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3 years ago

13

What does gravitational potential energy typically change into? Question 8 options: A-Electric energy B- Kinetic energy C- Radio

active energy D-Radioactive energy

2 answers:

MariettaO [177]3 years ago

7 0

The answer is “B”, Kinetic Energy

nika2105 [10]3 years ago

6 0

<em>Answer:</em>

<em>kinetic energy </em>

<em>Explanation:</em>

<em>i should know i had a test about it.</em>

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guajiro [1.7K]

Answer:

hello

Explanation:

3 0

3 years ago

What is the energy equivalent of an object with a mass of 1.83 kg?

xxMikexx [17]

To determine the energy equivalent of an object, we use the famous equation of Einstein which is E=mc^2 where m is the mass of the object and c is the speed of light (3x10^8 m/s). We calculate as follows:

E = mc^2
E = 1.83 kg (3x10^8 m/s)^2
E = 1.647x10^17 J

6 0

3 years ago

Read 2 more answers

Symptoms of excessive stress include all of the following EXCEPT:

sweet-ann [11.9K]

Answer:

Symptoms of excessive stress include all of the following EXCEPT: increased energy level.

6 0

2 years ago

Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

pantera1 [17]

Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$

The sum of the vertical components of F1 and F2 must equal the total force Ft

$\displaystyle -2F\ cos21^o+F\ sin\alpha =460$

Solving for $\alpha$ in the first equation

$\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o$

$\displaystyle cos\alpha =0.717=>\alpha =44.214^o$

$\displaystyle F(2cos21^o+sin\alpha)=460$

$\displaystyle F=\frac{460}{2cos21^o+sin\alpha}$

$\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}$

$\displaystyle F=179.37\ N$

The magnitude of F1 is

$|F1|=2*F=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

4 0

3 years ago

A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6

Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6 = V/0.0001

V = 2.0 × 10^6 * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q = 1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

5 0

2 years ago