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serious [3.7K]
3 years ago
8

Which kind of energy transformation occurs during skydiving? A) mechanical to gravitational B) gravitational to kinetic C) kinet

ic to potential D) mechanical to potential
Physics
2 answers:
Aleks [24]3 years ago
7 0

B) gravitational to kinetic

Explanation:

The skydiver, when he is located at a certain height h above the ground, possesses gravitational potential energy, equal to:

U=mgh

where m is the mass of the skydiver, g is the gravitational acceleration and h is the height above the ground. As he falls, its height h decreases, while his speed v increases, so part of the gravitational potential energy is converted into kinetic energy, which is given by

K=\frac{1}{2}mv^2

so, we see that as v increases, the kinetic energy increases. Therefore the correct answer is

B) gravitational to kinetic

Roman55 [17]3 years ago
6 0

I believe the answer is B). Gravitational energy to kinetic.

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At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial
elena55 [62]

Answer : The partial pressure of SO_3 is, 67.009 atm

Solution :  Given,

Partial pressure of SO_2 at equilibrium = 30.6 atm

Partial pressure of O_2 at equilibrium = 13.9 atm

Equilibrium constant = K_p=0.345

The given balanced equilibrium reaction is,

2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)

The expression of K_p will be,

K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}

Now put all the values of partial pressure, we get

0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}

p_{SO_3}=67.009atm

Therefore, the partial pressure of SO_3 is, 67.009 atm

6 0
3 years ago
The resistance of the body varies from approximately 500 kΩ (when it is very dry) to about 1.00kΩ (when it is wet). The maximum
Nady [450]

Answer:

you absolute buffoon Use Ohms' Law:   V = RI

V = (1x10^3)(5x10^-3) = 5 volts

Yes, this is in the range of normal household voltages.  

Explanation:

4 0
3 years ago
A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb
NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

\sum \tau = F*d

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

F*(27-6)= 6*600

F = \frac{6*600}{21}

F= 171.42 lb

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 0
3 years ago
True or False: A chemical reaction always happens when two substances are combined. (please help fast this is a test)
Makovka662 [10]

Answer:

no not always sometimes they react at all so false I hope I helped :)

6 0
3 years ago
Read 2 more answers
We know that the Moon revolves around Earth during a period of 27.3 days. The average distance from the center of Earth to the c
PtichkaEL [24]

Answer:

Explanation:

This is a circular motion questions

Where the oscillation is 27.3days

Given radius (r)=3.84×10^8m

Circular motion formulas

V=wr

a=v^2/r

w=θ/t

Now, the moon makes one complete oscillation for 27.3days

Then, one complete oscillation is 2πrad

Therefore, θ=2πrad

Then 27.3 days to secs

1day=24hrs

1hrs=3600sec

Therefore, 1day=24×3600secs

Now, 27.3days= 27.3×24×3600=2358720secs

t=2358720secs

Now,

w=θ/t

w=2π/2358720 rad/secs

Now,

V=wr

V=2π/2358720 ×3.84×10^8

V=1022.9m/s

Then,

a=v^2/r

a=1022.9^2/×3.84×10^8

a=0.0027m/s^2

3 0
3 years ago
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