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3 years ago

Will gravitational force change based on a change of mass or a change of distance between objects

Chemistry

1 answer:

Digiron [165]3 years ago

6 0

Answer:

The size of the gravitational force is proportional to the masses of the objects and weakens as the distance between them increases.

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A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

$Q_1=-Q_2$

Mass of steel= $m_1$

Specific heat capacity of steel = $c_1=0.452 J/g^oC$

Initial temperature of the steel = $T_1=2.00^oC$

Final temperature of the steel = $T_2=T=21.50^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2= 105 g$

Specific heat capacity of water= $c_2=4.18 J/g^oC$

Initial temperature of the water = $T_3=22.00^oC$

Final temperature of water = $T_2=T=21.50^oC$

$Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))$

On substituting all values:

$(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}$

<h3>25.0 grams is the mass of the steel bar.</h3>

3 0

3 years ago

Read 2 more answers

How many moles are in 525 g of ammonia, NH3?<br> 8940.75 M<br> 0.03 M<br> 30.83 M

victus00 [196]

Answer:

30.83 M

Explanation:

17.03052 re in one mole. So, if you multiply it by 30.83, you will get 535 g of ammonia.

In fact, the detailed answer is 30.827009392549122.

3 0

2 years ago

Read 2 more answers

A sample of chlorine gas occupies a volume of 707 ml at 2.90 atm. What is the new pressure in torr if the volume is compressed t

dolphi86 [110]

Answer:

The new pressure is 3850 torr.

Explanation:

The relation between volume and pressure is inverse as per Boyle's law. Its mathematical form is given by :

$P_1V_1=P_2V_2$

Here,

$P_1=2.9\ atm$

$V_1=707\ mL$

$V_2=0.533\ L$

Let $P_2$ is the new pressure. So using Boyle's law we get :

$P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{2.9\times 707}{0.533}\\\\P_2=3846.71\ \text{torr}$

$P_2=3850\ \text{torr}$

So, the new pressure is 3850 torr.

3 0

3 years ago

A certain first-order reaction has a rate constant of 2.65×10−2s−1 at 16 ∘C. What is the value of k at 63 ∘C if Ea = 88.5 kJ/mol

alisha [4.7K]

Answer:

chemistry ka mujy b krna☹️

5 0

3 years ago

The three basic components of an atom

Elza [17]

Proton neutron and electron

3 0

3 years ago