All categories

3 years ago

A polar molecule is one that exists only at low temperatures. true or false

Physics

1 answer:

bazaltina [42]3 years ago

4 0

Answer: no

Explanation:

It is not true

You might be interested in

how much current is in a circuit that includes a 9-volt battery and a bulb with a resistance of 3 ohms?

ExtremeBDS [4]

3 Amper 9 divided by 3

8 0

3 years ago

At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to (a) 2 times that of the Celsius and (b) 1/4 ti

castortr0y [4]

Answer:

(a) F = 320

(b) = F = -5.1625

Explanation:

The formula that converts degree Celsius (C) to degree Fahrenheit (F) is:

F = 1.8C + 32

Solving (a): F = 2C

Substitute 2C for F in the above equation

F = 1.8C + 32

2C = 1.8C + 32

Collect like terms

2C - 1.8C = 32

0.2C = 32

Multiply both sides by 5

5 * 0.2C = 32 * 5

C = 160

Recall that F = 2C

F = 2 * 160

F = 320

Solving (b): F = ¼C

Substitute ¼C for F in the above formula

F = 1.8C + 32

¼C = 1.8C + 32

Convert fraction to decimal

0.25C = 1.8C + 32

Collect like terms

0.25C - 1.8C = 32

-1.55C = 32

Divide both sides by -1.55

C = 32/(-1.55)

C = -32/1.55

C = -20.65

Recall that: F = ¼C

F = -¼ * 20.65

F = -5.1625

5 0

3 years ago

The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a

jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A $V_a=382\ V$

Potential at $V_c=785\ V$

when particle starts from A it reaches with velocity $v_b$ at Point while when it starts from C it reaches at point B with velocity $2v_b$

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

$q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1$

Change in Kinetic Energy of particle moving under the Potential From C to B

$q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2$

Divide 1 and 2 we get

$\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}$

on solving we get

$V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c$

$V_b=\frac{743}{3}=247.67\ V$

4 0

3 years ago

A torque of 0.77 N⋅m is applied to a bicycle wheel of radius 30 cm and mass 0.70 kg

Naddik [55]

Answer:

α = τ/I = 0.77 / (0.70(0.30²)) = 12.22222... = 12 rad/s²

Explanation:

4 0

3 years ago

Calculate the angular velocity of the neptune (orbital period of 165 earth period) in its orbit around the sun.

ludmilkaskok [199]

The correct answer is: Angular velocity = $1.208 * 10^{-9}$ rad/s

Explanation:
The angular velocity is given as:
ω = $\frac{2\pi}{T}$ --- (1)

Where T = 165 * (365 days) * (24 hours/day) * (60 minutes/hour) * (60 seconds/minute) = 5203440000 s

Plug in the value in (1):
ω = $\frac{2\pi}{5203440000} = 1.208 * 10^{-9}$ rad/s

7 0

3 years ago