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Inga [223]
2 years ago
13

A 1.0-kg block and a 2.0-kg block are pressed together on a horizontal frictionless surface with a compressed very light spring

between them. They are not attached to the spring. After they are released and have both moved free of the spring
the lighter block will have more kinetic energy than the heavier block.
the magnitude of the momentum of the heavier block will be greater than the magnitude of the momentum of the lighter block.
the heavier block will have more kinetic energy than the lighter block.
both blocks will both have the same amount of kinetic energy.
both blocks will have equal
Physics
2 answers:
kykrilka [37]2 years ago
8 0
After they are released and have both moved free of the spring <span>the lighter block will have more kinetic energy than the heavier block. The correct option among all the options that are given in the question is the first option. I hope that this is the answer that has actually come to your desired help.</span>
zubka84 [21]2 years ago
6 0

Answer:

<h2>The magnitude of the momentum of the heavier block will be greater than the magnitude of the momentum of the lighter block.</h2>

Explanation:

Remember that momentum is not the same as energy or speed. Momentum or quantity of motion can be defined as the product between the mass of the object and its speed. This means, if the object is heavy, it has a lot of mass, therefore it will have more momentum. On the other hand, if it's a light object, it won't have greater momentum.

So, in this context both blocks have different mass, and according to Momentum's definition, the block with more mass, will have more momentum.

Therefore, the answer is the second option.

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Suppose that a barometer was made using oil with rho=900 kg/m3. What is the height of the barometer at atmospheric pressure?
rewona [7]

Hey there!

The pressure under a liquid column can be , calculated using  the following formula :

P = p x g x h  

P atm = 1.013 x 10⁵ Pa

g = 9.8 m/s²

h = ?

h =  P / ( p x g ) =

h= ( 1.013 x 10⁵ Pa ) / ( 900 x 9.8 ) =

h = ( 1.013 x 10⁵ ) / ( 8820 ) =

h = 11.48 m ≈  11.50 m

Hope this helps!

5 0
2 years ago
If a freely suspended vertical spring is pulled in downward direction and then released, which type of wave is produced in the s
larisa [96]

Answer:

longitudinal wave

Explanation:

it is perpendicular to the direction of the wave

3 0
2 years ago
Read 2 more answers
Two objects have masses m and 5m, respectively. They both are placed side by side on a frictionless inclined plane and allowed t
poizon [28]

Answer:

(E) The two objects reach the bottom of the incline at the same time.

Explanation:

Given;

first object with mass, m

second object with mass, 5m

The acceleration of gravity for both object is the same = 9.8 m/s²

Since both objects have the same acceleration of gravity, and no external force due friction (frictionless inclined plane), they will reach bottom of the inclined at the time.

Thus, the acceleration due to gravity is constant for all objects regardless of their masses.

Therefore, the correct option is E;

(E) The two objects reach the bottom of the incline at the same time.

5 0
2 years ago
A steel cylinder at sea level contains air at a high pressure. Attached to the tank are two gauges, one that reads absolute pres
marishachu [46]

Answer:

C) The pressure reading stays the same.

Explanation:

3 0
2 years ago
Read 2 more answers
A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga
andreev551 [17]

Answer:

Given that

V= 0.06 m³

Cv= 2.5 R= 5/2 R

T₁=500 K

P₁=1 bar

Heat addition = 15000 J

We know that heat addition at constant volume process ( rigid vessel ) given as

Q = n Cv ΔT

We know that

P V = n R T

n=PV/RT

n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

So

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

Cp-Cv= R

Cp=7/2 R

Now by putting the values

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}

a)ΔS= 20.79 J/K

b)

If the process is adiabatic it means that heat transfer is zero.

So

ΔS= 20.79 J/K

We know that

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

Process is adiabatic

\Delta S_{surr}=0

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

\Delta S_{univ}= 20.79 +0

\Delta S_{univ}= 20.79

3 0
2 years ago
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