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Oxana [17]
3 years ago
7

Carl runs 100m West. He then runs 60m North. He then runs 20m East. The total distance travelled by Carl is ______m and his tota

l displacement is _____m.
Physics
1 answer:
drek231 [11]3 years ago
4 0

<u>Answer:</u>

   Displacement = 100 meter

   Distance covered = 180 meter.

<u>Explanation:</u>

   Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

  Carl runs 100 m West, Displacement = -100 i

  He then runs 60 m North, Displacement = 60 j

   He then runs 20 m East, Displacement = 20 i

  Total displacement of Carl's running = -100 i + 60 j + 20 i = (-80 i + 60 j) m (Vector)

  Magnitude of displacement =  \sqrt{(-80)^2+60^2} =100 m

  Distance traveled by Carl = 100 + 60 + 20 = 180 m ( Scalar)

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5. 3 women push a stalled car. Each woman pushes with a 400N force. What is the mass of the car if the car accelerates at 0.85 m
iris [78.8K]

Answer:

<h2>470.59 kg</h2>

Explanation:

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3 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
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There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
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