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3 years ago

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Two construction cranes are each able to lift a maximum load of 20000 N to a height of 250 m. However, one crane can lift that l

oad in 1 6 the time it takes the other. How much more power does the faster crane have?

1 answer:

Neporo4naja [7]3 years ago

5 0

Answer:

Explanation:

Given

Load $W=20000\ N$

height to which load is raised $h=250\ m$

Another crane take $\frac{1}{6}$ th time to lift the load

Energy required required to lift the Weight

$E=W\times h$

$E=20000\times 250$

$E=5,000,000\ J$

Suppose $P_1$ and $P_2$ is the Power required to lift the weight in t and $\frac{t}{6}$ time

$E=P_1\times t$

$E=P_2\times \frac{t}{6}$

$P_1\times t=P_2\times \frac{t}{6}$

thus

$P_2=6P_1$

Second Crane requires 6 times more power than the slow crane

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Number of miles that marker shows when passes through town= 160 miles.

Number of miles that marker shows currently to John = 115 miles.

We need to find the distance between town and John's current location.

For the problem, we can clearly see that Town is at 160 miles away but when John passes the marker shows 115 miles.

So, it's just the difference between 160 miles and 115 miles.

In order to find that difference, we need to subtract those two numbers.

160miles - 115miles = 45 miles.

So, we could say the distance between town and John's current location is 45 miles.

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A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi

malfutka [58]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

1)

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude $mg$ , where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

2)

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

$F_{net}=ma$

where

$F_{net}$ is the net force

m = 75.0 kg is the mass of the bicyclist

$a=2.20 m/s^2$ is its acceleration

Solving, we find the net force:

$F_{net}=(75.0)(2.20)=165 N$

3)

In this part, we basically want to find the forward force of push, F.

We can rewrite the net force acting on the bicyclist as

$F_{net}=F-R$

where:

F is the forward force of push

R is the air drag

We know that:

$F_{net}=165 N$ is the net force on the bicyclist

R = 60.0 N is the magnitude of the air drag

Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:

$F=F_{net}+R=165+60=225 N$

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3 years ago

A rectangular coil of 65 turns, dimensions 0.100 m by 0.200 m, and total resistance 10.0 ? rotates with angular speed 29.5 rad/s

vovikov84 [41]

Answer:

Explanation:

N = 65

Area, A = 0.1 x 0.2 = 0.02 m^2

R = 10 ohm

ω = 29.5 rad/s

B = 1 T

(a) at t = 0

e = N x B x A x ω

e = 65 x 1 x 0.02 x 29.5

e = 38.35 V

(b) The maximum rate of change of magnetic flux is equal to the maximum value of induced emf.

Ф = 38.35 Wb/s

(c) e = NBAω Sinωt

e = 65 x 1 x 0.02 x 29.5 x Sin (29.5 x 0.05)

e = 38.174 V

(d) Maximum torque

τ = M B Sin 90

τ = N i A B

τ = N e A B / R

τ = 65 x 38.35 x 0.02 x 1 / 10

τ = 5 Nm

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