Ask question

Ask question

All categories

3 years ago

5

Any Help??? Dued before 2:30pm and it's 9:30am by me... so please take yuh time and ans.... jus ans 1 if you want... This is for

course marks.... I appreciate all efforts❤

1 answer:

Kryger [21]3 years ago

5 0

Answer:

The methodology employed by Galileo contributed to the development of Physics by find moons of Jupiter. (I think)

sorry if it's wrong

You might be interested in

URGENT PHYSICS QUESTION WILL GIVE BRAINLIEST

Morgarella [4.7K]

Answer:

7.2V

Explanation:

Find the equivalent resistance:

Req = 10 ohms + 15 ohms = 25 ohms

Use ohm's law to find the current:

V = IR

12V = I(25 ohms)

I = .48 amps

Multiple the current with the value of R2 to get the voltage drop:

.48amps x 15 ohms = 7.2V

8 0

3 years ago

A wave with a frequency of 60 Hz is traveling along a string whose linear mass density is 230 g/m and whose tension is 65 N. If

matrenka [14]

To develop this problem we will use the concepts related to Speed in a string that is governed by Tension (T) and linear density (µ)

$V = \sqrt{\frac{T}{\mu}}$

Our values are given as:

$f = 60Hz\\\mu = 230 g/m = 0.230kg/m\\T = 65N\\P = 75w$

Replacing we have that the velocity is

$V = \sqrt{\frac{T}{\mu}}$

$V = \sqrt{\frac{65}{0.230}}$

$V = 16.81m/s$

From the theory of wave propagation the average power wave is given as

$P =\frac{1}{2} \mu \omega^2 A^2 V$

Where,

A = Amplitude

$\omega = 2\pi f \rightarrow$ Angular velocity

$A^2 = \frac{2P}{\mu \omega^2 V}$

$A^2 = \frac{2P}{\mu (2\pi f)^2 V}$

Replacing,

$A^2 = \sqrt{\frac{2(75)}{(0.230)(2\pi 60)^2(16.81)}}$

$A = 0.0165m$

Therefore the amplitude of the wave should be 0.0165m

8 0

3 years ago

___meaning two' sugar (or saccharide) units in a chain.

ludmilkaskok [199]

The answer is c

Explanation:

3 0

2 years ago

A torque of 50 N*m acts on a wheel of moment of inertia 25 kg*m^2 for 4 s and then is removed.

motikmotik

You have to find the calculate<span> the circumference first then you can just multiply the diameter by π, which is about 3.142. That gives you the distance for each </span>revolution<span>. Then you can multiply by the </span>number of revolutions<span> per minute.

</span>

3 0

2 years ago

a car travelling at 50m/h on a horizontal highway (a) if the coefficient of static friction between road and tyres on a rainy da

aleksandrvk [35]

Hope this helps!

-Lilly

3 0

3 years ago