?

Solution:

To solve this problem, we use the expression:

Acceleration = $\frac{v - u}{t}$

v is the final velocity

u is the initial velocity

t is the time taken

Acceleration = $\frac{4.5 - 3}{2.5}$ = 0.6m/s²

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The amount of work done against friction to slide a box in a straight line across a uniform, horizontal floor depends most on th

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Amount of force that is applied to the box

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3 years ago

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16 to 19 year old male and females are how much more likely to be involved in a crash?

Sedbober [7]

I say around 40% - 60%

https://www.dmv.ca.gov/portal/dmv/detail/teenweb/more_btn6/traffic/traffic
http://www.teendriversource.org/stats/support_teens/detail/57
http://www.rmiia.org/auto/teens/Teen_Driving_Statistics.asp

(I just corrected the question. Sorry if it is still incorrect.)

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4 years ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

For planet B the orbital radius $r_{2} =2r_{1}$ and mass $m_{2} = 3 m_{1}$

Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

$f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }$

$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

Hence the ratio is $\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }$

$=\frac{3}{4}$ [ ans]

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4 years ago

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In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e

Zanzabum

Answer:

The velocity is $v_t = 0.02175 \ m/s$

Explanation:

From the question we are told that

The mass of the bullet is $m_b = 0.024 \ kg$

The initial speed of the bullet is $u_b = 1200 \ m/s$

The mass of the target is $m_t = 320 \ kg$

The initial velocity of target is $u_t = 0 \ m/s$

The final velocity of the bullet is is $v_b = 910 \ m/s$

Generally according to the law of momentum conservation we have that

$m_b * u_b + m_t * u_t = m_b * v_b + m_t * v_t$

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4 years ago

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Likurg_2 [28]

Answer:

1

Explanation:

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3 years ago