The ratio of horizontal component and vertical component of

An object with a charge of -3.2 uC and a mass of 1.0Ã—10^(-2) kg experiences an upward electric force, due to a uniform electric

USPshnik [31]

Answer:

The magnitude of the acceleration is equal to 19.6m/s² and the acceleration is directed upwards though the magnitude of the charge has doubled. This is because the electric force is directed upwards and from newton's second law of motion the charge will have acceleration in the same direction as the electric force on the charge.

Explanation:

The detailed solution can be found in the attachment below.

Thank you for reading and I hope this is helpful to you.

6 0

3 years ago

Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .

aliya0001 [1]

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s

Let $S_{0}$ be the speed of the river's current given as 1.00 m/s.

Note that this speed is the magnitude of the velocity which is a vector quantity.

The direction of the swimmer is upstream.

Hence the resultant velocity is given as, $S_{R}$ = S — S 0 $S_{0}$

$S_{R}$ = 1.25 — 1

$S_{R}$ = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

6 0

3 years ago

The equation of a transverse wave traveling along a very long string is y 6.0 sin(0.020px 4.0pt), where x and y are expressed in

zhannawk [14.2K]

Answer:

given

y=6.0sin(0.020px + 4.0pt)

the general wave equation moving in the positive directionis

y(x,t) = ymsin(kx -?t)

a) the amplitude is

ym = 6.0cm

b)

we have the angular wave number as

k = 2p /?

or

? = 2p / 0.020p

=1.0*102cm

c)

the frequency is

f = ?/2p

= 4p/2p

= 2.0 Hz

d)

the wave speed is

v = f?

= (100cm)(2.0Hz)

= 2.0*102cm/s

e)

since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction

f)

the maximum transverse speed is

umax =2pfym

= 2p(2.0Hz)(6.0cm)

= 75cm/s

g)

we have

y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]

= -2.0cm

6 0

3 years ago

Help me with this review question please.

QveST [7]

Answer:

K E=( mv²)/2

=(60×3.5²)/2

=367.5J

6 0

3 years ago

"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e

notka56 [123]

Answer:

The Magnifying power of a telescope is $M = 109.26$

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective $f_{objective}$ = $\frac{R}{2}$

⇒ $f_{objective}$ = $\frac{590}{2}$

⇒ $f_{objective}$ = 295 cm

Focal length of eyepiece $f_{eyepiece}$ = 2.7 cm

Magnifying power of a telescope is given by,

$M = \frac{f_{objective} }{f_{eyepiece} }$

$M = \frac{295}{2.7}$

$M = 109.26$

therefore the Magnifying power of a telescope is $M = 109.26$

4 0

3 years ago