Find the speed of a wave (in meters/second) whose wavelength is 4 meters and

A certain copper wire has a resistance of 10.5 ω. at what fraction of the length l must the wire be cut so that the resistance o

Delvig [45]

The answer for that questions will be 759. Stars itir

6 0

3 years ago

Why is dose equvialent in rem or sv used to define the Effective Dose Equivalent and the Committed Dose Equivalent?

Komok [63]

Good question next question

8 0

3 years ago

I would like help with this physics problem

Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

$v_f=v_i+at$

We have $v_f=-v_i$ , so that

$-2v_i=at\implies-2\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)=\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=8\,\mathrm s$

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(c) The ball's average velocity is 0. Average velocity is given by $\dfrac{v_i+v_f}2$ , and we know that $v_f=-v_i$ .

(d) The position of the ball $x_f$ at time $t$ is given by

$x_f=x_i+v_it+\dfrac12at^2$

Take the starting position to be the origin, $x_i=0$ . Then after 6 seconds,

$x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m$

so the ball is 42 m away from where it started.

We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

$x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}$

Since the velocity is positive, the ball is still moving up the incline.

8 0

3 years ago

A motor-driven winch pulls a 50.0 kg student 5.00 m up the rope at a constant speed of 1.25 m/s. how much power does the motor u

nadya68 [22]

Power is the rate work done given by dividing work done by unit time. It is measured in watts equivalent to J/s.
In this case the force by the student is mg = 490 N (taking g as 9.8m/s²)
Work done is given by force × distance,
Therefore, Power =(force × distance)/ time, but velocity/speed =distance/time
Thus, Power = force × speed/velocity
= 490 N × 1.25
= 612.5 J/S (Watts)
Hence, power will be 612.5 Watts.

7 0

3 years ago

a ship travels a port p and travels 30 km due north. then it changes course and travels 20 km in a direction 30° east of north

liq [111]

When we represent what is given to us on a coordinate plane, we have a figure as shown in the attachment.

To find the distance between P and R, we have to find the Net Displacement of the ship (brown arrow in the figure).

For that, we use the rules for Vector addition.

We see that the first displacement $D_{1}$ = 30 km (blue arrow) is along the y-axis, but the second part of the ship's journey $D_{2}$ = 20 km (red arrow) is at an angle with reference to y-axis.