All categories

3 years ago

An object of the same mass has three different weights at different times. Which statement is possible?

1 answer:

7 0

If the object is at different places in the universe for the weight measurements then that's entirely possible. It would "weigh" less on the moon than on the earth. It would weigh differently on the planets. And, It might even weigh differently at different places on the earth, though my recollections of this are very vague. All of this assumes that bits don't fall off the mass, to make it, well "less massive".

You might be interested in

Give v=120m/s north and t = 10.2, calculate the displacement

Sedbober [7]

Answer:

Displacement from the starting position is 1224m

Explanation:

The object is moving at 120 meters per second, it does this for 10.2 seconds. Use multiplication to find the answer.

$120 \times 10.2 = 1224$

4 0

3 years ago

I need help asap plzz

jekas [21]

Its c). cause they are separate variables

7 0

3 years ago

Read 2 more answers

An AC circuit with a total resistance of R has an rms current of I. The circuit wraps around an iron core transistor with 50 loo

zvonat [6]

Answer:

current is 13.88 A

Explanation:

given data

no of loops = 50 loops

resistance = 40 Ω

R = 15.0 Ω

I = 5.0 amp

N = 18

to find out

rms current

solution

we will apply here transformed principle that is

v(s) / v(p) = I(p) / I(s) = N(s) / N(p) ....................1

here p is primary and s is secondary

put here all these value

5 / I(s) = 18 / 50

I(s) = 250 /18

I(s) = 13.88

so current is 13.88 A

6 0

4 years ago

What is the acceleration of a 5 kg box if 20 N of force is applied to it?

Alex787 [66]

Answer:

4 m/s²

Explanation:

Acceleration = Net Force/mass

A = $\frac{20 N}{5 kg}$ = $\frac{4 N}{kg}$ = $\frac{4 kg x m /s^{2} }{kg}$ = 4 m/s²

20 Newtons divided by 5 kg equals 4 N/kg. Newtons can also be written as kg·m/s², so the simplest units would be written as m/s².

3 0

3 years ago

The Andromeda galaxy is moving toward us at 3.01x10^5 m/s. By how much will the frequency of the 6.17x10^14 Hz hydrogen line be

just olya [345]

Answer:

f - f '= 6.18*10^11 Hz

Explanation:

The change in frequency is given by:

$\frac{1}{f'}=\frac{1}{f}\frac{\sqrt{1+v/c}}{\sqrt{1-v/c}}$

f': observed frequency

f: source frequency = 6.17*10^14 Hz

v: speed of the source = 3.01*10^55 m/s

c: speed of light = 3*10^8 m/s

By replacing all these values you obtain:

$\frac{1}{f'}=\frac{1}{6.17x10^{14} Hz}\frac{\sqrt{1+(3.01*10^5m/s)(3*10^8m/s)}}{\sqrt{1-(3.01*10^5m/s)(3*10^8m/s)}}=1.62*10^{-15}\\\\f'=6.16*10^{14}Hz$

hence, the change in frequency f-f' will be:

f - f '= 6.18*10^11 Hz

4 0

4 years ago