15 points! Answer FAST!

With Brainly, what happens if no one answers your question? Do you get points back?

Gre4nikov [31]

Explanation: When nobody answers your question, you cannot get your points back. Even if the moderators delete it. I'm sorry.

Hope this helps! :D

-TanqR

(Please mark me as brainliest)

4 0

2 years ago

Read 2 more answers

A question to think about on units: Suppose we wanted to exchange scientific information with a newly discovered species of inte

lubasha [3.4K]

Answer:

b. The speed of light is 2.99792458 x 10^8 meters/second.

Explanation:

Speed of light is a universal constant and its value is same throughout the universe . So alien living near Alpha Centauri will quickly understand about it . But other statements are not universal . Mass of electron can vary as per relativistic formula of Einstein . Similarly , mass of proton can also vary according to relativistic concept . It depends upon the velocity of particle . So, the ratio of mass of proton and mass of electron will also vary from one star to another .

7 0

2 years ago

A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.4 m/s and returns the shot with the ball

lora16 [44]

Explanation:

It is given that,

Mass of the ball, m = 0.06 kg

Initial speed of the ball, u = 50.4 m/s

Final speed of the ball, v = -37 m/s (As it returns)

(a) Let J is the magnitude of the impulse delivered to the ball by the racket. It can be calculated as the change in momentum as :

$J=m(v-u)$

$J=0.06\times (-37-(50.4))$

J = -5.24 kg-m/s

(b) Let W is the work done by the racket on the ball. It can be calculated as the change in kinetic energy of the object.

$W=\dfrac{1}{2}m(v^2-u^2)$

$W=\dfrac{1}{2}\times 0.06\times ((-37)^2-(50.4)^2)$

W = -35.1348 Joules

Hence, this is the required solution.

6 0

3 years ago

A playground merry-go-round of radius R = 1.40 m has a moment of inertia I = 265 kg · m2 and is rotating at 11.0 rev/min about a

Tom [10]

Answer:

The value of new value of angular speed of merry go round. $\omega_{2}$ = 0.96 $\frac{rad}{sec}$

Explanation:

Given data

r = 1.4 m

Moment of inertia $I_{1}$ = 265 kg - $m^{2}$

$N_{1} =$ 11 RPM

$\omega_{1} = \frac{2 \pi N}{60}$

$\omega_{1} = \frac{2 \pi (11)}{60}$

$\omega_{1}$ = 1.15 $\frac{rad}{sec}$

From conservation of momentum principal

$I_{1} \omega_{1} = I_{2} \omega_{2}$ ------- (1)

$I_{2} = m r^{2} + 265$

$I_{2} = 27 (1.4)^{2} + 265$

$I_{2} = 317.92 \ kg m^{2}$

Put all the values in equation (1)

265 × 1.15 = 317.92 × $\omega_{2}$

$\omega_{2}$ = 0.96 $\frac{rad}{sec}$

This is the value of new value of angular speed of merry go round.

6 0

3 years ago

If the diameter of a radar dish is doubled, what happens to its resolving power assuming that all other factors remain unchanged

kirill115 [55]

Answer:

θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

Explanation:

The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.

The first minimum occurs for m = 1, so the diffraction equation of a slit remains

a sin θ = λ

in general, the diffraction patterns occur at very small angles, so

sin θ = θ

θ = λ / a

in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.

θ = 1.22 λ /a

In this exercise we are told that the opening changes

a’ = 2 a

we substitute

θ ‘= 1.22 λ / 2a

θ' = (1.22 λ / a) 1/2

θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

8 0

3 years ago