Ask question

Ask question

All categories

3 years ago

9

A rubber band has a spring constant of 45 N/m. A classmate stretches it a length of 0.2 m before firing it at Mr. C. Ouch. Calcu

late the elastic
potential energy of the rubber band.

1 answer:

nata0808 [166]3 years ago

5 0

Answer:

<h2>0.9 Joules</h2>

Explanation:

Elastic potential energy of a spring= 1/2 × Spring constant × displacement²

following calculations you will get ur answer!!

You might be interested in

PLZ HELP MEEEEEEEEE ASAP

mafiozo [28]

The answer is “Impulse acting on it” according to the impulse-momentum theorem.

6 0

2 years ago

5. Find the mass of a car that is traveling at a velocity of 35 m/s West.

sattari [20]

Answer:

m = 9795.9 kg

Explanation:

v = 35 m/s

KE = 6,000,000 J

Plug those values into the following equation:

$KE = \frac{1}{2} mv^{2}$

6,000,000 J = (1/2)(35^2)m

---> m = 9795.9 kg

3 0

3 years ago

Homeostasis is the regulation of an organism’s internal environment to maintain conditions suitable for life. Homeostasis requir

kherson [118]

I think it would be B

5 0

3 years ago

Read 2 more answers

A long straight cylindrical shell has an inner radius R_i and an outer radius R_0. It carries a current i, uniformly distributed

Irina-Kira [14]

Answer:

Correct answer is option D

- Wire is on the cylinder axis and carries current i in the direction opposite to that of the current in the shell

Explanation:

- It cannot be Option E, because the magnetic field outside the wire would not be 0 due to the current carried by the conductor

-Also, the parallel wire cannot carry current in the same direction because, that would amplify the magnetic field created by the outer cylinder (since B is dir. proportional to the current) -and now, that leaves only option C and D. If, it is Option C, then that means one side of the cylinder would be more closer to the parallel wire than the other, so there would be different B fields on the two opposite sides of the cylinder. So, that means the answer is option D.

7 0

3 years ago

[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r

blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity $V_{f}$ for an object traveling distance h taking the initial vertical component of velocity as $V_{i}$ the kinematics equation is written as

$V_{f}^{2}=V_{i}^{2}+2ah$ where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

$V_{f}^{2}=V_{i}^{2}+2gh$

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

$V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}$ = 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

$V_{f}^{2}=V_{i}^{2}+2gh$

Since we have final velocity of 25 m/s

$V_{i}^{2}=2gh-V_{f}^{2}$

$V_{i}=\sqrt{(V_{f}^{2}-2gh)}$

$V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}$ = 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0

3 years ago