The earth orbits a start called the ___

You need to get to class, 300 meters away, and you arrive in 1 minute. What speed were you walking? (calculate in meters per sec

ella [17]

5 meters per second. 300 / 60(seconds)

3 0

2 years ago

A baseball has a mass of 0. 45 kg and is thrown with a speed of 25 m/s. What is the momentum of the baseball? 0. 018 kg â€˘ mete

aleksandr82 [10.1K]

Answer:

11 Kg.m/s

Explanation:

P=mv

P=0.45*25

P=11.25 Kg.m/s

3 0

2 years ago

When the two surfaces of a thin film are parallel and the light is incident at some angle what fringes are formed? A. Circular B

lord [1]

Answer:

The correct option is;

A. Circular

Explanation:

Some of the light that impinges on the surface are reflected and the rest are transmitted to a different medium

At the surface of the next medium also, some of the light are transmitted while the others are reflected and refracted through the first medium

The speed of light (and hence the wavelength and color) refracted through the thin film is changed as the distance the refracted light travels through the thin film is increased as we move away from the point directly in the front view to some distance as the reflected light path from those distance to the eye is increased due to their inclination giving them a different wavelength which are all equal at a radial distance from the eye hence forming a circular fringes.

8 0

2 years ago

What are the two varieties of friction?

murzikaleks [220]

There are two main types of friction, static friction and kinetic friction. Static friction operates between

6 0

3 years ago

Baseball player a bunts the ball by hitting it in such a way that it acquires an initial velocity of 2.4 m/s parallel to the gro

LUCKY_DIMON [66]

Let $\mathbf r_A$ denote the position vector of the ball hit by player A. Then this vector has components

$\begin{cases}r_{Ax}=\left(2.4\,\frac{\mathrm m}{\mathrm s}\right)t\\r_{Ay}=1.2\,\mathrm m-\frac12gt^2\end{cases}$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the magnitude of the acceleration due to gravity. Use the vertical component $r_{Ay}$ to find the time at which ball A reaches the ground:

$1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s$

The horizontal position of the ball after 0.49 seconds is

$\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m$

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector $\mathbf r_B$ of the ball hit by player B has