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2 years ago

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A sled is pulled at a constant velocity across a horizontal snow surface. If a force of 8.0 times 10^1 N is being applied to the

sled rope at an angle of 53 degrees to the grind, what is the force of friction between the sled and the snow?

1 answer:

Yanka [14]2 years ago

7 0

Answer:

48 Newtons

Explanation:

There is no acceleration and net forces.
$8*10x^{-1} N cos(53) = friction$
Answer: 48.15 Newtons

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klasskru [66]

Answer:

The magnitude of the car's acceleration as it slows during braking is 36.81 m/s²

Explanation:

From the question, the given values are as follows:

Initial velocity, u = 90 m/s

final velocity, v = 0 m/s

distance, s = 110 m

acceleration, a = ?

Using the equation of motion, v² = u² + 2as

(90)² + 2 * 110 * a = 0

8100 + 220a = 0

220a = -8100

a = -8100/220

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The value for acceleration is negative showing that car is decelerating to a stop. The magnitude of the car's acceleration as it slows during braking is therefore 36.81 m/s²

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3 years ago

A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washe

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To solve this exercise it is necessary to take into account the concepts related to thermal expansion.

The thermal expansion is given by the function,

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By quickly deducing the formula, we can realize that the greater the change in temperature, the greater the change in the length of the radius.

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A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed

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Answer:

Explanation:

Time taken to accelerate to 28 m /s

= 28 / 2 = 14 s

a ) Total length of time in motion

= 14 + 41 + 5

= 60 s .

b )

Distance covered while accelerating

s = ut + 1/2 at²

= 0 + .5 x 2 x 14²

= 196 m .

Distance covered while moving in uniform motion

= 28 x 41

= 1148 m

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v = u - at

0 = 28 - a x 5

a = 5.6 m / s²

v² = u² - 2 a s

0 = 28² - 2 x 5.6 x s

s = 28² / 2 x 5.6

= 70 m .

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3 years ago

In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn

kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

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Distance = 30 m

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Using equation of motion

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Put the value in the equation

$30=0+\dfrac{1}{2}\times1.6\times t_{A}^2$

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$t_{A}=6.12\ sec$

We need to calculate the time for B

Using equation of motion

$s=ut+\dfrac{1}{2}at_{B}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times2.0\times t_{B}^2$

$t_{B}=\sqrt{\dfrac{30\times2}{2.0}}$

$t_{B}=5.48\ sec$

We need to calculate the time for longest nap

Using formula for difference of time

$t'=t_{A}-t_{B}$

$t'=6.12-5.48$

$t'=0.64\ s$

Hence, B can take 0.64 sec for the longest nap .

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3 years ago