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-BARSIC- [3]
3 years ago
14

1. In what direction does an object move when affected by an unbalanced force?

Physics
1 answer:
solniwko [45]3 years ago
6 0

Answer: .............................................................

Explanation:

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To what potential should you charge a 2.0 μF capacitor to store 1.0 J of energy?
Bess [88]
E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V

You should charge it to 1000 volts to store 1.0 J of energy.
6 0
3 years ago
Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl
tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s

For Train 2 Velocity of the Source,

v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s

Frequency of Source,

f_{s}=500\ Hz

To Find:

Frequency of Observer,

f_{o}=?  (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source v_{s} and observer v_{o}, the original frequency of the sound waves f_{s} and the observed frequency of the sound waves f_{o},

The Equation is

f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

7 0
3 years ago
A man-made satellite orbits the earth in a circular orbit that has a radius of 32000 km. The mass of the Earth is 5.97e 24 kg. W
Gwar [14]

Answer:

= 3521m/s

The tangential speed is approximately 3500 m/s.

Explanation:

F = m * v² ÷ r

Fg = (G * M * m) ÷ r²

(m v²) / r = (G * M * m) / r²

v² = (G * M) / r

v = √( G * M ÷ r)

G * M = 6.67 * 10⁻¹¹ * 5.97 * 10²⁴ = 3.98199 * 10¹⁴

r = 32000km = 32 * 10⁶ meters  

G * M / r = 3.98199 * 10¹⁴ ÷ 32 * 10⁶

v = √1.24 * 10⁷  

v = 3521.36m/s

The tangential speed is approximately 3500 m/s.

8 0
4 years ago
Read 2 more answers
How does wavelength of radio wave compare to the wavelength of a visible light wave?
Lerok [7]

Answer:

The wavelength of radio waves is much longer than the wavelength of visible light waves

Explanation:

Electromagnetic waves are periodic oscillations of the electric and the magnetic fields, occuring in a plane perpendicular to the direction of motion the wave.

Electromagnetic waves are the only waves able to travel in a vacuum; they travel at the speed of light:

c=3.0\cdot 10^8 m/s

Electromagnetic waves are also classified into 7 different types, according to their wavelength/frequency. From longest to shortest wavelength, we have: (with the corresponding wavelength written in the brackets):

Radio waves (>1 m)

Microwaves (1 mm - 1 m)

Infrared (750 nm - 1 mm)

Visible light (380 nm - 750 nm)

Ultraviolet (10 nm - 380 nm)

X-rays (0.01 nm - 10 nm)

Gamma rays (<0.01 nm)

So from the table, we can see that:

The wavelength of radio waves is much longer than the wavelength of visible light waves.

6 0
3 years ago
A proton is located at the point (x = 1.0 nm, y = 0.0 nm) and an electron is located at the point (x = 0.0 nm, y = 4.0 nm). Find
OLga [1]

Answer:

C : 1.4*10^(-11) N.

Explanation:

q_1 = q_2 = 1.6 * 10^(-19)

R^2 = (1)^2 + (4)^2 = 1.7 * 10^(-17) m^2

The coulomb's law is as follows:

F_e = k*q_1*q_2 / R^2

F_e = k*q^2 / R^2

F_e = (9.0*10^9) * (1.6 * 10^(-19))^2 / 1.7 * 10^(-17)

F_e = 1.35 * 10^(-11) N

Hence, answer closes to obtained is option C : 1.4*10^(-11) N.

7 0
3 years ago
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