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Setler [38]
3 years ago
13

How deep can an object with 6360N hitting on a sponge get ???

Physics
1 answer:
borishaifa [10]3 years ago
5 0

Answer:

The sponge must go \dfrac{636}{m}\ \text{meter} deep

Explanation:

If F = 6360 N, then it is required to find how deep can an object with this force hitting on a sponge get.

We know that, F = mgh

m is mass

g is acceleration due to gravity

h=\dfrac{F}{mg}\\\\h=\dfrac{6360}{10m}\\\\h=\dfrac{636}{m}\ \text{meter}

So, the sponge must go \dfrac{636}{m}\ \text{meter} deep.

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How to find the total displacement of an object ?
OLga [1]

Answer:

it can be calculated by measuring the final distance away from a point, and then subtracting the initial distance

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Wave A has an amplitude of 2 and wave B has an amplitude of 2 as shown below. What will happen when the crest of wave A meets th
puteri [66]
Since the two waves have equal amplitudes, if the crest of one wave
meets the trough of the other one, they'll add to produce a level of zero
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6 0
3 years ago
Read 2 more answers
mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan
Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, v_e, for the planet is v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }

Where;

v_e = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, M_E

r = The radius of the planet = 3 × The radius of Earth, R_E

The escape velocity for Earth, v_e_E, is therefore given as follows;

v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }

\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } =  \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E

v_e = 16 × v_e_E

Given that the escape velocity for Earth, v_e_E ≈ 11,186 m/s, we have;

The escape velocity on the planet = v_e ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

3 0
2 years ago
a 280 nm thin film with index of refraction 1.6 floats on waterwhat is the largest wavelength of reflected light for which const
aksik [14]

Answer:

Inside the film the wavelength will be λ/n

For constructive interference to occur the film must be λf/4 thick where λf is the wavelength of the light in the film - there will be a 180 degree phase shift at the water/film interface since the index of refraction of the film is greater than that of water - and the light has to travel λ/2 inside the film for constructive interference to occur

280 nm / 1.6 * 4 = 700 nm is the greatest wavelength allowed

Note that 700 nm is also the upper wavelength of the visible spectrum

3 0
2 years ago
A ball is projected upward at time t = 0.0 s, from a point on a roof 70 m above the ground. The ball rises, then falls and strik
pentagon [3]

Answer:

The velocity of a ball will be "-70.13 m/s".

Explanation:

The given values are:

u = 70 m

t = 0.0 s

g = a = -9.8 m/s²

s = -1 m

v = ?

As we know,

The equation of motion will be:

⇒  v^2-u^2=2as

On substituting the estimated values, we get

⇒  v^2-(70)^2=2\times (-9.8)\times (-1)

⇒  v^2-4900=19.6

⇒  v^2=19.6+4900

⇒  v^2=4919.6

⇒  v=\sqrt{4919.6}

⇒  v=70.13 \ m/s

In downward direction, it will be:

⇒  v=-70.13 \ m/s

8 0
2 years ago
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