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4 years ago

WILL MARK BRAINLIEST PLS HELP

Physics

1 answer:

KiRa [710]4 years ago

4 0

Answer:

Our planet's rotation produces a force on all bodies moving relative to theEarth. Due to Earth's approximately spherical shape, this force is greatest at the poles and least at the Equator. The force, called the "Coriolis effect," causes the direction of winds and ocean currents to be deflected.

Hope this helps!

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A net force of 100 N is moving a mass with an acceleration of 5 m/s2. What is the mass of the object?

olya-2409 [2.1K]

Fnet=ma so m = fnet/a 100/5=20 the mass is 20

7 0

3 years ago

Can a magnet attract a pure substance?

Salsk061 [2.6K]

Answer:

Magnets strongly attract materials which already themselves have magnetic domains. They do not significantly attract many metals like gold, aluminum, silver, and even some types of high-chromium stainless-steel, which lack such domains. In fact, pure gold is slightly repelled.

Explanation:

3 0

3 years ago

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

${v}^{2} = {z}^{2} + 2ax$

$v = z + at$

First let's find the final velocity the plane will have at the end of the runway using the first equation:

${v}^{2} = {0}^{2} + 2(5)(1800)$

$v = 60 \sqrt{5}$

Now we can plug this into the second equation to find t:

$60 \sqrt{5} = 0 + 5t$

$t = 12 \sqrt{5}$

Then using 3 significant figures we round to 26.8 seconds

3 0

3 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s$

3 0

4 years ago

Please help ASAP!!

bixtya [17]

There is kinetic energy when it is sitting at the top, then as it goes towards the bottom, the kinetic energy is transformed into potential energy.

6 0

3 years ago