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3 years ago

BRAINLIESTTTT!!!!! THIS IS DUE IN 5 MINUTES!

Physics

1 answer:

Gnesinka [82]3 years ago

8 0

Equation: -10 40/4^2

Explanation: I multiplied the -10, but idk if that' what you do cause I've never done this. So if you do do that it's -1000, but that is probably wrong. Sorry, I tried.

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sing a rope that will snap if the tension in it exceeds 361 N, you need to lower a bundle of old roofing material weighing 455 N

yaroslaw [1]

Answer:

a)-2m/s^2

b)27.2m/s

Explanation:

Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)

W=455N=weight

W=mg

W=455N=weight

$m=\frac{W}{g} =\frac{455}{9.81}=46.38kg$

The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration

$a=\frac{T-W}{m} =\frac{361-455}{46.38kg} =-2m/s^2$

for point b we use the equations of motion with constant acceleration to find the velocity

$Vf=\sqrt{X(2)(a)+Vo^2}$

Where

Vf = final speed

Vo = Initial speed =0

A = acceleration =2m/s

X = displacement =6.8m

Solving

$Vf=\sqrt{(6.8)(2)(2)+0^2}=27.2m/s$

4 0

3 years ago

How is reflection of light used in research

mel-nik [20]

In order to read the publications of his peers, or read his own notes of the work
that he did on the previous day, or find his coffee mug on his desk in the lab, the
research scientist must arrange to have each of them illuminated with visible
wavelengths of light, and then he must catch the light reflected from each of them
with his eyes.

7 0

3 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago

A wave has a frequency of 875 hz and a wavelength of 352 m. At what speed is this wave traveling

Sonbull [250]

Answer:

308,000 or 30.8×10^3

Explanation:

v=f×lamda

v is ?, f is 875Hz, lamda is 352m

v=875×352

v=308,000

v=30.8×10^3 m/s

5 0

3 years ago

What is the maximum rated voltage for a 100 ohm 1/4 watt resistor?

Debora [2.8K]

This can be solve using the formula P = I^2 * Rwhere P is the powerI is the CurrentR is the resistanceP = I^2 * R
1/4 Watt = I^2 * 100 ohm solve for II^2 = 1/400 I = 0.05 amps then using the formula to solve for the voltage:V = I * RV = 0.05 amps * 100 ohms V = 5 volts

3 0

3 years ago

BRAINLIESTTTT!!!!! THIS IS DUE IN 5 MINUTES!​

BRAINLIESTTTT!!!!! THIS IS DUE IN 5 MINUTES!