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Paladinen [302]
3 years ago
13

A spring has a spring constant of 330 n/m. how far is the spring compressed when 150 newtons of force are used?

Physics
1 answer:
pishuonlain [190]3 years ago
5 0

Answer:

I think its B

Explanation:

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Hélène de Pourtalès she was the first women
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A car is traveling at 100 km/h when the driver sees an accident 80 m ahead and slams on the brakes. what minimum constant decele
Effectus [21]
Assuming the driver starts slamming the brakes immediately, the car moves by uniformly decelerated motion, so we can use the following relationship
2aS = v_f^2 - v_i^2 (1)
where 
a is the deleceration
S is the distance covered after a time t
v_f is the velocity at time t
v_i=100 km/h = 27.8 m/s is the initial speed of the car

The accident is 80 m ahead of the car, so the minimum deceleration required to avoid the accident is the value of a such that S=80 m and v_f=0 (the car should stop exactly at S=80 m to avoid the accident). Using these data, we can solve  the equation (1) to find a:
a=- \frac{v_i^2}{2 S}= -\frac{(27.8 m/s)^2}{2 \cdot 80 m} =-4.83 m/s^2
And the negative sign means it is a deceleration.

4 0
3 years ago
An ice skater glides for two meters across ice is work done or no work done
garik1379 [7]
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5 0
3 years ago
A railroad freight car rolls on a track at 3.50 m/s toward two identical coupled freight cars, which are rolling in the same dir
ladessa [460]

Answer:

Explanation:

Hello! To solve this problem we must be clear about the concept of energy conservation, and kinetic energy with the following sentence

The kinetic energy of the two cars (v = 1.2m / S) plus the kinetic energy of the third car (v = 3.5m / S) must be equal to the kinetic energy of the three cars together.

The kinetic energy is calculated by the following equation.

E=0.5mV^2

m= mass of the cars=26500kg

V=speed

E=kinetic energy

taking into account the above, the following equation is inferred

1=  the cars are separated

2= the cars are togheter

E1=E2

E1=0.5mV1^2+0.5mV1^2+0.5m(Va)^2

where

m= mass of each car

V1= 1.2m/s

Va=3.5,m/S

E2=0.5(3)(m)V^2

m= mass of each car

V=speed (in m/s) of the three coupled cars after the first couples with the other two

Solving

0.5mV1^2+0.5mV1^2+0.5m(Va)^2=0.5(3)(m)V^2

V1^2+V1^2+(Va)^2=(3)V^2.\\2V1^2+(Va)^2=(3)V^2\\V^2=\frac{2V1^2+(Va)^2}{3} \\

V=\sqrt{\frac{2V1^2+(Va)^2}{3}} \\V=\sqrt{\frac{2(1.2)^2+(3.5)^2}{3}} \\\\V=2.245m/s

the speed  of the three coupled cars after the first couples with the other two is 2.245m/s

7 0
3 years ago
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I believe the answer is Ca
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