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3 years ago

12

The moon and other satellites rotate around the earth. Identify the force that keeps these satellites in orbit. A) gravity B) fr

iction C) a strong force D) electromagnetic force

2 answers:

TEA [102]3 years ago

5 0

Answer:

gravity

Explanation:

RoseWind [281]3 years ago

4 0

The force is gravity

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Why will a struck tuning fork sound louder when it is held against a table?.

QveST [7]

Answer:
1.)the surface of the table is set into vibration
2.)the initial vibration was larger
3.)The entire table vibrates

Explanation:<u><em>hope this helps :3 have a good day xlXCherryColaXlx..(pls mark me brainliest)</em></u>

7 0

2 years ago

What is the mass of a car that weighs 19,000 N on earth?

Anna71 [15]

Divide by 9.8 I think
so <span>1938.77551</span>

8 0

3 years ago

Consider the following equilibrium at 979 K for the dissociation of molecular iodine into atoms of iodine. I2(g) equilibrium rea

GenaCL600 [577]

Answer: $I_2$ = 0.0050 M

$I$ = 0.0155 M

Explanation:

Initial moles of $I_2$ = 0.072 mole

Volume of container = 3.9 L

Initial concentration of $I_2=\frac{moles}{volume}=\frac{0.072moles}{3.9L}=0.018M$

The given balanced equilibrium reaction is,

$I_2(g)\rightleftharpoons 2I(g)$

Initial conc. 0.018 M 0

At eqm. conc. (0.018-x) M (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[I]^2}{[I_2]}$

$K_c=\frac{(2x)^2}{0.2-x}$

we are given : $K_c=1.60\times 10^{-3}$

Now put all the given values in this expression, we get :

$1.60\times 10^{-3}=\frac{(2x)^2}{(0.018-x)}$

$x=0.0025$

So, the concentrations for the components at equilibrium are:

$[I]=2\times x=2\times 0.0025=0.0050$

$[I_2]=0.018-x=0.018-0.0025=0.0155$

Hence, concentrations of $I_2$ and $I$ are 0.0050 M ad 0.0155 M respectively.

4 0

4 years ago

A car is travelling with a velocity of 10 m/s and has a mass of 550 kg. The<br> car has<br> energy.

Elenna [48]

Answer: The car has a kinetic energy (because it's in motion) of: $27500J$

Explanation:

$Formula: E=\frac{1}{2}mv^2$

$E=\frac{1}{2}mv^2$

$E=\frac{1}{2}(550kg)(10m/s)^2$

$E=\frac{1}{2}(550kg)(100m^2/s^2)$

$E=\frac{1}{2}(55000J)\\E=27500J$

3 0

3 years ago

A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 m length of massless string. The ball is struck, causing it

Delvig [45]

Answer

given,

mass of ball = 5.93 kg

length of the string = 2.35 m

revolve with velocity of 4.75 m/s

acceleration due to gravity = 9.81 m/s²

T cos θ = mg

T cos θ = $5.93\times 9.81$

T cos θ = 58.17

$T sin \theta =\dfrac{mv^2}{r}$

$T sin \theta =\dfrac{5.93\times 4.75^2}{2.35 sin \theta}$

$T sin^2 \theta =56.93$

$sin^2 \theta = 1 - cos^2 \theta$

$T (1 - cos^2 \theta) =56.93$

$T (1 - (\dfrac{58.17}{T})^2) =56.93$

T² - 56.93T - 3383.75 = 0

T = 93.22 N

$cos \theta = \dfrac{58.17}{93.22}$

θ = 51.39°

6 0

3 years ago