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3 years ago

For a projectile, why is the initial horizontal velocity equal to the final horizontal velocity?

1 answer:

4 0

I am assuming that the context is that of a thrown projectile (a ball, a bullet, etc.) in the gravitational field of the Earth.

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The electric current in a copper wire is composed of what

Tanzania [10]

Answer:

A copper wire current consists of electrons appropriately called conduction electrons.

Explanation:

This answer came from quizlet.com. I hope that this helps you and good luck!

8 0

2 years ago

a motorcycle starts from rest covers 200 meter distance in 6 second calculate final velocity and acceleration

Ivanshal [37]

Explanation:

s = ut + 1/2 a t^2

200 = 0 * 6 + 1/2 * a * (6)^2

200 = 1/2 * a * 36

200 = 18 a

a = 200/18

a= 11.1m/sec^2

v = u + at

v = 0 + 11.1 * 6

v = 66.6m/s

hope it helps you

3 0

2 years ago

A projectile is fired directly upward from the ground with an initial velocity of 112 f/sec its distance s above the ground afte

Travka [436]

Answer:

3.48 seconds

Explanation:

At maximum height Vf=0 m/s

Vf= Vi - g*t

⇒g*t= Vi

⇒t= Vi/g

⇒t= 112/32.17 sec

⇒ t= 3.48 s

so the projectile will achieve its maximum height in 3.48 seconds.

7 0

2 years ago

A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th

GREYUIT [131]

Answer:

$1.26\cdot 10^7 m/s$

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

$F=qvB$

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

$qvB=m\frac{v^2}{r}$

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

$v=\frac{qBr}{m}$

where in this problem we have:

$q=1.6\cdot 10^{-19}C$ is the magnitude of the charge of the electron

$B=208 G=208\cdot 10^{-4}T$ is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

$r=3.45 mm = 3.45\cdot 10^{-3} m$

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

So, the electron's speed is

$v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s$

6 0

3 years ago

Acceleration question plz help

andrezito [222]

For #5 It's helpful to draw a free body diagram so you know which way the forces are acting on the block.

the weight mg is acting downwards, and you need to find the vertical and horizontal components of mg using sin and cosine. so do 15x9.8xsin40 which is the force. Assuming no friction, this is the only force acting on the block, as the forces on the vertical plane cancel out i.e the normal force and weight of the block.

after, just do F=ma And since you know F and m, solve for a.

3 0

3 years ago