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ahrayia [7]
3 years ago
8

Use newton's first law of motion to explain why you feel tossed around whenever a roller coaster goes over a hill or through a l

oop.
Physics
2 answers:
REY [17]3 years ago
5 0

Explanation:

A roller coaster is hard to stop because it has a lot of inertia. Inertia can be simplified to, "objects want to keep doing what they are doing." It's when they are forced to do something different that they are doing that is why roller coaster riders experience the float or tossing around of the ride.

ikadub [295]3 years ago
4 0
Newton's first law states that an object at rest will stay at rest && an object in motion will remain in motion unless acted on by another force. With that being said you feel tossed around because you are initially at rest but when you ride the roller coaster you take it's course of motion. 

Hoped this helped!

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horrorfan [7]

Answer:

W= 4.89 KJ

Explanation:

Lets take

temperature of hot water T₁ = 100⁰C

T₁ = 373 K

Temperature of cold ice T₂= 0⁰C

T₂ = 273 K

The latent heat of ice LH= 334 KJ

The heat rejected by the engine Q= m .LH

Q₂= 0.04 x 334

Q₂= 13.36 KJ

Heat gain by engine = Q₁

For Carnot engine

Q_1=\dfrac{T_1}{T_2}Q_2

Q_1=\dfrac{373}{273}\times 13.36

Q₁  = 18.25 KJ

The work W= Q₁  - Q₂

W= 18.25 - 13.36 KJ

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7 0
3 years ago
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Sladkaya [172]

Ff=μN=μmg=0.255*20=5.1 N

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3 years ago
How much work is done if a force of 20 N moves an object a distance of 6 m?​
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Explanation:

W=F×s

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3 years ago
It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass
vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0
3 years ago
Find the instantaneous velocity at 1 s . can anyone help with c-h!!!
Lelu [443]
Rise over run at 1 second
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