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3 years ago

7

The magnitude of the normal force is

1 answer:

rjkz [21]3 years ago

7 0

45N
The normal is perpendicular to the forces that go up or down the slope. The downward force is just the weight .

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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:

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

3 0

3 years ago

Ill give 15 points to the one who helps me with this

alexandr402 [8]

Machenical number one

6 0

3 years ago

Some of the principal minerals commonly found in igneous rocks are

Verdich [7]

I believe that the answer is C. Hope this Helps:)))

3 0

3 years ago

Gravity on earth is 9.8 m/s squared, and gravity on the moon is 1.6 m/s squared. So if the mass of an object on earth is 40 kilo

garik1379 [7]

The mass of an object on Earth is the same as its mass on the Moon. The weight is different.

Weight = m * g

Weight ( Moon ) = 40 kg * 1.6 m/s² = 64 N

If the mass of an object on Earth is 40 kg, its mass on the Moon is 40 kg and its weight on the Moon is 64 N.

7 0

3 years ago

An object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object. Use

shtirl [24]

Answer:

-30m/s

Explanation:

Given:

Initial velocity of object = 200 feet/second

Final velocity of object = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration "a" is given by:

$a=\frac{v_f-v_i}{t}$

where vf represents final velocity, vi represents initial velocity and is time of travel.

Plugging in values to evaluate acceleration.

$a=\frac{50-200}{5}$

$a=\frac{-150}{5}$

$a= -30m/s$

The acceleration of the object is -30m/s

3 0

3 years ago