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Assoli18 [71]
3 years ago
10

A car travels 60 miles due West first then turns back and travels 120 miles due East in 3 hours. What is...

Physics
1 answer:
ella [17]3 years ago
7 0

Answer:

<h2>A. 180 miles</h2><h2>B. 60 miles</h2><h2 />

Explanation:

In this problem, we are required to solve for the total distance that the car travelled. and the displacement

A) the distance travelled by car

this can be gotten by summing all the distances the car has travelled.

i,e total distance= 60 miles+120 miles

total distance= 180 miles

B) the displacement of the car

the displacement can be gotten by  subtracting the final distance from the initial distance

final distance = 120 miles

initial distance= 60 miles

displacement= 120-60= 60 miles

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True or false : In crystalline solids the particles are not arranged in a regular pattern
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Two satellites are in circular orbits around Earth. Satellite A has speed vA. Satellite B has an orbital radius nine times that
Pani-rosa [81]

Answer:

option B

Explanation:

given,

Satellite B has an orbital radius nine times that of satellite A.

R' = 9 R

now, orbital velocity of the satellite A

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now, orbital velocity of satellite B

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8 0
2 years ago
A 500-g lump of clay is dropped onto a 1-kg cart moving at 60 cm/s. The clay is moving downward at 30 cm/s just before l dith t
viktelen [127]

Answer:

The speed of the cart and clay after the collision is 50 cm/s .

Explanation:

Given :

Mass of lump , m = 500 g = 0.5 kg .

Velocity of lump , v = 30 cm/s .

Mass of cart , M = 1 kg .

Velocity of cart , V = 60 cm/s .

We know by conservation of momentum :

mv+MV=(m+M)v'

Here , v' is the speed of the cart and clay after the collision .

Putting all value in above equation .

We get :

0.5\times 30+1\times 60=(0.5+1)\times v'\\\\v'=\dfrac{15+60}{1.5}\\\\v'=50\ cm/s

Hence , this is the required solution .

3 0
3 years ago
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
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