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lorasvet [3.4K]
2 years ago
6

A planet exerts a gravitational force of magnitude 4e22 N on a star. If the planet were 3 times closer to the star (that is, if

the distance between the star and the planet were 1/3 what is is now), what would be the magnitude of the force on the star due to the planet
Physics
1 answer:
Alex_Xolod [135]2 years ago
7 0

Answer:

3.6\times10^{23} N

Explanation:

F=\frac{GmM}{r^2}=4\times10^{22} N

F'=\frac{GmM}{(r/3)^2}=9\frac{GmM}{r^2}=9\times4\times10^{22}=3.6\times10^{23} N

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How much energy will be transferred to your eardrum while listening to this sound for 1.0 min?
Phoenix [80]

So E = 2x10^-3W/m^2*(π*(3.0x10^-3m)^2)*1min*60s... = 3.4x10^-6J

5 0
3 years ago
an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and
gavmur [86]

Answer:

Density of Sand is 2.653g/cm^{3}.

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle(w_{max})=48.4gm-23.5gm

w_{max}=24.9g

Since we know density of water d_{w} =1g/cm^{3} and w_{max}

We can calculate volume of empty space in the bottle(V).

w_{max}=d_{w}\timesV

V=\frac{w_{max} }{d_{w} }

V=\frac{24.9}{1}

V=24.9 g/cm^{3}

Now bottle is partially filled with sand,and weight of bottle is (w_{s})36.5gm

So,

Amount of sand added (m_{s})=36.5-Weight of the bottle

m_{s}=13g

After filling the bottle with water again,the weight of the bottle becomes (W_{2}=56.5g)

Therefore,

amount of water added to the bottle of sand in grams = W_{2}-36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/cm^{3}

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle(V_{w})=20cm^{3}

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V-V_{w}

V_{s}=24.9g-20g

V_{s}=4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = \frac{m_{s} }{V_{s} }

density of sand =\frac{13}{4.9}

density of sand = 2.653g/cm^{3}

8 0
3 years ago
Help plzzz!!! I only have 10 minutes to turn in
Lostsunrise [7]
  • The mechanic did 5406 Joules of work pushing the car.

That's the energy he put into the car.  When he stops pushing, all the energy he put into the car is now the car's kinetic energy.

  • Kinetic energy = (1/2) (mass) (speed²)

And there we have it

  • The car's mass is 3,600 kg.
  • Its speed is 'v' m/s .
  • (1/2) (mass) (v²) =  5,406 Joules

(1/2) (3600 kg) (v²) = 5406 joules

1800 kg (v²) = 5406 joules

v² = (5406 joules) / (1800 kg)

v² = (5406/1800) (joules/kg)

= = = = = This section is just to work out the units of the answer:

  • v² = (5406/1800) (Newton-meter/kg)
  • v² = (5406/1800) (kg-m²/s²  /  kg)
  • v² = (5406/1800)  (m²/s²)

= = = = =

v = √(5406/1800)  m/s

<em>v = 1.733 m/s</em>

4 0
3 years ago
A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The system is released with an init
AURORKA [14]

a) 19.4 cm

b) 3.2 m/s

Explanation:

a)

A horizontal spring-mass system has a motion called simple harmonic motion, in which the mass oscillates following a periodic function (sine or cosine) around an equilibrium position.

As the system oscillates back and forth, its total mechanical energy (sum of elastic potential energy and kinetic energy) will remain conserved (since we consider friction negligible). The elastic potential energy at any point is given by:

U=\frac{1}{2}kx^2

where

k is the spring constant

x is the displacement of the system

While the kinetic energy at any point is

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

So the total mechanical energy of the system is

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

For this system, when it is initially released,

m = 0.6 kg

k = 165 N/m

x = 7 cm = 0.07 m

v = 3 m/s

So the total energy is

E=\frac{1}{2}(0.6)(3)^2+\frac{1}{2}(165)(0.07)^2=3.1 J

Since friction is negligible, this total energy remains constant. Therefore, when the system reaches its maximum stretch during the motion, the kinetic energy will be zero and all the mechanical energy will be elastic potential energy; so we will have:

E=U=\frac{1}{2}kx_{max}^2

where x_{max} is the maximum stretch. Solving for x_{max},

x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(3.1)}{165}}=0.194 m

So, 19.4 cm.

b)

The maximum speed in a spring-mass oscillating system is reached when the kinetic energy is maximum, and therefore, since the total energy is conserved, when the elastic potential energy is zero:

U=0

which means when the displacement is zero:

x = 0

So, when the system is transiting through the equilibrium position.

Therefore, the total mechanical energy is equal to the maximum kinetic energy:

E=K=\frac{1}{2}mv_{max}^2

where

m is the mass

v_{max} is the maximum speed

Here we have:

E = 3.1 J

m = 0.6 kg

Therefore, solving for the maximum speed,

v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(3.1)}{0.6}}=3.2 m/s

6 0
3 years ago
Coulomb’s Law relates which of the following?
Verdich [7]
There's nothing on the list you provided that's related to Coulomb's Law.
8 0
3 years ago
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