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8090 [49]
4 years ago
13

Calculate the potential difference across a 25-Ohm. resistor if a 0.3-A current is flowing through it.

Physics
2 answers:
Sveta_85 [38]4 years ago
6 0

Answer:

7.5 V

Explanation:

nikdorinn [45]4 years ago
4 0

Answer:7.5V

Explanation:

Ohm's law, V=IR

so, V=0.3×25

V=7.5V

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Label the producers and consumers, then label the different types of consumers.
sdas [7]
The base of the pyramid has the producers and everything else above the base falls under the consumers category i.e the locusts,frogs and the snake. The grass is the producer, the locust is a consumer, the frog is a special type of omnivore, termed the "life-history omnivore" since they eat both plants and animals but at different times in their lives. In this case they are just omnivores and lastly, the snake is a carnivore.
3 0
3 years ago
If a car accelerates uniformly from rest to 15 meters
Talja [164]

Answer:

1.125m/s^2

Explanation:

Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically

v^2= u^2+2as

Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.

a = ?

u = 0m/s

v = 15m/s

s = 100m

Substituting the values into the formula above

v^2= u^2+2as

15^2=0^2+2×a×100

225= 0+200a

225= 200a

Divide both sides by 200

225/200 = 200a/200

a= 1.125m/s^2

Hence the acceleration of the car is 1.125m/s^2.

Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s

8 0
3 years ago
Read 2 more answers
Hiii please help i’ll give brainliest if you give a correct answer please thanks!
mrs_skeptik [129]

Answer:

25 to the right

Explanation:

there you go friend your awsome

4 0
3 years ago
Light waves have some similarities with water and sound waves, but they are not exactly the same. Describe all the differences y
makkiz [27]

<u>Answer:</u>


<h2>All the waves are pertubations that propagate (transport) energy.</h2><h2></h2>

Nevertheless, they have some differences:


1. Light waves are<u> electromagnetic waves</u>, while sound and water waves are <u>mechanical waves</u>, this is the first and principal difference.  

2. Electromagnetic waves can<u> propagate in vacuum</u> (they do not need a medium or material), but mechanical waves obligatory need a material to propagate

3. Light waves are always <u>transversal waves</u>, this means <u>the oscillatory movement is in a direction that is perpendicular to the propagation</u>; but mechanical waves may be both: <u>longitudinal waves</u> (the oscillation occurs in the same direction as the propagation) or transversal waves.

4. Electromagnetic waves propagates at a <u>constant velocity</u> (Light velocity) while the velocity of mechanical waves will depend on the type of wave and the <u>density</u> of the medium or material.

5. <u>Mechanical waves</u> are characterized by the regular variation of a single magnitude, while <u>electromagnetic waves</u> are characterized by the variation of two magnitudes: the electric field and the magnetic field

6. <u>Water waves</u> are 2-dimensional waves, while the <u>light and the sound</u> are tridimensional spherical waves

7. Light waves <u>transports energy in the form of </u><u>radiation</u>, while mechanical waves t<u>ransport energy with </u><u>material</u>


3 0
3 years ago
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally h
zysi [14]

Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

Number of turns, N = 500-turn

time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

where;

A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s =  60 rev/s

Induced peak emf = NABω

                               = 500 x (π x 0.25²) x 0.425 x 376.738

                               = 15721.16 V

                               = 15.721 kV

Therefore, the peak emf generated by the coil is 15.721 kV

5 0
3 years ago
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