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Harman [31]
4 years ago
14

A charge is moving in a magnetic field that points to the

Physics
1 answer:
padilas [110]4 years ago
3 0
Up, left, into the screen
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An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next sec
Viefleur [7K]

Answer:

E. 3h

Explanation:

We know that

u = 0 m/s.

velocity after t = 1s

v = u+gt = 0+9.81 x 1s= 9.81 m/s

distance covered in 1st sec

= =>> ut+0.5 x g x t²

=>>0 + 0.5x 9.81 x 1 = 4.90m

Let 4.90 be h

distance travelled in 2nd second will now be used

So velocity after t = 1s

=>>1 x t+ 0.5 x g x t²

=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90

So since h= 4.90

Then the ans is 3x h = 3h

3 0
3 years ago
Which element most likely has a shiny luster​
ruslelena [56]

..

Answer:

Rhodium. This extremely rare, valuable and silvery-colored metal is commonly used for its reflective properties. ...

Platinum.

Gold.

Iridium.

Osmium.

Palladium.

Rhenium

Explanation:

4 0
3 years ago
Read 2 more answers
Which gives the kinetic energy of a descending yo-yo?
matrenka [14]
A five pushing and letting go of the yoyo
8 0
3 years ago
TRUE OR FALSE. if an object does not change its position at a given time interval, then it is at rest or its speed is zero or no
Andrew [12]

Answer:

true

Explanation:

a body can only be accelerating and have speed if it's in motion

3 0
3 years ago
A bob of mass m = 0.250 kg is suspended from a fixed point with a massless string of length L = 22.0 cm. You will investigate th
katen-ka-za [31]

To solve the problem, it is necessary to use the concepts of gravitational force, centripetal force and trigonometric components that can be extrapolated from the statement.

By definition we know that the Force of Gravity is given by

F_g=mg

Where,

m= Mass

g = Gravitational Acceleration

The centripetal force is given by,

F_c = \frac{mv^2}{R}

Where,

m = Mass

v = Velocity

R = Radius

For the case described in the problem, the Force of gravity the net component would be given by sin?, While for the centripetal force the net component is in the horizontal direction, therefore it corresponds to the cos\theta

Then,

F_g = mg sin\theta

F_c = \frac{mv^2}{r}cos\theta

From the radius we have its length but not the net height, which would be given by

r = L sin\theta

So equating the equations we have to

F_g = F_c

mg sin\theta=\frac{mv^2}{r}cos\theta

mg sin\theta=\frac{mv^2}{Lsin\theta}cos\theta

Re-arrange to find v,

v = \sqrt{\frac{gLsin^2\theta}{cos\theta}}

Replacing with our values

v = \sqrt{\frac{(9.8)(22*10^{-2})(sin^2 24)}{cos24}}

v = 0.624

Therefore the tangential velocity of the mass is 0.624m/s

5 0
4 years ago
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