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zhuklara [117]
3 years ago
15

The formula for Ohm’s law can be written: V equals a. l x R b. l/R c. R/l

Physics
1 answer:
stellarik [79]3 years ago
7 0

The formula for Ohm’s law can be written: V equals  l x R

Answer: Option A

<u>Explanation:</u>

Ohm's law defines that an electric current flowing through the conductor between two ends is directly proportionate to the voltage at these two points. The introduction of a constant proportionality, resistance, gives a simple mathematical equation describing this relationship. Particularly, Ohm's law also mentions that R is constant in this respect i.e. independent on current.

                        \text {voltage }(V)=\text { current }(I) \times \text { Resistance }(R)

Where,  

I is the current conduction in amperes

V is the voltage calculated by the conductor in volts

R is the conductor’s resistance in ohms.

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Read 2 more answers
2. A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.5 . At 30.0 s after blast
harina [27]

Answer:

a)The highest point reached by the rocket is 1412 m

b)The rocket crashes after 54.7 s

Explanation:

Hi there!

The equations of height and velocity of the rocket are the following:

h = h0 + v0 · t + 1/2 · a · t² (while the engines work).

h = h0 + v0 · t + 1/2 · g · t² (when the rocket is in free fall).

v = v0 + a · t (while the engines work).

v = v0 + g · t (when the rocket is in free fall).

Where:

h = height of the rocket at a time t.

h0 = initial height of the rocket.

v0 = initial velocity.

t = time.

a = acceleration due to the engines.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the rocket at a time t.

First, let's find the velocity and height reached by the rocket until the engines fail:

h = h0 + v0 · t + 1/2 · a · t²

Let's set the origin of the frame of reference at the launching point so that h0 = 0. Since the rocket starts from rest, v0 = 0. So after 30.0 s the height of the rocket will be:

h = 1/2 · a · t²

h = 1/2 · 2.5 m/s² · (30.0 s)²

h = 1125 m

Now let's find the velocity of the rocket at t = 30.0 s:

v = v0 + a · t (v0 = 0)

v = 2.5 m/s² · 30.0 s

v = 75 m/s

After 30.0s the rocket will continue to ascend with a velocity of 75 m/s. This velocity will be gradually reduced due to the acceleration of gravity. When the velocity is zero, the rocket will start to fall. At that time, the rocket is at its maximum height. So, let's find the time at which the velocity of the rocket is zero:

v = v0 + g · t

0 = 75 m/s - 9.8 m/s² · t (v0 = 75 m/s because the rocket begins its free-fall motion with that velocity).

-75 m/s / -9.8 m/s² = t

t = 7.7 s

Now, let's find the height of the rocket 7.7 s after the engines fail:

h = h0 + v0 · t + 1/2 · g · t²

The rocket begins its free fall at a height of 1125 m and with a velocity 75 m/s, then, h0 = 1125 m and v0 = 75 m/s:

h = 1125 m + 75 m/s · 7.7 s - 1/2 · 9.8 m/s² · (7.7 s)²

h = 1412 m

The highest point reached by the rocket is 1412 m

b) Now, let's calculate how much time it takes the rocket to reach a height of zero (i.e. to crash) from a height of 1412 m.

h = h0 + v0 · t + 1/2 · g · t² (v0 = 0 because at the maximum height the velocity is zero)

0 = 1412 m - 1/2 · 9.8 m/s² · t²

-1412 m / -4.9 m/s² = t²

t = 17 s

The rocket goes up for 30.0 s with an acceleration of 2.5 m/s².

Then, it goes up for 7.7 s with an acceleration of -9.8 m/s².

Finally, the rocket falls for 17 s with an acceleration of -9.8 m/s²

The rocket crashes after (30.0 s + 7.7 s + 17 s) 54.7 s

6 0
3 years ago
A uniform rod of length 0.7 m and mass 10 kg rotates freely about a horizontal axis passing through one end of the rod a bullet
mars1129 [50]

Answer:

<u>ω = 1.7 rad/s</u>

Explanation:

Conservation of angular momentum

Assuming the rod is initially hanging vertically at rest.

Initial angular momentum is carried by the bullet only

L = Iω = (mR²)(v/R) = mvR = 0.020(200)(0.7) = 2.8 kg•m²/s

the same angular momentum exists after impact, only the moment of inertia has increased by that of the rod. I = ⅓mR²

2.8 = (⅓(10)(0.7²) + 0.020(0.7²))ω

2.8 = (1.64313333...)ω

ω = 1.70406134...

3 0
3 years ago
Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 2.6 m long, its mass is 0.5
Irina-Kira [14]

Answer:

15.13 m/s

Explanation:

The wave speed of the stretched rope can be calculated using the following formula

v = \sqrt{\frac{F_T}{\mu}}

where F_T = 44N is the tension on the rope and \mu = m/L = 0.5 / 2.6 = 0.1923 kg/m is the density of the rope per unit length

v = \sqrt{\frac{44}{0.1923}} = \sqrt{228.8} = 15.13 m/s

6 0
3 years ago
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