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OleMash [197]
3 years ago
10

Someone please help I’m so confused

Physics
1 answer:
Alexeev081 [22]3 years ago
3 0

Explanation:

the switch and the battery

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When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5
Scilla [17]

Answer:

1.6 x 10⁻¹⁹ C

Explanation:

Let us arrange the charges in the ascending order and round them off as follows :-

1.53 x 10⁻¹⁹ C   → 1.6x 10⁻¹⁹ C

3.26 x 10⁻¹⁹C   → 3.2 x 10⁻¹⁹ C

4.66 x 10⁻¹⁹C   → 4.8 x 10⁻¹⁹ C

5.09 x 10⁻¹⁹C   → 4.8 x 10⁻¹⁹ C

6.39 x 10⁻¹⁹C   → 6.4 x 10⁻¹⁹ C

The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.

Here we observe that

2 nd charge is almost twice the first charge

3 rd and 4 th charges are almost 3 times the first charge

5 th charge is almost 4 times the first charge.

This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first  charge , 2nd to 5 th charges can be  written as 2e,  3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of  1.6 x 10⁻¹⁹ C  exists.

3 0
3 years ago
A moving object must have which type of energy
xxMikexx [17]

Answer:

Kinetic Energy

Explanation:

Kinetic energy is the energy an object has because of its motion. If we want to accelerate an object, then we must apply a force. ... Kinetic energy can be transferred between objects and transformed into other kinds of energy. For example, a flying squirrel might collide with a stationary chipmunk.

7 0
4 years ago
Halleys comet has period of 75.3 years. Using Kepler’s third law, find it’s semimajor axis expressed in astronomical units?
natta225 [31]

Answer: 17.83 AU

Explanation:

According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.  </em>

T^{2}\propto a^{3}  (1)

Talking in general, this law states a relation between the <u>orbital period</u> T of a body (moon, planet, satellite, comet) orbiting a greater body in space with the <u>size</u> a of its orbit.

However, if T is measured in <u>years</u>, and a is measured in <u>astronomical units</u> (equivalent to the distance between the Sun and the Earth: 1AU=1.5(10)^{8}km), equation (1) becomes:

T^{2}=a^{3}  (2)

This means that now both sides of the equation are equal.

Knowing T=75.3years and isolating a from (2):

a=\sqrt[3]{T^{2}}=T^{2/3}  (3)

a=(75.3years)^{2/3}  (4)

Finally:

a=17.83AU  (5)

4 0
3 years ago
Singly charged uranium-238 ions are accelerated through a potential difference of 2.90 kV and enter a uniform magnetic field of
Lera25 [3.4K]

Answer: 0.091 m

Explanation:

r = 1/B * √(2mV/e), where

r = radius of their circular path

B = magnitude of magnetic field = 1.29 T

m = mass of Uranium -238 ion = 238 * amu = 238 * 1.6*10^-27 kg

V = potential difference = 2.9 kV

e = charge of the Uranium -238 ion = 1.6*10^-19 C

r = 1/1.29 * √[(2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]

r = 1/1.29 * √(2.21*10^-21 / 1.6*10^-19)

r = 1/1.29 * √0.0138

r = 1/1.29 * 0.117

r = 0.091 m

Therefore, the radius of their circular path is 0.091 m

6 0
3 years ago
Calculate the current through the ammeter (look at photo)
AleksAgata [21]

Answer:

6 amps

Explanation:by Kirchhoff's loop rule the current at any point in the loop must be equal or charge would be building up. The current at the ammeter is equally to the total current through the sun of the paths in parallel which it is in series with

8 0
2 years ago
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