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3 years ago

I can't type, so... please make brainliest answer.

Physics

1 answer:

Damm [24]3 years ago

6 0

Im not sure because I didn’t see the passage but I feel like it would either be A or B sorry if this didn’t help at all

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What units must the constant G have in order for the u it’s F to be newtons (N)?

EleoNora [17]

The units for G must be $[N][m^2][kg^{-2}]$

Explanation:

The magnitude of the gravitational force between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

F is the force

G is the gravitational constant

$m_1, m_2$ are the masses of the two objects

$r$ is the separation between the objects

We know that:

The units of F are Newtons (N)
The units of $m_1,m_2$ are kilograms (kg)
The units of $r$ are metres (m)

So, we can rewrite the equation in terms of G, to find its units:

$G=\frac{Fr^2}{m_1 m_2}=\frac{[N][m]^2}{[kg][kg]}=[N][m^2][kg^{-2}]$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

5 0

3 years ago

PLZ HELP WILL GIVE BRAINLIEST

xxTIMURxx [149]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of the diver = 77kg

Height = 8.18m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we use one of the motion equations.

v² = u² + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

v² = 0² + (2 x 9.8 x 8.18)

v² = 160.3

v = 12.7m/s

7 0

3 years ago

Which of the following is an impact of trampling of beach grass by humans and vehicles?

grigory [225]

Increase in sea water pollution

6 0

3 years ago

Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te

seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy 5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant $\gamma =1.4$

specific heat is given as $=\frac{\gamma R}{\gamma -1}$

gas constant =287 J⋅kg−1⋅K−1

$Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}$

specific heat at constant volume

$Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}$

change in internal energy $= Cv(T_2 -T_1)$

$\Delta U = 717.5 (800-295) = 3.62*10^5 J kg^{-1}$

change in enthalapy $\Delta H = Cp(T_2 -T_1)$

$\Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}$

change in entropy

$\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})$

$\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})$

$\Delta S = 382.79 J kg^{-1} K^{-1}$

7 0

3 years ago

A 5.00 L air sample at a temperature of -50 °C has a pressure of 107 kPa. What will be the new pressure if the temperature is ra

Serjik [45]

Its simple use formuila ,
PV=nRT
n,R is constant as the both have same moles.
so,
(p1v1)/T1 = (p2v2)/T2
so, 128.53338kpa

4 0

3 years ago