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3 years ago

Please answer this question only if you know it!! 17 points and brainliest :)

Physics

2 answers:

patriot [66]3 years ago

7 0

The answer is 15. I hope my answer is correct. Please tell me if not. Goodluck!

Hope this helpss :)

Scrat [10]3 years ago

5 0

Power = (voltage) x (current)

90 Watts = (6 volts) x (current)

Divide each side by (6 v):

Current = (90 W) / (6v)

Current = 15 Amperes

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3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between

Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

$v^2 -u^2 = 2ad$ (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

$F=ma=-\mu mg$

where

m = 1000 kg is the mass of the car

$\mu = 0.80$ is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

$a=-\mu g$

And we can substitute it into eq.(1) to find d:

$v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m$

7 0

3 years ago

How could glacial marks on one continent be similar to glacial marks on another continent far away?

BigorU [14]

Answer:

Plate Tectonics

Explanation:

Plate tectonics is the scientific theory that the continents move around due to due plate tectonics and the movement of magma under the crust. This explains why some fossils of the same kind are found on other continents, and same goes for the Glacial Marks.

6 0

3 years ago

Two cars A and B are 100m apart moving towards each other with

maxonik [38]

Let car A's starting position be the origin, so that its position at time t is

A: x = (40 m/s) t

and car B has position at time t of

B: x = 100 m - (60 m/s) t

They meet when their positions are equal:

(40 m/s) t = 100 m - (60 m/s) t

(100 m/s) t = 100 m

t = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) t - (100 m - (60 m/s) t) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) t = 200 m

t = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

3 0

3 years ago

A man pushes a heavy cart with a force exerted of 250 Newtons to keep it moving at a constant velocity. What is the kinetic fric

miv72 [106K]

It can't be less than 250 N or the cart wouldn't move at all. That means there is only 1 answer. It's between not enough info or 250 N. The answer is 250 N. If it was any more, there would be acceleration.

3 0

3 years ago

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago