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2 years ago

If your mass is 63.7 kg, and you are standing 7.5 m away from a boulder with

Physics

1 answer:

murzikaleks [220]2 years ago

4 0

Answer:

Explanation:

F= $\frac{Gm1m2}{r^2}$

F=(6.67 x 10^-11) $\frac{63.7X9,750.6}{7.5^2}$

F=7.36502254e-7

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a car takes 125 ft to brake from 60 to 0 mph. Assume that the acceleration of the Prius is constant while braking. Find how long

max2010maxim [7]

Answer:

2.84 seconds

Explanation:

t = ?

distance = 125

Velocity origianal = 60 m/hr = 88 ft/s

AVERAGE velocity = 88/2 = 44 ft/s

44 t = 125

t = 125/44 = 2.84 s

7 0

2 years ago

What is the name of the scale used to measure hurricanes

shusha [124]

Yea it’s called the Saffir-Simpson Hurricane scale, made in 1960s and further developed in 1970s

8 0

3 years ago

The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t

ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

$r=\frac{mv}{qB}$ (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

$\omega=\frac{v}{r}$

Then, you solve the equation (1) in order to obtain v/r:

$\frac{v}{r}=\omega=\frac{qB}{m}$

Finally, you replace the values of the parameters:

$\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}$

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

$f=2\pi \omega=5.5*10^7Hz$

5 0

3 years ago

A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica

Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed $\omega =40rad/s$

mass of $m_2=0.8 kg$ dropped at r=0.3 m from center

let $\omega _2$ be the final angular velocity of cylinder

Conserving Angular momentum

$L_1=L_2$

$\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2$

$\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2$

$26.01\times 40=26.082\times \omega _2$

$\omega _2=39.88 rad/s$

3 0

3 years ago

What is the speed of a wave in (m/s) with a 5 meter wavelength and a period of 20 seconds?

arlik [135]

Answer: 0.25 m/s

Explanation: Speed = wavelengt · frequency

v = λf and frequency is 1/period f = 1/T

Then v = λ/T = 5 m / 20 s = 0.25 m/s

6 0

3 years ago