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babymother [125]
3 years ago
11

Calculate the current if a charge of 5.60C passes through a point in a conductor in 15.4s

Physics
1 answer:
Ulleksa [173]3 years ago
7 0
Electric current is defined as charge per time in units of Coulombs/second
So we have
I=5.6C/15.4s = 0.3636A
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Compare and contrast microscopic and macroscopic energy transfer Give least three comparisons for each
Eduardwww [97]

Answer:

Differences

microscopic refers to substances visible to the naked eye

macroscopic are substances invisible to naked eye

Similarities

both refer to different scales that are useful to determining the size to different compounds.

Explanation:

Ima find more

3 0
3 years ago
Suppose a ball is dropped from shoulder height, falls, makes a perfectly elastic collision with the floor, and rebounds to shoul
Bezzdna [24]

Answer:Same magnitude

Explanation:

When ball is dropped from shoulder height h then velocity at the bottom is given by

v_1=\sqrt{2gh}

if it makes elastic collision then it will acquire the same velocity and riser up to the same height

If m is the mass of ball then impulse imparted is given by

J=m(v_2-v_1)

J=2m\sqrt{2gh}

Thus impulse imparted by gravity and Floor will have same magnitude of impulse but direction will be opposite to each other.

7 0
3 years ago
Which of the following identifies the number of protons in an atom?
Lunna [17]

Answer:

The number of protons can be found by looking at the atomic number

Explanation:

Its at the very top of the little element box

8 0
3 years ago
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to
Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0
3 years ago
A:10i - 2j -4k and B: i +7j - k. Determine |A-B| ​
maksim [4K]

<em>A</em> - <em>B</em> = (10<em>i</em> - 2<em>j</em> - 4<em>k</em>) - (<em>i</em> + 7<em>j</em> - <em>k</em>)

<em>A</em> - <em>B</em> = 9<em>i</em> - 9<em>j</em> - 3<em>k</em>

|<em>A</em> - <em>B</em>| = √(9² + (-9)² + (-3)²) = √189 = 3√19

8 0
2 years ago
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