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Svetllana [295]
3 years ago
7

The planet Krypton has a mass of 7.6 × 1023 kg and radius of 1.7 × 106 m. What is the acceleration of an object in free fall nea

r the surface of Krypton? The gravitational constant is 6.6726 × 10−11 N · m2 /kg2 . Answer in units of m/s 2 .
Physics
1 answer:
olga55 [171]3 years ago
5 0

Answer:

17.55 m/s²

Explanation:

Parameters given:

Mass of Krypton, M = 7.6 * 10^23 kg

Radius, R = 1.7 * 10^6 m

Gravitational constant, G = 6.6726 * 10^(-11) Nm²/kg²

Acceleration due to gravity of planet of mass M is given as:

g = GM/R²

Since the object is close to the surface of Krypton, we can say that the distance from the Centre of Krypton is the radius of the planet Krypton.

Therefore,

g = (6.6726 * 10^(-11) * 7.6 * 10^23)/(1.7 * 10^6)²

g = 17.55 m/s²

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The siren on an ambulance is emitting a sound whose frequency is 2450 Hz. The speed of sound is 343 m/s. If the ambulance is sta
lyudmila [28]

Answer:

The wavelength is 0.14 m

Explanation:

Given that,

Frequency = 2450 Hz

Speed of sound = 343 m/s

We need to calculate the wavelength

Using formula of wavelength

v=  f\lambda

Where, v = speed of sound

f = frequency

Put the value into the formula

\lambda=\dfrac{v}{f}

\lambda=\dfrac{343}{2450}

\lambda=0.14\ m

Hence,  The wavelength is 0.14 m

5 0
3 years ago
Read 2 more answers
Desde que altura debes de lanzar una canica de 50g para que adquiera una energia de 100j
AnnZ [28]

Answer:

204 m

Explanation:

When the marble is dropped from a certain height, its gravitational potential energy converts into kinetic energy. So the kinetic energy gained is equal to the variation of gravitational potential energy:

\Delta U=mg\Delta h

where

m is the mass of the marble

g = 9.8 m/s^2 is the acceleration of gravity

\Delta h is the change in height

In this problem, we have

m = 50 g = 0.05 kg

\Delta U = 100 J

Solving the formula for \Delta h, we find the necessary height from which the marble should be dropped:

\Delta h = \frac{\Delta U}{mg}=\frac{100}{(0.05)(9.8)}=204 m

7 0
2 years ago
A 6 kg block initially at rest is pulled to East along a horizontal surface with coefficient of kinetic friction μk=0.15 by a co
ozzi

Answer:

1.8 m/s

Explanation:

Draw a free body diagram of the block.  There are four forces:

Normal force Fn up.

Weight force mg down.

Applied force F to the east.

Friction force Fn μ to the west.

Sum the forces in the y direction:

∑F = ma

Fn − mg = 0

Fn = mg

Sum the forces in the x direction:

F − Fn μ = ma

F − mg μ = ma

a = (F − mg μ) / m

a = (12 N − 6 kg × 9.8 m/s² × 0.15) / 6 kg

a = 0.53 m/s²

Given:

Δx = 3 m

v₀ = 0 m/s

a = 0.53 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (0.53 m/s²) (3 m)

v = 1.8 m/s

8 0
3 years ago
When water in a lake freezes, the ice that forms floats on top of any water that is still liquid. Why does the ice float?
ddd [48]

Answer:

it floats because the lake is cold at that moment so when part of the lake freezes it still remains solid and floats because of the lake and the surrounding of the lake is still cold

5 0
3 years ago
Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh
mel-nik [20]

Answer:u=\frac{v}{2}\sqrt{5-4sin\phi }

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

m(2v)+m(-vsin\phi )=2m(ucos\theta )

2ucos\theta =v(2-sin\theta )------1

Conserving momentum in x direction

mvcos\phi =2musin\theta-----2

squaring and adding 1 &2

(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2

4u^2=4v^2+v^2-4v^2sin\phi

4u^2=5v^2-4v^2sin\phi

u=\frac{v}{2}\sqrt{5-4sin\phi }

7 0
2 years ago
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