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BlackZzzverrR [31]
2 years ago
5

PLEASE HELP FOR PHYSICS!

Physics
1 answer:
Stolb23 [73]2 years ago
8 0

Hi there!

Recall Newton's Law of Universal Gravitation:

\large\boxed{F_g = G\frac{m_1m_2}{r^2}}

Where:

Fg = Force of gravity (N)

G = Gravitational Constant

m1, m2 = masses of objects (kg)

r = distance between objects (m)

Plug in the given values stated in the problem:

F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}

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Evaluate the expression a+b when a=23 and b=45 . Write in simplest form
Viefleur [7K]
If a = 23 and b = 45

a + b = ?

23 + 45 = 68
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2 years ago
5.8 meters/second + 1.2 meters/second
Alex Ar [27]

Answer:7m/s

Explanation:

5.8+1.2

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3 years ago
Read 2 more answers
A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune wi
USPshnik [31]

Answer:

a.3Hz

b.0.0034m

Explanation:

First, we know the flute is an open pipe, because open pipe as both end open and a close organ pipe as only one end close.

The formula relating the length and he frequency is giving as

f=\frac{nv}{2l}\\.

a.we first determine the length of the flute at the fundamental frequency i.e when <em>n</em>=1 and when the speed is in the 342m/s

Hence from

f=\frac{nv}{2l}\\\\l=\frac{342}{2*440}\\ l=0.389m\\.

since the value of the length will remain constant, we now use the value to determine the frequency when the air becomes hotter and the speed becomes 345m/s.

f=\frac{nv}{2l} \\f=\frac{345}{2*0.389}\\f=443.4Hz

Hence the require beat is

B=/f_{1}-f_{2}/\\B=/440-443/\\B=3Hz.

b. since the length is dependent also on the speed and frequency, we determine the new length when she plays with a fundamental frequency when the speed of sound is 345m/s

using the formula

L_{new}=\frac{v}{2f}\\\\L_{new}=\frac{345}{2*440}\\\\L_{new}=0.39204

Now to determine the extension,

L_{extend}=L_{new}-L_{old}\\L_{extend}=0.39204- 0.38864\\L_{extend}=0.0034m\\

4 0
3 years ago
A force, 10 N drags a mass 10 kg on a horizontal table with an acceleration of 0.2
kifflom [539]

Answer:

20&£+)##&843&()-_££-()&_2+0&&-£_!)

4 0
2 years ago
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal
galben [10]
Hello

1) First of all, since we know the radius of the wire (r=1.2~mm=0.0012~m), we can calculate its cross-sectional area
A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2

2)  Then, we can calculate the current density J inside the wire. Since we know the current, I=3~A, and the area calculated at the previous step, we have
J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity \sigma=3.6\cdot10^7~ \frac{A}{Vm} of the aluminium, the electric field is given by
E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m

4 0
3 years ago
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