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Elis [28]
3 years ago
9

An athlete jumping vertically on a trampoline leaves the surface with a velocity of 8.5 m/s upward. what maximum height does she

reach?
Physics
1 answer:
Mumz [18]3 years ago
8 0
<span>Her center of mass will rise 3.7 meters. First, let's calculate how long it takes to reach the peak. Just divide by the local gravitational acceleration, so 8.5 m / 9.8 m/s^2 = 0.867346939 s And the distance a object under constant acceleration travels is d = 0.5 A T^2 Substituting known values, gives d = 0.5 9.8 m/s^2 (0.867346939 s)^2 d = 4.9 m/s^2 * 0.752290712 s^2 d = 3.68622449 m Rounded to 2 significant figures gives 3.7 meters. Note, that 3.7 meters is how much higher her center of mass will rise after leaving the trampoline. It does not specify how far above the trampoline the lowest part of her body will reach. For instance, she could be in an upright position upon leaving the trampoline with her feet about 1 meter below her center of mass. And during the accent, she could tuck, roll, or otherwise change her orientation so she's horizontal at her peak altitude and the lowest part of her body being a decimeter or so below her center of mass. So it would look like she jumped almost a meter higher than 3.7 meters.</span>
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Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan
Drupady [299]

Answer:

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

Explanation:

We are given that

Gravitational force=F_g=\frac{Gm_Emr}{R^3_E}

r=0,U(0)=0

We know that

Gravitational potential energy=-\int F_gdr

U(r)=-\int\frac{Gm_Emr}{R^3_E}dr

U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C

Substitute r=0 ,U(0)=0

0=0+C

C=0

Substitute the value

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

4 0
3 years ago
(II) You buy a 75-W lightbulb in Europe, where electricity is delivered at 240 V. If you use the bulb in the United States at 12
Elodia [21]

Answer:

Explanation:

You are looking for the resistance to start with

W = E * E/R

75 = 240 * 240 / R

75 * R = 240 * 240

R = 240 * 240 / 75

R = 57600 / 75

R = 768

Now let's see what happens when you try putting this into 110

W = E^2 / R

W = 120^2 / 768

W = 18.75

So the wattage is rated at 75. 18.75 is a far cry from that. I think they intend you to set up a ratio of

18.75 / 75 = 0.25

This is the long sure way of solving it. The quick way is to realize that the voltage is the only thing that is going to change. 120 * 120 / (240 * 240) = 1/2*1/2 = 1/4 = 0.25

4 0
3 years ago
Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin
likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0
3 years ago
The electrons of an atom store nuclear energy.
Sladkaya [172]
False. The nuclear energy is found within the nucleus. Electrons are located outside the nucleus.
4 0
3 years ago
Read 2 more answers
Parasaurolophus was a dinosaur whose distinguishing feature was a hollow crest on the head. The 1.5-m-long hollow tube in the cr
Oksi-84 [34.3K]

Answer:

f1 = 58.3Hz, f2 = 175Hz, f3 = 291.6Hz

Explanation:

lets assume speed of sound is 350 m/s.

frequencies of a standing wave modes of an open-close tube of length L

fm = m(v/4L)

where m is 1,3,5,7......

and fm = mf1

where f1 = fundamental frequency

so therefore: f1 = 350 x 4 / 1.5

f1 = 58.3Hz

f2 = 3 x 58.3

f2 = 175Hz

f3 = 5 x 58.3

f3 = 291.6Hz

5 0
3 years ago
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