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3 years ago

What symbol is used when representing momentum in equations?

1 answer:

6 0

Almost in every Physics books, momentum is denoted by P with small arrow(→) pointing towards right, placed on top of P.
I am guessing you asked about linear momentum.

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When the starter motor on a car is engaged, there is a 310 A current in the wires between the battery and the motor. Suppose the

Blababa [14]

Answer:

Diameter will be 351.42 mm

Explanation:

We have given current flowing in the copper wire i = 310 A

Voltage drop across the wire V = 0.55 volt

We know that resistance is given by $R=\frac{V}{i}=\frac{0.55}{310}=0.00177\Omega$

Length of the copper wire l = 1 m

Resisitivity of the copper wire $\rho =1.72\times 10^{-8}\Omega m$

We know that resistance $R=\frac{\rho l}{A}$

$0.00177=\frac{1.72\times 10^{-8}\times 1}{A}$

$A=969.4545\times 10^{-8}m^2$

As area $A=\pi r^2$

$3.14\times r^2=969.4545\times 10^{-8}$

$r=17.57\times 10^{-4}m$

So diameter $d=17.57\times 10^{-4}\times 2=35.142\times \times 10^{-4}m$ = 351.42 mm

3 0

3 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

3 years ago

What is the frequency of a wave that has a wavelength of 0.39 m and a speed

gogolik [260]

Answer:

<h3>The answer is option B</h3>

Explanation:

The frequency of a wave can be found by using the formula

$f = \frac{c}{ \lambda} \\$

where

c is the velocity

From the question

wavelength = 0.39 m

c = 86 m/s

We have

$f = \frac{86}{0.39} \\ = 220.512820...$

We have the final answer as

Hope this helps you

7 0

3 years ago

PLZZ ANSWER THE QUESTION

mixas84 [53]

Its Displacement and Time for sure.
Thank You!
Pls mark Brainliest!

7 0

3 years ago

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final

Serhud [2]

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B = -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

$x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s$

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

5 0

2 years ago