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3 years ago

7

Starting a fire to

1 answer:

docker41 [41]3 years ago

8 0

I’m pretty sure that it’s D

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if a torque of 55.0 N/m is required and the largest force that can be exerted by you is 135 N what is th e length of the lever a

Whitepunk [10]

Answer:

$r=0.41m$

Explanation:

Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

$\tau=r\times F$

Due to the definition of cross product, the magnitude of the torque is given by:

$\tau=rFsin\theta$

Where $\theta$ is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when $sin\theta$ is equal to one, solving for r:

$r=\frac{\tau}{F}\\r=\frac{55\frac{N}{m}}{135N}\\r=0.41m$

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3 years ago

What is an independent or manipulated variable?

Paha777 [63]

Is the variable you change, independent, I, something I change.

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3 years ago

If a refrigerator is a heat pump that follows the first law of thermodynamics, how much heat was removed from food inside of the

Anna007 [38]

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4 years ago

Read 2 more answers

Please do number 11 need this question bad

ArbitrLikvidat [17]

Gas always expands or contracts to exactly fill whatever you put it in.

So to measure the volume of a gas, just measure the volume of the jar, the tank, the bottle, the can, or the balloon that the gas is in.

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3 years ago

Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25

Sveta_85 [38]

Answer:

8.6 miles

Explanation:

We need to calculate the components of the total displacement along the east-west and north-south directions first.

In the first part, Erica moves 5.2 miles at 25∘ north of east. So the components of this displacement along the two directions are:

East: $d_{1x} = 5.2 cos 25^{\circ}=4.7 mi$

North: $d_{1y} = 5.2 sin 25^{\circ}=2.2 mi$

In the second part, Erica moves 5.0 miles north. So, the components of this displacement are:

East: $d_{2x}=0$

North: $d_{2y} = 5.0 mi$

So the components of the total displacement are

East: $d_x = d_{1x}+d_{2x}=4.7 + 0 = 4.7 mi$

North: $d_y = d_{1y}+d_{2y}= 2.2 + 5.0 = 7.2 mi$

Therefore the magnitude of the displacement, which is the straight-line distance from the starting point to the end of the race, is

$d=\sqrt{d_x^2 +d_y^2}=\sqrt{4.7^2+7.2^2}=8.6 mi$

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4 years ago