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wariber [46]
2 years ago
11

PLS HELP ASAP I REALLY NEED HELP AND WILL GIVE BRAINLIEST <2

Physics
1 answer:
Vedmedyk [2.9K]2 years ago
6 0

Answer:

6. Paula is correct. Think about buildings, buildings are constantly pushed by different forces like wind or gravity. The buildings aren’t moving but this doesn’t mean that they aren’t under force

Explanation:

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When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo
Sliva [168]

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s  first law, the net force on gymnast,

F_{net} =F-W=ma

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast (9\times g acceleration due to gravity)  

Therefore,

F= ma+W OR F=ma+mg=m(g+a)

Given m = 30 kg anda=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}

Substituting these values in above formula and calculate the force exerted by the gymnast,  

F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )

F=3.537\times10^{3}N

6 0
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
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
3 years ago
An objects speed is the distance it travels____the amount of times it takes?
beks73 [17]
Multiplied by; speed = distance x time
8 0
3 years ago
A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect
zaharov [31]

Answer:

X=X_o+\dfrac{1}{2}gt^2

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

\dfrac{1}{2}mV^2=mg(X-X_o)

So

V=\sqrt{2g(X-X_o)}

We know that

\dfrac{dX}{dt}=V

\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}

\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX

At t= 0 ,X=Xo

So we can say that

X=X_o+\dfrac{1}{2}gt^2

So the length of cable after time t

X=X_o+\dfrac{1}{2}gt^2

6 0
3 years ago
Please help only have a short amount of time left!
VARVARA [1.3K]

Answer:

I think all will display the same value

Explanation:

Sorry if this is wrong

6 0
2 years ago
Read 2 more answers
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