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Marysya12 [62]
2 years ago
9

When the mass of an object increases, the forcé of gravity​

Physics
2 answers:
diamong [38]2 years ago
7 0

Answer:

increace

Explanation:

they are both going up

slava [35]2 years ago
3 0
Hi I hope you have a good day today
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The volume occupied by a sample of gas is 480 mL when the pressure is 115 kPa.What pressure must be applied to the gas to make i
balandron [24]

Answer:

The answer is

<h2>84.9 kPa</h2>

Explanation:

Using Boyle's law to find the final pressure

That's

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the final pressure

P_2 =  \frac{P_1V_1}{V_2}

From the question

P1 = 115 kPa

V1 = 480 mL

V2 = 650 ml

So we have

P_2 =  \frac{115000 \times 480}{650}  = \frac{55200000}{650}  \\  = 84923.076923...

We have the final answer as

<h3>84.9 kPa</h3>

Hope this helps you

7 0
3 years ago
All of the waves in the electromagnetic spectrum are _______ waves.
user100 [1]
Transverse, I think. I may be wrong.
5 0
3 years ago
Read 2 more answers
A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two p
Anettt [7]

Answer:

A   u = 0.36c      B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

     u ’= (u-v) / (1- uv / c²)

Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ’= 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

      u’ (1- u v / c²) = u -v

      u + u ’uv / c² = v - u’

      u (1 + u ’v / c²) = v - u’

      u = (v-u ’) / (1+ u’ v / c²)

Let's calculate

      u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)

      u = 0.51c / (1 + 0.4042)

      u = 0.36c

We repeat the calculation for the other piece

In this case u ’= - 0.35c

We calculate

       u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)

       u = 1.29c / (1- 0.329)

       u = 0.961c

6 0
3 years ago
Question 2 of 5 Instructions:Select the correct answer. What do scientists use to determine the temperature of a star? thermomet
Lena [83]
Scientists use the color of a star to determine its temperature.
PreviousReset C:\format dir\..<span />
4 0
3 years ago
Read 2 more answers
A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

5 0
3 years ago
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