Ask question

Ask question

All categories

2 years ago

14

Let the period of evolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star

, its period of revolution will be √8 T.

1 answer:

CaHeK987 [17]2 years ago

7 0

Explanation:

8T BR 45 Yeuslilshshbe

You might be interested in

Big Al stands on a skateboard at a rest and throws a 0.5-kilogram rock at a velocity of 10.0 m /sec. Big Al moves back at 0.05 m

Vlada [557]

Let
his and skateboard's combined mass is x
we know
F1=-F2
0.5kg*10m/s=-0.05m/s xkg
5=0.05x(minus cancel with the m/s as it represented the opposite direction of velocity and now there is no velocity in this equation.. so minus is avoidable. and the kgs cancel out)
x=10
so their combined mass is a 100 kg.. (I hope I didn't mess thing up for you)

4 0

3 years ago

1.2 Define the following terms and in each case give the symbol and the unit: 1.2.1 wavelength (4) ·

Alex777 [14]

Wavelength is represented by lambda( $\lambda$ )

$\\ \bull\tt\longmapsto \lambda=\dfrac{c}{v}$

C is speed of light in air.
v is the frequency.

It has unit as meter(m)

8 0

2 years ago

What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed

tekilochka [14]

Answer:

charge will be equal to $2.03\times 10^{-5}C$

Explanation:

We have given mass of the particle m = 1.45 gram = 0.00145 kg

Acceleration due to gravity $g=9.8m/sec^2$

Electric field E = 700 N/C

Electric force will be equal to $F=qE$ , here q is charge and E is electric field

For particle to be stationary this force must be equal to force due to gravity , that is mg force

So qE = mg

$q=\frac{mg}{E}=\frac{0.00145\times 9.8}{700}=2.03\times 10^{-5}C$

So charge will be equal to $2.03\times 10^{-5}C$

8 0

2 years ago

What is the frequency of a photon with an energy of 4. 56 x 10^-19 j

Sauron [17]

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

$\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma = 6.88 \times 10^{14}\ s^{-1}$

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

Learn more about Frequency:

brainly.com/question/5102661

#SPJ4

5 0

1 year ago

Read 2 more answers

A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is

HACTEHA [7]

Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

Given;

mass of the person in cart, m₁ = 45 kg

mass of the cart, m₂ = 43 kg

acceleration due to gravity, g = 9.8 m/s²

final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

$mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m$

Therefore, the maximum vertical height the person in the cart can reach is 18.42 m

5 0

2 years ago