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Mariulka [41]
3 years ago
9

How is an electromagnet different from a bar magnet?

Physics
2 answers:
salantis [7]3 years ago
6 0

Answer:

this was wrong

Explanation:

Rus_ich [418]3 years ago
5 0
The answer is c) an electromagnet is temporary
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What is the collective noun of chocolate​
Vedmedyk [2.9K]

Answer:

creamy with milk and chocolate

4 0
3 years ago
On a journey of 600 km, a train was delayed 1 hour and 30 min after having covered 1 4 of the way. To arrive on time at the dest
mars1129 [50]

Answer:

The train travelled 10hours.

Explanation:

Using speed= distance/time ...eq1

Let the time taken by train to cover the journey be t.

Let the speed of train be s

Time= distance/speed ...eq2

Time t =600/s ...eq3

The train is delayed for 1 1/2 hours=3/2 hours

Train increased by 15km/hr.

Train travelled 1/4 of 600k.= 150km.

Speed increased by s + 15 to cover the remaining 450km

t = 150/s + 400/(s + 15) + 3/2 ..eq4

Equating eq3 and 4

600/s= 150/s + 460/(s + 15) + 3/2

450/s = 3/2 + 450/(s + 15)

3s^2 + 45s -13500=0

Solving the quadratic equation

S= -45 +-sqrt(45^2-4(3×13500)/2×3

S= 60 , -75

Hence speed of train is 60km/hr

Using eq 2 distance/speed=time

600/60=10

t= 10hours

5 0
4 years ago
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's a
s2008m [1.1K]

Answer:

A. Power generated by meteor = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Workdone = 981000 J

Power required = 19620 Watts

Note: The question is incomplete. A similar complete question is given below:

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?

Explanation:

A. Power = workdone / time taken

Workdone = Kinetic energy of the meteor

Kinetic energy = mass × velocity² / 2

Mass of meteor = 1.5 g = 0.0015 kg;

Velocity of meteor = 50 km/s = 50000 m/s

Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J

Power generated = 1875000/2.1 = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Work done by elevator against gravity = mass × acceleration due to gravity × height

Work done = 1000 kg × 9.81 m/s² × 100 m

Workdone = 981000 J

Power required = workdone / time

Power = 981000 J / 50 s

Power required = 19620 Watts

Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.

6 0
3 years ago
2. A Plate 0.02 mm distance from a fixed Plate moves at a velocity Of 0.6mls and requires a force of 1.962 N Per unit area to ma
Marizza181 [45]

Answer:

6.54 × 10⁻⁵ Pa-s

Explanation:

Since the shear force, F = μAu/y where μ = viscosity of fluid between plates, A = area of plates, u = velocity of fluid = 0.6 m/s and y = separation of plates = 0.02 mm = 2 × 10⁻⁵ m

Since F = μAu/y

F/A = μu/y where F/A = force per unit area

Since we are given force per unit area, F/A = 1.962 N per unit area = 1.962 N/m²

So,  μ = F/A ÷ u/y

substituting the values of the variables into the equation, we have

μ = F/A ÷ u/y

μ = 1.962 N/m² ÷ 0.6 m/s/2 × 10⁻⁵ m

μ = 1.962 N/m² ÷ 0.3 × 10⁵ /s

μ = 6.54 × 10⁻⁵ Ns/m²

μ = 6.54 × 10⁻⁵ Pa-s

5 0
3 years ago
A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.64
Galina-37 [17]

Answer:

The downwards acceleration is 3.53 m/s2.

Explanation:

Let the true weight is m g.

The reading of the balance, R = 0.64 mg

Let the acceleration is a.

As the apparent weight is less than the true weight so the elevator goes down wards with some acceleration.

Use Newton's second law

m g - R = m a

m g - 0.64 m g = m a

0.36 g = a

a = 3.53 m/s2

6 0
4 years ago
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