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Ket [755]
1 year ago
14

A spring-powered dart gun is unstretched and has a spring constant 16.0 N/m. The spring is compressed by 8.0 cm and a 5.0 gram p

rojectile is placed in the gun. The kinetic energy of the projectile when it is shot from the gun is
Physics
1 answer:
stepladder [879]1 year ago
4 0

Answer:

Explanation:

Given that,

Spring constant = 16N/m

Extension of spring

x = 8cm = 0.08m

Mass

m = 5g =5/1000 = 0.005 kg

The ball will leave with a speed that makes its kinetic energy equal to the potential energy of the compressed spring.

So, Using conservation of energy

Energy in spring is converted to kinectic energy

So, Ux = K.E

Ux = ½ kx²

Then,

Ux = ½ × 16 × 0.08m²

Ux = 0.64 J

Since, K.E = Ux

K.E = 0.64 J

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yawa3891 [41]

Answer:

the answer the correct  is 3

Explanation:

Let's use the relationship between momentum and momentum

         I = Δp

         I = m v_{f} - m v₀

     

Let's calculate

         I = 0.4 5.0 - 0

         I = 2.0 N s

By Newton's law of action and reaction the force on the ball is equal to the force that the ball exerts on the foot, therefore the impulse on the foot of equal magnitude, but in the opposite direction

        I = 2.0 Ns with 60°

When reviewing the answer the correct  is 3

4 0
2 years ago
Bill drives and sees a red light. He slows down to a stop. A graph of his velocity over time is shown below.
antoniya [11.8K]

Answer:

-2 m/s^2

Explanation:

Acceleration is equal to the slope of the graph. You just find the slope of that section. The rise is -20 and the run is 10, so you get -2.

5 0
1 year ago
Read 2 more answers
If the surface temperature of that person's skin is 30∘C (that's a little lower than healthy internal body temperature becaus
anastassius [24]

Answer:

E=477.92\ W.m^{-2}

Explanation:

Given that:

Absolute temperature of the body, T=273+30=303\ K

  • emissivity of the body, \epsilon=1

<u>Using Stefan Boltzmann Law of thermal radiation:</u>

E=\epsilon. \sigma.T^4

where:

\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}   (Stefan Boltzmann constant)

Now putting the respective values:

E=1\times 5.67\times 10^{-8}\times 303^4

E=477.92\ W.m^{-2}

5 0
1 year ago
A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol
qaws [65]

Answer:

Part a)

v = \sqrt{v_o^2 + g^2t^2}

Part b)

d = \frac{1}{2}gt^2

Part c)

v_f = v_o - gt

Part d)

d = \frac{1}{2}gt^2

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

v_y = -gt

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{v_o^2 + g^2t^2}

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

d = \frac{1}{2}gt^2

Part c)

If helicopter is rising upwards with uniform speed

then final speed of the package after time t is given as

v_f = v_i + at

v_f = v_o - gt

Part d)

distance from helicopter

d = \frac{1}{2}gt^2

8 0
2 years ago
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12*8A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 13 feet. The ball is started in
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Answer:

See attached pictures.

Explanation:

See attachments for explanation.

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