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ale4655 [162]
2 years ago
9

If an object is dropped from 10 m above the ground, what is the height at which its kinetic energy and potential energy are equa

l?​
Physics
1 answer:
Mama L [17]2 years ago
5 0

potential energy is converted to kinetic energy as the ball drops.


they will be equal at the halfway point.


answer: 5 m

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find the time taken, if the speed of a train increased from 72 km/hr to 90 km/hr for 234 km. leave your answer in seconds
Airida [17]

Answer:

Time taken = 10400 s

Explanation:

Given:

Initial speed of the train, u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}

Final speed of the train, v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}

Displacement of the train, S=234\textrm{ km}=234\times 1000=234000\textrm{ m}

Using Newton's equation of motion,

v - u = at\\a=\frac{v-u}{t}

Now, using Newton's equation of motion for displacement,

v^{2}-u^{2}=2aS

Now, plug in the value of a=\frac{v-u}{t} in the above equation. This gives,

v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}

Now, plug in 234000 m for S, 25 m/s for v and 20 m/s for u. Solve for t.

t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}

Therefore, the time taken by the train is 10400 s.

3 0
4 years ago
Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−370 kJ X2+2Z2⟶2XZ2ΔH2=−120 kJ 2Y2+Z2⟶2Y2ZΔH3=−270 kJ Calculate the change in
Alchen [17]

Answer : The change in enthalpy of the reaction is, -310 kJ

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

4XY_3+7Z_2\rightarrow 6Y_2Z+4XZ_2    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) X_2+3Y_2\rightarrow 2XY_3     \Delta H_1=-370kJ

(2) X_2+2Z_2\rightarrow 2XZ_2    \Delta H_2=-120kJ

(3) 2Y_2+Z_2\rightarrow 2Y_2Z    \Delta H_3=-270kJ

Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :

(1) 4XY_3\rightarrow 2X_2+6Y_2     \Delta H_1=2\times (+370kJ)=740kJ

(2) 2X_2+4Z_2\rightarrow 4XZ_2    \Delta H_2=2\times (-120kJ)=-240kJ

(3) 6Y_2+3Z_2\rightarrow 6Y_2Z    \Delta H_3=3\times (-270kJ)=-810kJ

The expression for enthalpy of formation of CH_4 will be,

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+740kJ)+(-240kJ)+(-810kJ)

\Delta H=-310kJ

Therefore, the change in enthalpy of the reaction is, -310 kJ

5 0
3 years ago
A ball has a diameter of 3.79 cm and average density of 0.0838 g/cm3.
suter [353]

Answer: 0.258 N

Explanation:

As the density of the object is much less than the density of water, it’s clear that the buoyant force, is greater than the weight of the object, which means that in normal conditions, it would float in water.

So, in order to get the ball submerged in water, we need to add a downward force, that add to the weight, in order to compensate the buoyant force, as follows:

F = Fb – Fg

Fb= δH20* 4/3*π*(d/2)³  * g

Fg = δb* 4/3*π*(d/2)³ *g

F= (δH20- δb) * 4/3*π*(d/2)³*g

Replacing by the values of the densities, and the ball diameter, we finally get:

F= 0.258 N

3 0
4 years ago
A box of mass m is pushed at an angle 0 by a force Fp along a frictionless surface. The box travels to right with acceleration a
Genrish500 [490]

Answer:A block rests on a horizontal, frictionless surface. A string is attached to the block, and is pulled with a force of 45.0 N at an…

Explanation:

5 0
3 years ago
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Without completing the calculations, determine what the new volume will be in the problem below. Also, explain how you were able
Talja [164]
Boyles law

Pressure and volume are inversely proportional as the new variable changes from the known.

Double the pressure equals 1/2 of original volume, assuming temperature remains the same.

So 40.0 mL is the new volume as it is compressed.
7 0
4 years ago
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