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Alex787 [66]
3 years ago
13

Question 3 of 25 What is the period of a wave that has a frequency of 3 Hz?​

Physics
1 answer:
Katarina [22]3 years ago
8 0

Answer:

use the formula

Explanation:

F = 1/ T

where F stands for frequency and T stands for period

sub in 3 Hz into frequency

3 = 1/T

0.333 = T

or in a fraction form

1/ 3

hope this helps

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Explain the why a constant speed has a slope of 0 on a graph of speed v.time.
Studentka2010 [4]
Because if I do 0*time you get the answer as 0
6 0
3 years ago
WILL MARK AS BRAINLIEST IF CORRECT ANSWERS PLZ HELP 15 PTS!!!!!!
nlexa [21]

Question 9 is 100 N.

Question 10 is No, the bike's motion is not changing so it could be at rest or moving at a constant velocity.

Question 11 is Be doubled.

Question 12 is Ella is correct.

Hope i helped.

6 0
3 years ago
Read 2 more answers
1. 1500j of work was done to move a box 20m. What force was applied to the box ?
Fantom [35]

Answer:

1. 75N

2. 67,983 J (=67.98 kJ)

Explanation:

1. Work = Force x Distance

we are given that Work = 1,500J and Distance = 20m

hence,

Work = Force x Distance

1,500 = Force x 20

Force = 1,500 ÷ 20 = 75N

2. Potential Energy, PE = mass x gravity x change in height

we are given that mass = 165 kg and change in height = 42m

assuming that gravity, g = 9.81 m/s²

Potential Energy, PE = mass x gravity x change in height

Potential Energy, PE = 165 x 9.81 x 42 = 67,983 J (=67.98 kJ)

4 0
3 years ago
The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p
beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

v(t)=ate^{-6t}   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a

hence, the maximum speed is v_max = ((1/6)e^-1)a

5 0
3 years ago
Assume that the equator of the Earth is 24,000 miles long (that’s its circumference). Now, pretend that you are standing somewhe
slega [8]

Answer:

1000 mph

Explanation:

P = Perimeter = 24000 mi

r = Radius of the equator

t = Time taken to complete one rotation = 24 h

Perimeter of a circle is given by

P=2\pi r\\\Rightarrow r=\dfrac{P}{2\pi}\\\Rightarrow r=\dfrac{24000}{2\pi}\ mi

Angular speed is given by

\omega=\dfrac{2\pi}{t}\\\Rightarrow \omega=\dfrac{2\pi}{24}

Velocity if given by

v=r\omega\\\Rightarrow v=\dfrac{24000}{2\pi}\times \dfrac{2\pi}{24}\\\Rightarrow v=1000\ mph

The person would be going at a speed of 1000 mph

3 0
3 years ago
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