<u> Ohms law: </u> This law relates voltage difference between two points. Mathematically, the law states that V=IR;
Where
V = voltage difference ; in volts
I = Current ; in Amperes
R = Resistance ; in ohms
<u>1. Answer : </u> given that R = 10 ; V= 12 V ; I = ?
From ohms law, I = V/R
= 12/10
= 1.2 Amp.
<u>2. Answer:</u> given that R = 10 ; V= ? ; I = 5
From ohms law, V = IR
= 10×5 = 50 V
<u>3 . Answer:</u> given that R = ? ; V= 120 ; I = 5
From ohms law, R = V/I
= 120/5
= 24 Ω
<u>4 . Answer:</u> given that R = ? ; V= 10 ; I = 20
From ohms law, R = V/I
= 10/20
= 0.5 Ω
<u>5 . Answer:</u> given that R = 480 ; V= 24 ; I = ?
From ohms law, I = V/R
= 24/480
= 0.05 A
<u>6. Answer:</u> given that R = 150 ; V= ? ; I = 1
From ohms law, V = IR
= 1 × 150
= 150 V
Ok, I’ll try to help, but I’d need to see the picture of the positions of the sun, earth, and moon to help you fully.
So, the first thing to note is that gravity is an attractive force, meaning that; something that has mass, call the m1 will “pull toward” another mass, call it m2. The two objects pull on each other, mutually.
If an object has more mass it pull more, and if an object has less mass, it pulls less.
Another thing to note is that distances matter. The closer the objects are to each other, the more pull they’ll “feel”.
So, the ocean tides are the effect of ocean water responding to a gravitational gradient, the moon plays a larger role in creating tides than the sun does. But the sun's gravitational gradient across the earth is significant and it does contribute to tides as well.
So, when the bulge of the ocean caused by the sun’s gravity, partially cancels out the bulge of the ocean caused by the moon gravity. This produces moderate tides known as the neap tides, meaning that high tides are a little lower and low tides are a little higher than average.
I hope that helps.
Answer:
Class 1 Levers: Moving the fulcrum closer to the load will increase the mechanical advantage. Moving the effort farther from the fulcrum will increase the mechanical advantage.
Explanation:
Hope it helps you
The new frequency of oscillation when the car bounces on its springs is 0.447 Hz
<h3>Frequency of oscillation of spring</h3>
The frequency of oscillation of the spring is given by f = (1/2π)√(k/m) where
- k = spring constant and
- m = mass on spring
Now since k is constant, and f ∝ 1/√m.
So, we have f₂/f₁ = √(m₁/m₂) where
- f₁ = initial frequency of spring = 1.0 Hz,
- m₁ = mass of driver,
- f₂ = final frequency of spring and
- m₂ = mass on spring when driver is joined by 4 friends = 5m₁
So, making f₂ subject of the formula, we have
f₂ = [√(m₁/m₂)]f₁
Substituting the values of the variables into the equation, we have
f₂ = [√(m₁/m₂)]f₁
f₂ = [√(m₁/5m₁)]1.0 Hz
f₂ = [√(1/5)]1.0 Hz
f₂ = 1.0 Hz/√5
f₂ = 1.0 Hz/2.236
f₂ = 0.447 Hz
So, the new frequency of oscillation when the car bounces on its springs is 0.447 Hz
Learn more about frequency of oscillation of spring here:
brainly.com/question/15318845
Earthquakes can also occur far from the edges of plates, along faults. Faults<span> are cracks in the earth where sections of a plate (or two plates) are moving in different directions. Faults are caused by all that bumping and sliding the plates do. They are more common near the edges of the plates.</span>
