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Lorico [155]
3 years ago
6

Varg sees a spring that has a spring constant of 4 N/m that is stretched 5 m. He stretches the spring an additional 5 m. Conside

r the system to be the spring. How much elastic energy does the spring have after Varg stretches the spring?
Physics
2 answers:
AysviL [449]3 years ago
8 0

Answer:

Elastic potential energy, E = 200 J

Explanation:

It is given that,

Spring constant, K = 4 N/m

initial stretching in the spring, x = 5 m

Finally, it is stretched an additional 5 m i.e. x' = 5 m        

Let E is the elastic energy in the spring after Varg stretches the spring. it is given by :

E=\dfrac{1}{2}k(x+x')^2

E=\dfrac{1}{2}\times 4\times (10)^2

E = 200 J

So, the elastic energy in the spring after Varg stretches the spring is 200 J. hence, this is the required solution.

otez555 [7]3 years ago
3 0

Answer:

200 J

Explanation:

Spring constant, K = 4 N/m

initial extension, x' = 5 m

Another extension, x'' = 5 m

Total extension, x = 5 + 5 = 10 m

The elastic potential energy is given by

U = \frac{1}{2}kx^{2}

U = 0.5 x 4 x 10 x 10

U = 200 J

Thus, the elastic potential energy is 200 J.

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A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m
Svetach [21]

Answer:

the impulse experienced by the passenger is 630.47 kg

Explanation:

Given;

initial velocity of the car, u = 0

final velocity of the car, v = 9.41 m/s

time of motion of the car, t = 4.24 s

mass of the passenger in the car, m = 67 kg

The impulse experienced by the passenger is calculated as;

J = ΔP = mv - mu = m(v - u)

           = 67(9.41 - 0)

           = 67 x 9.41

           = 630.47 kg

Therefore, the impulse experienced by the passenger is 630.47 kg

8 0
3 years ago
Sally and Juan’s science teacher has directed the class to build a paper airplane. The challenge is to build an airplane that fl
devlian [24]
Evaluating a solution
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A(n) 30 kg boy rides a roller coaster. The acceleration of gravity is 9.8 m/s 2 . With what force does he press against the seat
forsale [732]

Answer:

Force, F = 187.42 N

Explanation:

It is given that,

Mass of boy, m = 30 kg

Acceleration due to gravity, a=9.8\ m/s^2

Radius of curvature of the roller coaster, r = 15 m

Speed of the car, v = 7.3 m/s

The force acting on the boy are force of gravity and the centripetal force. The net force acting on him is as follows :

F=mg-\dfrac{mv^2}{r}

F=m(g-\dfrac{v^2}{r})

F=30\times (9.8-\dfrac{(7.3)^2}{15})

F = 187.42 N

So, he press against the seat with a force is 187.42 N. Hence, this is the required solution.

5 0
3 years ago
A regular polygon has angkes of size 150° each.how many side has the polygon​
emmasim [6.3K]

Answer: 12

Explanation:

ıf each interior is 150 degrees, then each exterior angle is 180–150 or 30 degrees. Hence the polygon has 360/30 = 12 sides

5 0
3 years ago
One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing
Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

m = m_{1}=m_{2}

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Along X- axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}

v_{1}=u_{1}+u_{2}

Put the value into the formula

v_{1}=2.2i-0.80i

v_{1}=1.4i\ m/s

Along Y-axis

0=m_{1}v_{1}+m_{2}v_{2}

m_{1}v_{1}=-m_{2}v_{2}

v_{1}=-v_{2}

Put the value into the formula

v_{1}=-1.36j\ m/s

Then the final speed of the first ball

v_{1}=\sqrt{(1.4)^2+(1.36)^2}

v_{1}=1.95\ m/s

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

\tan\theta=\dfrac{v_{2}}{v_{1}}

\tan\theta=\dfrac{-1.36}{1.4}

\theta=\tan^{-1}\dfrac{-1.36}{1.4}

\theta=-44.16^{\circ}

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

7 0
3 years ago
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