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4 years ago

Give three practical uses of electromagnets

1 answer:

7 0

Three practical uses of electromagnets would be
Light. The waves
Electric locks. The bolts that are used by the
coils
Cranes. The electric ones that are found in
scraps
Cars computers washing machines all use
Electromagnets . Even CD players DVDs
Players

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Which mathematically describes the wave properties of electrons?.

Afina-wow [57]

The Quantum Theory

Hope this helps

8 0

2 years ago

Read 2 more answers

Traveling at an average rateaverage rate of between 50 and 60 miles per hour for 4 to 7 hours select the best estimate for the d

borishaifa [10]

Answer:

best close estimate distance is 315 miles

Explanation:

given data

speed v1 = 50 miles per hour

speed v2 = 60 miles per hour

time t1 = 4 hr

time t2 = 7 hr

to find out

best estimate for the distance traveled

solution

we know lower end distance

distance = v1 × t1

distance = 50 × 4 = 200 miles

and

upper end distance

distance = v2 × t2

distance = 60 × 7 = 420 miles

so middle value distance is

v = 55 miles per hour and t = 5.5 hours

distance = v × t

distance = 55 × 5.5

distance = 302.5 miles

so best close estimate distance is 315 miles

3 0

3 years ago

You are an engineer helping to design a roller coaster that carries passengers down a steep track and around a vertical loop. Th

vova2212 [387]

Answer:

h >5/2r

Explanation:

This problem involves the application of the concepts of force and the work-energy theorem.

The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.

Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)

So from newton's second law of motion,

W – N = mv²/r

N = normal force = 0

W = mg

mg = ma = mv²/r

mg = mv²/r

v²= rg

v = √(rg)

The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop

ΔPE = ΔKE = 1/2mv²

The height at the roller coaster starts is usually higher than the top of the loop by design. So

ΔPE =mgh - mg×2r = mg(h – 2r)

2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.

In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.

So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.

ΔPE > ΔKE

mg(h–2r) > 1/2mv²

g(h–2r) > 1/2(√(rg))²

g(h–2r) > 1/2×rg

h–2r > 1/2×r

h > 2r + 1/2r