A substance with no definite volume and no definite shape is classified as a

An Amtrak going 250m/s comes to a stop in 12s. What is the<br> acceleration?

astraxan [27]

Answer:

$a=\frac{v-u}{t} \\a = \frac{0-250}{12} = -20.83 m/s$

Explanation:

you mean deceleration right ? because the acceleration is 250m/s

7 0

3 years ago

When a car turns a corner on a level road, which force provides the necessary centripetal acceleration?

Vinil7 [7]

Answer:

The frictional force between the tire made with the road

Explanation:

This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

5 0

3 years ago

10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr

12345 [234]

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

$F_{g0}=G \frac{mM_{E0}}{r^{2}}$

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=2M_{E0}$

so:

$F_{gf}=G \frac{2mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=2$

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=3M_{E0}$

so:

$F_{gf}=G \frac{3mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=3$

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

$M_{Ef}=\frac{M_{E0}}{4}$

so:

$F_{gf}=G \frac{mM_{E0}}{4r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=\frac{1}{4}$

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

4 0

3 years ago

Why are magnetic fields evidence of sea floor spreading

xenn [34]

Eruptions of molten material, magnetic stripes in the rock of the ocean floor, and the ages of the rocks themselves.

8 0

3 years ago

A basketball player throws the ball at a 47 angle above the horizontal to a hoop which is located a horizontal distance L = 5.0

FromTheMoon [43]

Answer:

$v_0 =$ 1.71

Explanation:

the parabolic movment is described by the following equation:

$y = tan(a)x-\frac{1}{2v_0^2(cos(a))^2}gx^2$

where y is the height of the ball, a is the angle of launch, $v_0$ the initial velocity, g the gravity and x is the horizontal distance of the ball.

So, if we want that the ball reach the hood, we will replace values on the equation as:

$0.8 = tan(47)(5)-\frac{1}{2v_0^2(cos(47))^2}(9.8)(5)^2$

Finally, solving for $v_0$ , we get:

$v_0=\sqrt{\frac{-9.8(5)^2}{(0.8-tan(47)(5))2cos^2(47)}}$

$v_0 =$ 1.71

4 0

4 years ago

A substance with no definite volume and no definite shape is classified as a —