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3 years ago

11

Which of the following laws relates emf, potential, current, and resistance in a circuit? Ohm’s Law Newton’s Law of Gravity Coul

omb’s Law Conservation of Charge

2 answers:

lisabon 2012 [21]3 years ago

3 0

I think it’s Ohm’s Law.

joja [24]3 years ago

3 0

Answer:

Ohm's law

Explanation:

As per ohm's law we know that the potential difference across the resistance is applied then the current flows through is given by

$i = \frac{V}{R}$

here we know that

V = EMF or potential difference across the resistance

R = Resistance

i = electric current

So ohm's law relates the potential difference across the resistance and current flow through the resistance with the resistance of the wire

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How to find the gradient of a velocity time graph

Sidana [21]

The area-

The area under the line in a velocity-time graph represents the distance travelled. To find the distance travelled in the graph above, we need to find the area of the light-blue triangle and the dark-blue rectangle.

<span><span>Area of light-blue triangle -
<span>The width of the triangle is 4 seconds and the height is 8 meters per second. To find the area, you use the equation: <span>area of triangle = 1⁄2 × base × height </span><span>so the area of the light-blue triangle is 1⁄2 × 8 × 4 = 16m. </span></span></span><span> Area of dark-blue rectangle The width of the rectangle is 6 seconds and the height is 8 meters per second. So the area is 8 × 6 = 48m.</span><span> Area under the whole graph <span>The area of the light-blue triangle plus the area of the dark-blue rectangle is:16 + 48 = 64m.<span>This is the total area under the distance-time graph. This area represents the distance covered.</span></span></span></span>

7 0

3 years ago

A satellite of mass m is moving in a circular orbit around the earth at a constant speed v and at an altitude h above the earth'

qwelly [4]

The satellite executes a rotation motion around the earth, because Earth's force of attraction plays the role of centripetal force:

Fa=Fcp=>k*Mp*m/(Rp+r)²=mv²/(Rp+r)=>v=√(k*Mp/(Rp+r))=√(6.67*10⁻¹¹*5.98*10²⁴/(6371*10³+1000*10³))=√(39.88*10¹³/(7371*10³))=√(5.41*10⁷)=7355.53 m/s

Check the calculations again !

7 0

3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 4.6 m from the center of the ride. then the operator turns on the rid

ioda

<span>Radius = 4.6 m
Time for one complete rotation t = 5.5 s.
Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
Force of the cat Fc = 6m, m being the mass.
Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116 So coefficient of static friction = 0.6116</span>

4 0

3 years ago

umka21 [38]

Answer:

Part a)

$a = 3.68 m/s^2$

Part b)

$a = 11.8 m/s^2$

Explanation:

Part a)

For force conditions of two blocks we will have

$m_1g - T = m_1 a$

$T - m_2g = m_2 a$

now from above equations we have

$(m_1 - m_2) g = (m_1 + m_2) a$

$a = \frac{m_1 - m_2}{m_1 + m_2} g$

now we know that

$m_1 = \frac{908}{9.8} = 92.65 kg$

$m_2 = \frac{412}{9.8} = 42 kg$

now from above equation we have

$a = \frac{92.65 - 42}{92.65 + 42}(9.8)$

$a = 3.68 m/s^2$

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

$F - mg = ma$

$908 - 412 = 42 a$

$a = 11.8 m/s^2$

8 0

3 years ago

Before starting this problem, review Conceptual Example 3 in your text. Suppose that the hail described there comes straight dow

bulgar [2K]

Answer:

0.9 N

Explanation:

The force exerted on an object is related to its change in momentum by:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force exerted

$\Delta p$ is the change in momentum

$\Delta t$ is the time interval

The change in momentum can be rewritten as

$\Delta p = m(v-u)$

where

m is the mass

u is the initial velocity

v is the final velocity

So the formula can be rewritten as

$F=\frac{m(v-u)}{\Delta t}$

In this problem we have:

$\frac{m}{\Delta t}=0.030 kg/s$ is the mass rate

$u=-15 m/s$ is the initial velocity

$v=+15 m/s$ is the final velocity

Therefore, the force exerted by the hail on the roof is:

$F=(0.030)(+15-(-15))=0.9 N$

6 0

3 years ago