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scZoUnD [109]
3 years ago
10

A current of 4 ampere flows in a wire of resistivity 60 ohm calculate power

Physics
2 answers:
Reil [10]3 years ago
7 0
The answer is 240 Volt
ohaa [14]3 years ago
4 0
Plz see the attachment....

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Scientists take scientific measurements carefully in order to ensure that reliability and validity. What is the difference betwe
Gala2k [10]
Accuracy describes how close a measurement is to the truest value while precision shows how close the measurements are to one another when repeated analysis of the same measurements are performed. Precision most likely is free from any any variation.
7 0
3 years ago
7) A crazy cat (yes, this is redundant) is running along the roof of a 60 m tall building. The cat is moving at a constant veloc
gregori [183]

Answer:

The distance from the base of the building to the landing site is 154 m.

The total flight time is 3.5 s.

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

Explanation:

The equations for the position and velocity vectors of the cat are as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector of the cat at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity

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

v = velocity vector of the cat at time t.

Please, see the attached figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the launching point so that x0 and y0 = 0. In a horizontal launch, initially there is no vertical velocity, then, v0y = 0.

When the cat reaches the ground, the position vector of the cat will be r1 (see figure). The vertical component of r1 is -60 m and the horizontal component will be the horizontal distance traveled by the cat (r1x). Then, using the equation of the y-component of the position vector, we can obtain the time of flight and with that time we can obtain the horizontal distance traveled by the cat:

r1y = y0 + v0y · t + 1/2 · g · t²

-60 m = 0 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

- 60 m = -4.9 m/s² · t²

-60 m / - 4.9 m/s² = t²

t = 3.5 s

The cat reaches the ground in 3.5 s

Now, we can calculate the horizontal component of r1:

r1x = x0 + v0 · t

r1x = 0 m + 44 m/s · 3.5 s

r1x = 154 m

The distance from the base of the building to the landing site is 154 m.

The total flight time was already calculated and is 3.5 s.

The velocity vector of the cat when it reaches the ground will be:

v = (v0x, v0y + g · t)

v = (44 m/s, 0 m/s - 9.8 m/s² · 3.5 s)

v = (44 m/s, -34.3 m/s)

The magintude of the vector "v" is calculated as follows:

|v| = \sqrt{(44 m/s)^{2}+(-34.3 m/s)^{2}} = 55.8 m/s

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

6 0
3 years ago
A greyhound's velocity changes from 10 meters per second to 15 meters per second in 0.5 second. What is the greyhound's average
alexgriva [62]
I think it is 5 m/s  
4 0
3 years ago
Read 2 more answers
What is the weight of a rock with a mass of 3.6 kilograms
Marysya12 [62]
3.85 pounds is the answer
4 0
3 years ago
What is the magnitude of the change in potential energy of the block-spring system when it travels from its lowest vertical posi
maksim [4K]

Answer:

 ΔU = 2 mg h

Explanation:

In a spring mass system the potential energy is U = m g h

where h is measured from the equilibrium point of the spring

the potential energy at the highest point is

         U₁ = m g h

the potential energy at the lowest point is

         U₂ = m g (-h)

instead in this energy it is

          ΔU = 2 mg h

In this two points the kinetic energy is zero, but there is elastic potential energy that has the same value in the two points, so its change is zero

4 0
3 years ago
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