What is the acceleration of a baseball when it is moving downward after being tossed directly upward?

the picture below shows two parallel wires carrying current in the same direction. what is the net force on wire 2 due to wire 1

erastovalidia [21]

Answer:

I dont see a picture where is it?

Explanation:

I cannot see anything L question Luestion

3 0

2 years ago

2. A car is moving south. It is getting faster. In which direction is the acceleration?

notka56 [123]

South it is the same direction the car is already moving

7 0

3 years ago

Kinematics practice problems Answers: 4. A race car is traveling at +76 m/s when is slows down at -9 m/s2 for 4

arlik [135]

The new velocity after 4 s is 40 m/s

The height of the spaceship above the ground after 5 seconds is 1,127.5 m

The given parameters for the first question;

initial velocity of the car, u = 76 m/s

acceleration of the car, a = - 9 m/s²

time of motion, t = 4 s

The new velocity after 4 s is calculated as;

v = u + at

v = 76 + (-9)(4)

v = 76 - 36

v = 40 m/s

(5)

The given parameters;

height above the ground, h = 500 m

velocity of spaceship, u = 150 m/s

time of motion, t = 5

The height of the spaceship above the ground after 5 seconds is calculated as;

$h_y = h_0 + ut - \frac{1}{2}gt^2\\\\h_y = 500 + (150\times 5) - (0.5\times 9.8 \times 5^2)\\\\h_y = 1,127.5 \ m$

Learn more here: brainly.com/question/24527971

5 0

2 years ago

would the phases of the moon be affected if the moon didn't make one rotation for each revolution of earth?

Aleks [24]

As long as the moon orbits the Earth in the same way, the phases will be the same.

4 0

3 years ago

Water is falling on a surface, wetting a circular area that is expanding at a rate of 4 mm2 /s. How fast is the radius of the we

STALIN [3.7K]

Answer:

The radius of the wetter area expands at a rate of $4.244\times 10^{-3}$ milimeters per second when radius is 150 milimeters.

Explanation:

From Geometry we remember that area of a circle is described by this expression:

$A =\pi\cdot r^{2}$ (Eq. 1)

Where:

$r$ - Radius of the circle, measured in milimeters.

$A$ - Area of the circle, measured in square milimeters.

Then, the rate of change of the area in time is derived by concept of rate of change, that is:

$\frac{dA}{dt} = 2\pi\cdot r\cdot \frac{dr}{dt}$ (Eq. 2)

Where:

$\frac{dr}{dt}$ - Rate of change of radius in time, measured in milimeters per second.

$\frac{dA}{dt}$ - Rate of change of area in time, measured in square milimeters per second.

Now the rate of change of radius in time is cleared within equation above:

$\frac{dr}{dt} = \left(\frac{1}{2\pi\cdot r}\right)\cdot \frac{dA}{dt}$

If we know that $r = 150\,mm$ and $\frac{dA}{dt} = 4\,\frac{mm^{2}}{s}$ , then the rate of change of radius in time is:

$\frac{dr}{dt} = \left[\frac{1}{2\pi\cdot (150\,m)} \right] \cdot \left(4\,\frac{mm^{2}}{s} \right)$

$\frac{dr}{dt}\approx 4.244\times 10^{-3}\,\frac{mm}{s}$

The radius of the wetter area expands at a rate of $4.244\times 10^{-3}$ milimeters per second when radius is 150 milimeters.

7 0

3 years ago