All categories

4 years ago

PLS PLS PLS PLS PLS PLS PLS PLS PLS HELP FREE POINTS

Physics

1 answer:

Ne4ueva [31]4 years ago

3 0

Answer:

see below

Explanation:

First: Leave a couple inches of wire loose at one end and wrap most of the rest of the wire around iron u-shaped bar and make sure not to overlap the wires.

Second:Cut the wire (if needed) so that there is about a couple inches loose at the other end too.

Third: Now remove about an inch of the plastic coating from both ends of the wire and connect the one wire to one end of a battery and the other wire to the other end of the battery.

You might be interested in

A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th

xz_007 [3.2K]

Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

$v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}$

(b) The tension, $T$ , is given by

$v =\sqrt{\dfrac{T}{\mu}}$

where $v$ is the speed, $T$ is the tension and $\mu$ is the mass per unit length.

Hence,

$T = \mu\cdot v^{2}$

To determine $\mu$ , we need to know the mass of the cable. We use the density formula:

$\rho = \dfrac{m}{V}$

where $m$ is the mass and $V$ is the volume.

$m=\rho\cdot V$

If the length is denoted by $l$ , then

$\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}$

$T = \dfrac{\rho\cdot V}{l} v^{2}$

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

$V = \pi \dfrac{d^{2}}{4} l$

$T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2$

$T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2$

$T = 11159.4186\ldots \text{ N} = 11000 \text{ N}$

4 0

4 years ago

20 points Please And WILL mark a as a brainlest

boyakko [2]

Probably the earth traveling around the sun

3 0

4 years ago

A 98.0 N grocery cart is pushed 12.0 m along an aisle by a shopper who exerts a constant horizontal force of 40.0 N. If all fric

Digiron [165]

Answer:

9.8 m/s

Explanation:

The work done by the force pushing the cart is equal to the kinetic energy gained by the cart:

$W=K_f -K_i$

where

W is the work done

$K_f$ is the final kinetic energy of the cart

$K_i$ is the initial kinetic energy of the cart, which is zero because the cart starts from rest, so we can write:

$W=K_f$

But the work is equal to the product between the pushing force F and the displacement, so

$W=Fd=(40.0 N)(12.0 m)=480 J$

So, the final kinetic energy of the cart is 480 J. The formula for the kinetic energy is

$K_f=\frac{1}{2}mv^2$ (1)

where m is the mass of the cart and v its final speed.

We can find the mass because we know the weight of the cart, 98.0 N:

$m=\frac{F_g}{g}=\frac{98.0 N}{9.8 m/s^2}=10 kg$

Therefore, we can now re-arrange eq.(1) to find the final speed of the cart:

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(480 J)}{10 kg}}=9.8 m/s$

7 0

3 years ago

Emeka carried out a reaction in which a gas was given off. He followed the progress of the reaction by measuring the mass of the

pickupchik [31]

Answer:

17.5

1.1 g/min

I know it's one of these, try getting a second opinion

6 0

3 years ago

The reaction mixture represented above is at equilibrium at 298 K. The value of the equilibrium constant for the reaction is 16

andre [41]

Answer : The equilibrium concentration of T(g) is 0.5 M

Solution :

Let us assume that the equilibrium reaction be:

The given equilibrium reaction is,

$R(g)+2T(g)\rightleftharpoons 2X(g)+Z(g)$

The expression of $K_c$ will be,

$K_c=\frac{[Z][X]^2}{[R][T]^2}$

where,

$K_c$ = equilibrium constant = 16

[Z] = concentration of Z at equilibrium = 2.0 M

[R] = concentration of R at equilibrium = 2.0 M

[X] = concentration of X at equilibrium = 2.0 M

[T] = concentration of T at equilibrium = ?

Now put all the given values in the above expression, we get:

$16=\frac{(2.0)\times (2.0)^2}{(2.0)\times [T]^2}$

$[T]=0.5M$

Therefore, the equilibrium concentration of T(g) is 0.5 M

8 0

4 years ago