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2 years ago

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at maximum extension a bungee cord stores 2.0 X 10^6 J of energy. A 10 kg mass extends the bungee cord 1.3m. what is the maximum

extension of the bungee cord.

1 answer:

Gnoma [55]2 years ago

4 0

Explanation:

A k E increases While P E Decreases

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Please help really easy worth 20

vovangra [49]

Answer:

Stationary

20N

Explanation:

From the graph, we see that the body traveling is on a fixed position. Therefore, it is a stationary body.

The graph given is a position - time curve.

This curve depict a body changing position with given time.

Since the line of the curve is on a single position, the body is not changing position with the passage of time therefore, it is a stationary object.

B. 20N

From Newton's third law of motion we understand that "action and reaction force are equal but oppositely directed".

Since the person is exerting a force of 20N on the balance.

So, the reaction force by the balance is 20N upward.

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2 years ago

Consider the circuit below, which is powered by a 6-v battery. switch s is opened at t = 0 after having been closed for a long t

Semmy [17]

The answer should be B

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2 years ago

How many zebras automatically survive the first interaction with the lions in Generation 1?

lesya692 [45]

I think the answer is c. but I think it depends on how many zebras you have

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3 years ago

Mixing of sand in water is an irreversible change true aur false

USPshnik [31]

Answer:

<h3>true</h3>

Explanation:

<h3>hope it helps you ❤️</h3><h3>happy to help</h3>

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2 years ago

determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund

wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is $a_G= 6.78 m/s^2$

<u>Explanation</u>:

$N_A+N_B-Mg = 0$

$-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0$

$0.8N_B +0.8N_A = 975a_G$

$N_A+N_B = 9564.75$ -------------(1)

$-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0$

$-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0$

$-2.26N_A -0.06N_B= 0$ ----------------(2)

Solving the equation (1) and(2)

$N_A + N_B = 9564.75$

$-2.26N_A-0.06N_B=0$

$N_A = -260.85N$

$N_B = 9825.60N$

$\mu_s N_B + \mu_s N_A = 975a_G$

$0.8(9825.60)+0.8(-260.85) = 975a_G$ $a_G=\frac{7651.8}{975}$ $a_G_1=7.4848m/s^2$

Next lets assume that the front wheels contact with the ground N_A = 0

$F_B = Ma_G$

$N_B = M_g$

$N_B - M_g = 0$

$N_B(b-a) –F_Bh = 0$

$F_B = 975a_G$

$N_B-975(9.8) = 0$

$N_B=9564.75N$

$9564.75(2.20 -1.82) -F_B(0.55)=0$

$\frac{3634.605}{0.55}=F_B$

$F_B = 6608.3$

$F_B = Ma_G$

$6608.3 = 975a_G$

$a_G = 6.7778 m/s^2$

$a_G_2 = 6.78m/s^2$

Choosing the critical case

$a_G = min(a_G_1 ,a_G_2)$

$a_G = min(7.848, 6.78)$

$a_G= 6.78 m/s^2$

3 0

2 years ago