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2 years ago

Can someone help me with these two? Pls

Physics

1 answer:

Gelneren [198K]2 years ago

8 0

For the first one: A falling object has higher gravitational potential energy and as it falls this decreases but kinetic energy increases. As the object hits the floor, energy is transferred to heat and sound.

42*9.8*12= 4939.2
Kinetic energy = 4939.2 OR 4939

To find the final speed use the kinetic energy formula;

Ek = 1/2 MV2

4939=1/2*42*v2
V2= 4939 divided by (1/2 * 42)
V2= 4939 / 21
V2= 235.1904761904762
Find square root
V= 15.3359211067
= 15.

Reach the ground at 15m s -¹

Note:
Formula To calculate kinetic energy:
Ek = 1/2 MV2
Ek= kinetic energy of an object
M= mass of object
V= velocity/speed of object

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an airplane flies at a speed of 100 m/s and starts to accelerate constantly at a rate of 50 m/s2. how fast is the plane flying a

mart [117]

Answer:

331.7m/s

Explanation:

Given parameters:

Initial velocity = 100m/s

Acceleration = 50m/s²

Distance = 1km = 1000m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we have to apply the right motion equation shown below;

v² = u² + 2aS

v is the final velocity

u is the initial velocity

a is the acceleration

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Now insert the parameters and solve;

v² = 100² + (2 x 50 x 1000)

v² = 110000

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2 years ago

Research shows that Mount Everest Grows an Average of 4 millimeters each year. How many will it take to grow 720 millimeters.?

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Time = 720 / 4 =180years

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3 years ago

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Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the

anygoal [31]

Answer:

The value is the temperature of the air inside the tire $T_{2} =$ 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet $P_{1}$ = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet $P_{2}$ = 3.2 + 1 = 4.2 atm

Temperature at inlet $T_{1}$ = 300 K

(a) Volume of the system is constant so pressure is directly proportional to the temperature.

$\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }$

$\frac{T_{2} }{300} = \frac{4.2}{3.7}$

$T_{2} =$ 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

$m_{1} T_{1} = m_{2} T_{2}$

$\frac{m_{2} }{m_{1} } = \frac{T_{1} }{T_{2} }$

$\frac{m_{2} }{m_{1} } = \frac{300}{354.54}$

$\frac{m_{2} }{m_{1} } =0.00294$

value of the original mass of air in the tire should be released is $\frac{m_{2} - m_{1}}{m_{1}}$

$\frac{0.00294-1}{1}$

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

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When four atomic orbitals are mixed to form hybrid orbitals, how many hybrid orbitals are formed?

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