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MrMuchimi
3 years ago
15

A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put

on the launcher, the launcher spring compresses by 10.0 cm. The watermelon is then pushed down by an additional 30.0 cm and it’s ready to go. Just before the launch, how much energy is stored in the spring?
Physics
1 answer:
olga_2 [115]3 years ago
6 0

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

F_s = F_g

kx = mg

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

k = \frac{mg}{x}

k = \frac{8*9.8}{0.1}

k = 784N/m

Just before launch the compression is 40cm, then from Potential Elastic Energy definition

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

PE =\frac{1}{2} 784*0.4^2

PE = 63.72J

Therefore the energy stored in the spring is 63.72J

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An engine moves a motorboat through water at a constant velocity of 22 meters/second. If the force exerted by the motor on the b
trapecia [35]

Answer: Option B: 1.3×10⁵ W

Explanation:

Power = \frac{Work \hspace{1mm} done}{Time}

P=\frac{W}{t}

Work Done, W= F.s

Where s is displacement in the direction of force and F is force.

\Rightarrow P = \frac{F.s}{t} =F \times \frac{s}{t}=F.v

where, v is the velocity.

It is given that, F = 5.75 × 10³N

v = 22 m/s

P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W

Thus, the correct option is B

6 0
2 years ago
an electrically charged particle with mass m, velocity v, and charge q moves in a circle in magnetic field b. what is the formul
kolbaska11 [484]

Answer:

r=\frac {mv}{qb}

Explanation:

In this case,  since the charged particle moves in circular motion,  the centripetal force is equivalent to the magnetic force.

\frac {mv^{2}}{r}=qvb\\\frac {mv}{r}=qb\\r=\frac {mv}{qb}

5 0
3 years ago
The tongue weight of a trailer should be what percent of the gross trailer weight rating
mario62 [17]

Answer:

between 10 and 15 percent

Explanation:

How to put your load

- First load the heavy

The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.

In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.

-  Hold your load

After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.

- Trailer weight

To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.

GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.

7 0
2 years ago
Match these items.
Nitella [24]

Explanation :

(1) Involuntary muscles are the muscles that are not controlled by our will.

(2) Tendons are the connective tissues that join the muscle to bones. Tendons are tissues that have fibers.

(3) Cardiac muscle is also involuntary muscles. For example heart muscle. It shows contraction and relaxation throughout life.

(4) Voluntary muscle is the muscles that are not controlled by our will.  

(5) Biceps are the arm muscles.

Hence, this the required explanation as per options.

4 0
3 years ago
Find the work w1 done on the block by the force of magnitude f1 = 95.0 n as the block moves from xi = -5.00 cm to xf = 4.00 cm .
Vlad [161]
<h3><u>Answer;</u></h3>

= 8.55 Joules

<h3><u>Explanation;</u></h3>

Work done is the product of force and the distance moved by an object.

Work done = Force × distance

Force = 95 Newtons

Distance = X2 -X1

               = 4 - (-5)

               = 9 cm

Thus;

work done = 95 × 9/100

                  <u>= 8.55 Joules </u>

5 0
2 years ago
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