A motor is used to produce

1. A ball initially rolling at 10m/s comes to a stop in 25 seconds. Assuming the ball has

LiRa [457]

Answer:

B) 125 m

Explanation:

$s = \frac{u + v}{2} t \\ s \: = \frac{10 + 0}{2} (25) \\ s \: = 125m$

5 0

3 years ago

Why is a camera lens round but the pictures come out square

sergey [27]

It is round or say spherical to widen the range of photography and it is designed to focus the image as the ray are coming from infinity so !!

I am not an expert of camera but ya it is the main theme !!

6 0

3 years ago

Read 2 more answers

Find Vxl and Vyl of a pumpkin launched at a velocity of 55 m/s at an angle of 20 degrees

Vinvika [58]

Answer:

Explanation:

is A projectile is any object on which the only force acting is gravity and air resistance (drag).

Examples of projectiles are:

baseballs and softballs in the air after being hit by the bat

golf balls hit by a club

objects dropped from aircraft, such as people (skydivers), bombs, crates of food being dropped to refugees

objects launched by cannons, such as cannonballs, shells, and circus performers

Once the baseball, softball, golf ball, skydiver, bomb, crate, cannonball, shell, or clown are no longer touching the bat, club, aircraft, or cannon, and are in the air with only gravity and slight air resistance acting on it, then it is a projectile.

Here is an online projectile motion applets to play with, just for fun.

Unless otherwise stated in a particular problem or discussion, we will be ignoring the effects of air resistance.

The key to understanding the motion of projectiles is that the horizontal motion and the vertical motion of the projectile are independent of each other. So we can write separate equations for the displacement of the projectile in the horizontal (x) and vertical (y) directions.

The only common variable between these two equations is t, the time. Because in projectile problems there is usually no acceleration (i.e. we ignore air resistance) in the horizontal direction, we can write

The velocity components follow the same equations we used for one-dimensional motion.

Because there is usually no acceleration in the x direction, the x-velocity is constant.

3 0

3 years ago

A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find

pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0=4\pi \cdot 10^{-7} H/m$ is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

$I=1.41 A$ (current in the wire)

$B=5.61\mu T=5.61\cdot 10^{-6} T$ (strength of magnetic field)

Solving for r, we find the distance from the wire:

$r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m$

4 0

3 years ago

It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used

stepladder [879]

Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

$a_{c}$ = rω²

ω² = $a_{c}$ / r

ω = √( $a_{c}$ / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a $a_{c}$ = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²

ω = 0.0500647 ≈ 0.05 rad/s

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM

1 rpm = 60 rph

so

ω = 0.478082 × 60

ω = 28.6849 revolutions per hour

Therefore, the required revolution per hour is 28.6849

7 0

3 years ago