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tangare [24]
3 years ago
13

Can someone complete this assignment will give brainliest! The assignment is attached

Physics
1 answer:
Musya8 [376]3 years ago
3 0
... I can’t see the attached assignment that you put on here.
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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
iVinArrow [24]
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
7 0
4 years ago
How are feathers heavy as steel?
Alekssandra [29.7K]

Answer: This is because they both weigh a pound. As the density of steel is much greater than the density of feathers, a pound of feathers would make up a much greater volume.

Explanation: brainliest plz :)

6 0
3 years ago
A -1.12 μC charge is placed at the center of a conducting spherical shell, and a total charge of +8.65 μC is placed on the shell
Lisa [10]

Answer: 7.53 μC

Explanation: In order to explain this problem  we have to use the gaussian law so we have:

∫E.dS=Qinside/εo we consider a gaussian surface inside the conducting spherical shell so E=0

Q inside= 0 = q+ Qinner surface=0

Q inner surface= 1.12μC so in the outer surface the charge is (8.65-1.12)μC=7.53μC

7 0
4 years ago
Can someone please give me the (Answers) to this? ... please ...<br><br> I need help….
IceJOKER [234]

#1.

<em>Car </em>1<em> weighs </em>300 kilograms<em> and is moving right at </em>3 meters per second (m/s)

  • v1 (before) = 3 m/s

  • v2 (before) = -1 m/s

  • v1 (after) = 0.5 m/s

#2.

Law of conservation of momentum

momentum before collorion = momentim after collosion

MV + mv = MV' + mv'

1500x25+ 1000x5

37500 + 15000

6 0
2 years ago
2. A test reveals that 150 J of work is required to lift an object 3 m at a
Nuetrik [128]

Answer:

50N

Explanation:

W=Fd

150=F(3)

50N=F

7 0
3 years ago
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