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Rasek [7]
3 years ago
11

Newtons third law of motion describes how forces act in pairs true or false HELPPPP!!!!

Physics
2 answers:
marin [14]3 years ago
8 0

Answer:

True

Explanation:

Hope this helps! Add Brainliest

Elanso [62]3 years ago
5 0

Answer:

true

Explanation:

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It is very important that experiments are documented and<br> conducted with a procedure that can be?
Ksenya-84 [330]

Answer:

Repeated

Explanation:

An experiment must be done multiple times to prove its validity.

hope this helps- Cam ♡

5 0
3 years ago
Read 2 more answers
1. A rocket is fired vertically from the launch pad with a
zavuch27 [327]

Answer:

Max height= 36000 meters

Total Time = 120 seconds

Explanation:

0 = U - at

U = at

U= 20*60

U= 1200 m/s

MAX altitude would be

(U²Sin²tita)/2g

Max height= 1200² *( SIN90)²/(2*20)

Time of FLIGHT

2 * 1200/20

2400/20

120 sec onds

7 0
3 years ago
We know the frequency range of certain sounds are: 400-560 Hz, what are the ranges of wavelength in meters when the signal trans
Ksivusya [100]

Answer:

Range of wavelength will be 5.35\times 10^5m to 7.5\times 10^5m

Explanation:

We have given range of frequency is 400-560 Hz

Speed of the light c=3\times 10^8m/sec

We have to find the range of the wavelength of signal transmitted

Ween know that velocity is given by v=\lambda f, here \lambda is wavelength and f is frequency

So for 400 Hz frequency wavelength will be \lambda =\frac{c}{f}=\frac{3\times 10^8}{400}=7.5\times 10^5m

And wavelength for frequency 560 Hz \lambda =\frac{c}{f}=\frac{3\times 10^8}{560}=5.35\times 10^5m

So range of wavelength will be 5.35\times 10^5m to 7.5\times 10^5m

8 0
3 years ago
A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st
Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r

where,

\tau= shear stress at a distance 'r' from the center

T = is the applied torque

I_{p} = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

\tau _{max}=\frac{4}{3}\times \frac{V}{A}

Applying values we get

\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa

3 0
2 years ago
~~~~~NEED HELP ASAP~~~~~
Romashka-Z-Leto [24]

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

8 0
2 years ago
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