What is the current if a charge of 25 coulombs passes in 5 seconds?

Three students have been studying relative motion and decide to do an experiment to demonstrate their knowledge. The experiment

miskamm [114]

Answer with Explanation:

We are given that

Constant speed of Jane=12.6 m/s

a.When Fred can throw the ball 30 m/s

We have to find the angle relative to the horizontal when he throw the ball in order for Sue to see the ball travel vertically upward.

Let $\theta$ be the angle .

Therefore,

$30 cos\theta=12.6$

$cos\theta=\frac{12.6}{30}=0.42$

$\theta=cos^{-1}(0.42)=65.165^{\circ}$

b.We have to find the height to which ball reach.

$v^2-v^2_0=2aS$

$S=\frac{v^2-v^2_0}{2a}=\frac{0-(30 sin65.165)^2}{2(-9.81)}$

$S=37.78 m$

7 0

3 years ago

What is the average acceleration of the particle between 0 seconds and 4 seconds? A. 0 meters/second2 B. 0.04 meters/second2 C.

lisov135 [29]

Answer

D. 0.25 meters/second2

Explanation

The average acceleration is the ratio of change in velocity to the change in time of travel.Taking in this case that the change of velocity is a unit, then Average acceleration is given by;

Aacc=Vf-Vi/Tf-Ti

where Vf=final velocity,Vi=initial velocity' Tf=final time, Ti=initial time

Vf-Vi=1m/s

Tf-Ti=4-0=4seconds

Avacc=1/4=0.25m/s2

6 0

3 years ago

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses $x_1$ distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1$

When speed doubles

$\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2$

divide 1 and 2

$\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}$

$x_2=2x_1$

Therefore spring compresses twice the initial value

7 0

3 years ago

The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from

grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

$\epsilon=-N\dfrac{d\phi}{dt}$

Where

$\dfrac{d\phi}{dt}$ is the rate of change of magnetic flux,

And $\phi=BA$

$\epsilon=-NA\dfrac{dB}{dt}$

$\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}$

$\epsilon=28.32\ V$

Using Ohm's law, $\epsilon=I\times R$

Induced current, $I=\dfrac{\epsilon}{R}$

$I=\dfrac{28.32}{1.5}$

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0

3 years ago

hergy A Sprinter runs at a a forward velocity of 10,9 m/s. If the sprinter has a mass of 72.5 kg, What is the Sprinter's Kinetic

Nina [5.8K]

Answer:

4306.8625 J or 344549/80J (fraction form)

Explanation:

kinetic energy= $\frac{1}{2\\}$ mv²

= $\frac{1}{2}$ ×72.5×10.9²

= 4306.8625 J or 344549/80 J

7 0

1 year ago