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Arturiano [62]
3 years ago
7

3. A ball thrown vertically upward returns to its starting point in 4s. Find its initial speed. [4]

Physics
1 answer:
Alinara [238K]3 years ago
6 0

Answer:

9.8 ×4 equal 39.2 m/s This is v intial

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Question 11 of 11 | Page 11 of 11
KiRa [710]

Answer:

Decreases the time period of revolution

Explanation:

The time period of Cygnus X-1 orbiting a massive star is 5.6 days.

The orbital velocity of a planet is given by the formula,

                                        v = √[GM/(R + h)]

In the case of rotational motion, v = (R +h)ω

                                         ω = √[GM/(R + h)] /(R +h)

Where 'ω' is the angular velocity of the planet

The time period of rotational motion is,

                                     T = 2π/ω

By substitution,

                                      <em>T = 2π(R +h)√[(R + h)/GM] </em>

Hence, from the above equation, if the mass of the star is greater, the gravitational force between them is greater. This would reduce the time period of revolution of the planet.

3 0
3 years ago
Pls help Which one of the following is NOT acceleration
ch4aika [34]

Answer:

C

Explanation:

Am increase in velocity?

3 0
2 years ago
Read 2 more answers
A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se
Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

v = \frac{Bqr}{m}

v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}

v = 661098.9 = 661099 m/s

(b)

\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}

e / m = 1.76 x 10^14 C / kg

(c) Let K be the kinetic energy

K = 0.5 x mv²

K = 0.5 x 9.1 x 10^-31 x 661099 x 661099

K = 1.99 x 10^-19 J

K = 1.24 eV

So, the potential difference is

V = 1.24 V

(d) if the acceleration voltage is doubled

V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J

Let v is the speed

K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

r=\frac{mv}{Bq}

r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}

r = 0.0113 m = 1.13 cm

7 0
3 years ago
A technique in which people use machines to learn how to control their bodies is known as __________.
Debora [2.8K]

Answer:

The technique in which people use machines to learn how to control their bodies is known as D, Biofeedback.

Explanation:

Biofeedback is a variety of different machines that help people learn how to control their bodies depending on their specific needs, varying from things like scalp sensors, electrocardiographs, electromyographs and more.

7 0
3 years ago
Read 2 more answers
14. The average speed of a car was 60 m/s by the time it reached the finish line. The car moved in a straight line and traveled
Leni [432]
7.5m. Have a good day :)
4 0
2 years ago
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