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3 years ago

What is the Gravitational Potential Energy of a 30 kg box lifted 1.5 meters off the ground?

Physics

1 answer:

juin [17]3 years ago

6 0

Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 30 × 9.8 × 1.5

We have the final answer as

Hope this helps you

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There are two different alleles for the number of fingers on the hands: 5 finger allele and 6 finger allele. When both the 5 fin

padilas [110]

Answer:

The 6 fingers allele is dominant

Explanation:

We are told that the the individual is genotypically heterozygous, that is the have both types of the finger allele: the 5 finger allele and the 6 fingers allele however phenotypically, 6 fingers are observed. From this we can conclude that the 6 fingers allele is the one that is dominant because it is the one that is expressed phenotypically.

8 0

3 years ago

Add these measurements, using significant digit rules:<br><br> 1.0090 cm + 0.02 cm = cm

marin [14]

Answer:

1.029

Explanation:

1.0090 can also be looked at as "1.009"

0.02 can also be looked at as "0.020"

I think of it as 20+9 which is 29. There for your answer should be 1.029. There are no measurement rules applying to this equation since they are both in centimeters. So you don't have to convert anything.

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3 years ago

What is the equator ?????e define

pav-90 [236]

$\bold{ANSWER}$

EQUATOR IS THE LINE THAT DIVIDES EARTH INTO TWO HEMISPHERES .

7 0

3 years ago

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

Vladimir79 [104]

Answer:

The answer is " $1.01 \times 10^{-13}$ "

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

$KE=U=\frac{kqq'}{r}\\\\$

Calculating the closest distance:

$\to r=\frac{kqq'}{KE}\\\\$

$=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\$

$=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}$

5 0

3 years ago

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeof

il63 [147K]

<h2> The potential and kinetic energy of airplane are affected by these factors </h2>

Explanation:

When airplane rises up , it requires potential energy . This potential energy can be taken from the kinetic energy of airplane .

Thus if the speed of wind is larger , it can either oppose the motion of velocity or can favour the velocity of airplane . By which its kinetic energy is effected .

If the weight of airplane is changed , it will effect the potential energy required . Thus heavier plane requires higher potential energy for attaining the same height .

Thus these two factor has important role in the flight of airplane .

6 0

3 years ago