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goldfiish [28.3K]
3 years ago
10

Gravity is measured using?

Physics
1 answer:
vlada-n [284]3 years ago
4 0
Gravity is measured with an instrument called a Gravimeter. Gravity is expressed using a special unit called a Newton. Although both of these answers would be technically correct, I believe the answer to this question is Newtons.
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If y(t)y(t) describes the position with time, what is the proper formula for velocity with time? (Recall velocity is related to
PilotLPTM [1.2K]

Answer:

Although this question is not complete, I would give a general solution to this kind of problems.

If y(t) describes the position of a body with time such that

y(t) = at^(n) + bt^(m) + C

Then

V(t) = dy(t)/dt = ant^(n-1) + bmt^(m-1)

Explanation:

As an example supplies the position of a particle is given by

y(t) = 4t³- 3t² + 9

V(t) = 4x3t²- 3x2t¹

V(t) = d(t)/dt = 12t² - 6t.

Another example,

If y(t) = 15t³ - 2t² + 30t -80

V(t) = d(t)/dt = 15x3t² - 4t +30 = 45t² + 4t + 30.

Basically, in the equations above the powers of t reduces by one when computing the velocity function from y(t) by differentiation (calculating the derivative of y(t)). The constant term C (9 and 80 in the functions of y(t) in examples 1and 2 above) reduces to zero because the derivative of a constant (and ordinary number without the t attached to it) is always zero.

One last example,

y(t) = 2t^6 -3t²

V(t) = d(t)/dt = 12t^5 - 6t

6 0
4 years ago
Calculate the rate of heat generation in kW due to the burning of the fuel when you drive a car rated at 34 MPG (miles per gallo
alexira [117]

Answer:

Explanation:

Let us calculate gallon used in one hour .

It travels 70 miles in one hour

in 70 miles it uses 70 / 34 gallons of fuel

70 / 34 gallons = 70 / 34 x 3.7854 kg

= 7.8 kg

heat generated = 7.8 x 44 x 10⁶ J

= 343.2 x 10⁶ J

This is heat generated in one hour

heat generated in one second = 343.2 x 10⁶ / 60 x 60 J/s

= 95.33 x 10³ J /s

= 95.33 kW.

4 0
3 years ago
Neutrons incident on a heavy nucleus with spin J 0 show a resonance at an incident energy ER = 250 eV in the total cross-section
ivolga24 [154]

Answer:

elastic partial width is 2.49 eV

Explanation:

given data

ER  E = 250 eV

spin J = 0

cross-section magnitude σ = 1300 barns

peak P = 20ev

to find out

elastic partial width W

solution

we know here that

σ = λ²× W /  ( E × π × P )     ...................1

put here all value

σ = (0.286)² × W  × 10^{-16}/  ( 250 × π × 20 )

1300 × 10^{-24} = (0.286)² × W  × 10^{-16}/  ( 250 × π × 20 )

solve it and we get W

W = 249.56 × 10^{-2}

so elastic partial width is 2.49 eV

8 0
4 years ago
I know what dark matter is, but what exactly is dark energy and what does it do?
aksik [14]
If you know what dark matter is, my friend, then there are several thousand Astrophysicists and Cosmologists at universities around the world who would pay dearly for the opportunity to buy you a cup of coffee, and chat while you sip it.
7 0
4 years ago
A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di
Lyrx [107]

Answer:

D

Explanation:

From the information given:

The angular speed for the block \omega = 50 \ rad/s

Disk radius (r) = 0.2 m

The block Initial velocity is:

v = r \omega \\ \\  v = (0.2  \times 50) \\ \\  v= 10 \ m/s

Change in the block's angular speed is:

\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s

However, on the disk, moment of inertIa is:

I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2

The time t = 10s

∴

Frictional torques by the wall on the disk is:

T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10})  \\ \\ =0.6 \ N.m

Finally, the frictional force is calculated as:

F = \dfrac{T}r{}

F= \dfrac{0.6}{0.2} \\ \\ F = 3N

8 0
3 years ago
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