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4 years ago

11

You should always wear your seat belt just in case the car comes to an abrupt stop. The seat belt will hold you in place so that

your body does not continue moving when the car stops.

2 answers:

Schach [20]4 years ago

7 0

Answer:

Inertia, first

Explanation:

did the assignment

Gwar [14]4 years ago

5 0

Newtons 1st law of motion states that the object will continue to move at its present speed and direction until an outside force acts upon it.

So unless the objects inside the car are restrained, they will continue moving at whatever speed the car is traveling at, even if the car is stopped by a crash.

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A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s . The coefficient o

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2.04 meters distance is traveled by the sled before stopping.

Mass of the sled = m

The initial speed of the sled = 2 m/s

Coefficient of kinetic friction between sled and ice = 0.100

Let the distance the sled moves before it stops be d.

Gravity = 9.8 m/ s²

Let the initial kinetic energy sled be

$= K _{i}$

$K_{i} = \frac{1}{2} mv ^{2}$

The work done by the frictional force is,

$Work \: done \: by \: frictional \: force =W_{f}$

$W _{f} = μ_{k}mgd$

Work done by frictional force= Initial kinetic energy of the sled

$W_{f} = K_{i}$

$μ_{k}mgd= \frac{1}{2} mv ^{2}$

So, the distance traveled by the sled before stopping is

$d= \frac{1mv ^{2} }{2 \:μ_{k}mg}$

$d= \frac{1v ^{2} }{2 \:μ_{k}g}$

$d= \frac{2^{2} }{2 \times \:0.100 \times 9.8}$

$d= 2.04 \: m$

Therefore, the distance traveled by the sled before stopping is 2.04 meters.

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2 years ago

What would an automobile engineer MOST LIKELY suggest to improve the efficiency of a car's engine? A) Remove the engine's coolin

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Answer:

D) Reduce the internal friction of the engine's parts.

Explanation:

To increase the efficiency of the car's engine, an engineer most like has to reduce the friction within the internal engine parts. Reducing these friction reduces the useful energy lost as heat in these internal parts of the engine, when the engine parts do work against friction to move. In most everyday activities, engines and machines, energy is usually lost as heat due to frictional forces arising from two or more surfaces in contact.

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4 years ago

1. My grass is dying, and I believe it's because it is not getting enough water. Sol

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Answer: I actually need the same answer

Explanation:

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3 years ago

A resultant force of magnitude 15 N acts on a body of mass 250 g. Calculate the magnitude of the acceleration

lakkis [162]

Answer:

a=60

Explanation:

F=ma

F=15N

m=250g convert to kg

m=250/1000

m=0.25kg

applying F=ma

15=0.25*a

a=15/0.25

a=60

4 0

3 years ago

Your starship lands on a mysterious planet. As chief scientist-engineer, you make the following measurements: A 2.50-kg stone th

olga55 [171]

Answer:

Part a)

$M = 6.08 \times 10^{19} kg$

Part b)

$T = 4510 hours$

Explanation:

As we know that stone is thrown upwards with speed

$v_i = 12 m/s$

Now it returns back to the surface of Earth after t = 6 s

so the displacement of the stone is zero

$\Delta y = 0 = v t + \frac{1}{2}at^2$

$0 = 12 t - \frac{1}{2}g t^2$

$g = \frac{2(12)}{t}$

$g = 4 m/s^2$

Part a)

Now we know that the circumference of the planet at the equator is of length

$L = 2 \times 100 km$

$2\pi R = 2\times 10^5 m$

$R = 3.2 \times 10^4 m$

Now we have formula of acceleration due to gravity as

$g = \frac{GM}{R^2}$

$4 = \frac{6.67 \times 10^{-11} M}{(3.2 \times 10^4)^2}$

$M = 6.08 \times 10^{19} kg$

Part b)

Time to complete one revolution around the planet is given as

$T = 2\pi\sqrt{\frac{r^3}{GM}}$

here we know that

r = distance from center of the planet

$r = 3.2 \times 10^4 + 3\times 10^7 = 3.003 \times 10^7 m$

now we have

$T = 2\pi\sqrt{\frac{(3.003\times 10^7)^3}{(6.67 \times 10^{-11})(6.08\times 10^{19})}}$

$T = 4510 hours$

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3 years ago