What moon phase happens when the earth is between the moon and the sun?

A car speeds up from 14 meters per second to 21 meters per second in 6 seconds. Whats the acceleration and the distance passed w

Solnce55 [7]

Answer:

a = 1.666... m/s²

Explanation:

a = v2 - v1 / t2 - t1

a = 21m/s - 14m/s / 6s - 0s

a = 7m/s / 6s

a = 1.666... m/s²

7 0

2 years ago

Two elements in Period 4 that are most similar to Cobalt?

lisabon 2012 [21]

Iron and nickel are the elements most similar to cobalt

5 0

3 years ago

the kinetic energy of an object with mass m moving with a velocity of 5 m/s is 25 j what will be its kinetic energy when its vel

Dmitriy789 [7]

Answer:

100 J, 225 J

Explanation:

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the velocity of the object

In this problem, the initial kinetic energy of the object is

K = 25 J

Then, the velocity is doubled, which means

v' = 2v

Therefore, the new kinetic energy will be

$K'=\frac{1}{2}m(2v)^2 = 4(\frac{1}{2}mv^2)=4K$

Therefore, the kinetic energy has quadrupled:

$K' = 4(25)=100 J$

Later, the velocity is tripled, which means

v'' = 3v

Therefore, the new kinetic energy will be

$K''=\frac{1}{2}m(3v)^2 = 9(\frac{1}{2}mv^2)=9K$

Therefore, the kinetic energy has increased by a factor of 9:

$K' = 9(25)=225 J$

4 0

3 years ago

Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax

kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$