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Firdavs [7]
3 years ago
10

Please help quick please its a test

Physics
1 answer:
navik [9.2K]3 years ago
6 0

Answer:

Im pretty sure its y!

Explanation:

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A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p
Andrei [34K]

Answer:

 13.4 x 10 raise to power -19  C

Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

. The uniform electric field is  E = 214 N/M

, The decrease in electrical potential energy is   d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

which is given by the equation

d(P.E) =qEd

51.63 x 10 power -19 = q(214)(0.018)

51.63 x 10 power -19 =3.852q

by making q the formular,

q = 13.4 x 10 power -19 C  

5 0
3 years ago
Two application of heat energy​
densk [106]

Answer:

Heat is very important in our daily life in warming the house, cooking, heating the water, and drying the washed clothes. The heat has many usages in the industry as making and processing the food and manufacture of the glass, the paper, the textile, and etc.

Explanation:

4 0
3 years ago
What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then
Sedbober [7]

Answer:

x=(0.088m)\cos(\sqrt{\frac{k}{m} }  t)

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

\omega =\sqrt{\frac{k}{m}}

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

x=A\cos(\omega t)

Finally, the equation of the motion of the system is:

x=(0.088m)\cos(\omega t)

or

x=(0.088m)\cos(\sqrt{\frac{k}{m} }  t)

7 0
4 years ago
Write any two uses of simple machines.
Nana76 [90]
Simple machines could be used to reduce effort or extend the ability of people to perform tasks beyond their normal capabilities.
Examples include pulley, lever, and incline plane
4 0
3 years ago
A steel cable with Cross Sectional Area 3.00cm² has an elastic limit of 2.40 x 10^8pascals. Find the maximum upward acceleration
bazaltina [42]

Answer:

Stress = F / A       force per unit area

A = 3.00 cm^2 = 3 E-4  m^2

F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N    max force applied

F/3 = 2.4E4 N  if force not to exceed limit   (= f)

f = M a

a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2      about 2 g

3 0
3 years ago
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