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2 years ago

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The speed of a boat in still water is 20 mph. it travels from one pier to another with the current in 4 hours and back against t

he current in 6 hours and 40 minutes. if the distance between piers is 100 miles, what is the speed of the current

1 answer:

saw5 [17]2 years ago

3 0

The answer is, "the speed of the current is 5 miles per hour."

To calculate the speed of the current,
let's assume speed of current = xmph. Time taken to travel from one pier to another with the current = 100/(20+x)h

But the time taken to travel from one pier to another with the current, which is given is = 4 hours. So, 4=100/(20+x) 80+4x = 100

4x = 20

x = 5 Thus, the speed of the current is 5 miles per hour.

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The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?

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The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
$\rho = \frac{AR}{L}$
where
$\rho$ is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
$A=\pi r^2$
where r is the radius of the wire. Substituting in the previous equation ,we find
$\rho = \frac{\pi r^2 R}{L}$

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
$\rho' = \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4 \frac{\pi r^2 R}{L} = 4 \rho$
Therefore, the new resistivity must be 4 times the original one.

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2 years ago

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The energy consumed by a home during a month is 90 kWh, how many Joules are we talking about? a good explanation please is for t

andreev551 [17]

Answer:

3.24×10⁸ J, or 324 MJ

Explanation:

"kWh" is a kilowatt-hour. It's the energy used by 1 kilowatt of power after one hour.

A kilowatt is a kilojoule per second.

90 kWh

= 90 kW × 1 hr

= 90 kJ/s × 1 hr

= 90 kJ/s × 3600 s

= 324,000 kJ

= 324,000,000 J

The energy is 3.24×10⁸ J, or 324 megajoules.

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List the four outer planets from smallest to largest.<br> WILL MARK BRAINLIEST! ASAP

dusya [7]

Answer: Mercury, Mars, Venus, Earth, Neptune, Uranus, Saturn, and Jupiter.

Explanation:

That's all of the planets if you need them. Hope this helps!

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2 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$

where

m = 950 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\Delta h$ is the change in height, which is

$\Delta h = L sin 30^{\circ}$

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

$\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J$

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On Earth, it's (24 hours)/(sine of latitude).

On the moon, it would be (27.32 days)/(sine of latitude).

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