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bearhunter [10]

3 years ago

12

a square loop of wire side 3.0 cm carries 3.0 A of current. A uniform magnetic field of magnitude 0.67 T makes an angle of 37 de

grees with the plane of the loop. What is the magnitude of the torque on the loop

1 answer:

Varvara68 [4.7K]3 years ago

8 0

Explanation:

Below is an attachment containing the solution.

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A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$

where

$F_{net}$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_{net}=(60)(3.1)=186 N$

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_{net}=F_a-R$

where:

$F_{net}$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_{net}=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_{net}+R=186+60=246 N$

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qwelly [4]

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You launch a ball at an angle of 35 degrees above the horizontal with an initial velocity of 38 m/s. What is the time the ball w

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0=38+(-9.8)(?)

?=38-0+(-9.8)

?=28.2 s

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No. A mirror works because of reflection.

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A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by υ⃗ =[5.00m/

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