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madam [21]
3 years ago
5

A frog leaps with a displacement equal to vector u and then leaps with a displacement equal to vector v, as

Physics
1 answer:
denpristay [2]3 years ago
7 0

Answer:

The answer is B on khan academy

Explanation:

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Jasper and Gemma are going to play on a teeter totter. Gemma gets on first. When Jasper gets on, Gemma moves into the air, but s
Nezavi [6.7K]

Answer:

it will most likely be a: the forces are balanced because jasper weighs the same as gemma.

Explanation:

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3 years ago
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g A 4-foot spring is elongated167feet long after a mass weighing 16 pounds is attached to it. The medium throughwhich the mass m
marta [7]

Answer: hello question b is incomplete attached below is the missing question

a) attached below

b) V = 0.336 ft/s

Explanation:

Elongation ( Xo)  = 16/ 7 feet

mass attached to 4-foot spring = 16 pounds

medium has 9/2 times instanteous velocity

<u>a) Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 2 ft/s</u>

The motion is an underdamped motion because the value of β < Wo

Wo = 3.741 s^-1

attached below is a detailed solution of the question

3 0
3 years ago
A thin film of oil of thickness t is floating on water. The oil has index of refraction no = 1.4. There is air above the oil. Wh
kkurt [141]

Answer:

t = 120.5 nm

Explanation:

given,    

refractive index of the oil = 1.4

wavelength of the red light = 675 nm

minimum thickness of film = ?

formula used for the constructive interference

2 n t = (m+\dfrac{1}{2})\lambda

where n is the refractive index of oil

t is thickness of film

for minimum thickness

m = 0

2 \times 1.4 \times t = (0+\dfrac{1}{2})\times 675

t = \dfrac{0.5\times 675}{2\times 1.4}

        t = 120.5 nm

hence, the thickness of the oil is t = 120.5 nm

7 0
3 years ago
A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting
ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

T=\dfrac{38}{8}\\\\T=4.75\ s

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

k = spring stiffness constant

So,

k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m

When the cord is in air,

mg=kx

x = the extension in the cord

x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

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