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german
4 years ago
12

A rock with a mass of 0.3 kg falls from the top of a cliff. If it takes the rock 2.5 s to reach the ground, what was the impulse

on the rock during the fall?
Physics
1 answer:
kaheart [24]4 years ago
4 0
Impulse=force x time
force=mass x acceleration due to gravity
force=
300 \times 10 = 3000
impulse =3000 x 2.5= ( sorry i don't have a calculator right now so you must calculate this yourself)
I converted from kg to g because it is the standard.
Hope this helps you.
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What two things does light provide?
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Answer:

Light provides brightness to see and also light carries energy

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3 years ago
The type of function that describes the amplitude of damped oscillatory motion is _______. The type of function that describes t
Salsk061 [2.6K]

Answer:

exponential

Explanation:

type of function that describes the amplitude of damped oscillatory motion is exponential because as we know that here function is

y = A × e^{\frac{-bt}{2m}}  × cos(ωt + ∅ )    ..................................... ( 1 )          

here function A × e^{\frac{-bt}{2m}}   is amplitude

as per equation ( 1 )it is exponential

so that we can say that amplitude of damped oscillatory motion is exponential

8 0
3 years ago
As a rubber band and moves forward, which of the following is true
Yuki888 [10]

Answer:

It can go back to it's original shape

Explanation:

7 0
3 years ago
A long coaxial cable (Fig. 2.26) carries a uniform volume charge density rho on the inner cylinder (radius a), and a uniform sur
Yuki888 [10]

Answer:

a) E = ρ / e0

b) E = ρ*a / (e0 * r)

c) E = 0

Explanation:

Because of the geometry, the electric field lines will all have a radial direction.

Using Gauss law

Q/e0 = \int \int E * dA

Using a Gaussian surface that is cylinder concentric to the cable, the side walls will have a flux of zero, because the electric field lines will be perpendicular. The round wall of the cylinder will have the electric field lines normal to it.

We can make this cylinder of different radii to evaluate the electric field at different points.

Then:

A = 2*π*r (area of cylinder per unit of length)

Q/e0 = 2*π*r*E

E = Q / (2*π*e0*r)

Where Q is the charge contained inside the cylinder.

Inside the cable core:

There is a uniform charge density ρ

Q(r) = ρ * 2*π*r

Then

E = ρ * 2*π*r / (2*π*e0*r)

E = ρ / e0 (electric field is constant inside the charged cylinder.

Between ther inner cilinder and the tube:

Q = ρ * 2*π*a

E = ρ * 2*π*a / (2*π*e0*r)

E = ρ*a / (e0 * r)

Outside the tube, the charges of the core cancel each other.

E=0

4 0
3 years ago
A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica
Evgen [1.6K]

This question is incomplete, the complete question is;

A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick. What is the maximum charge?

(The dielectric constant of mica is 5.4, and its dielectric strength is 1.00×10⁸ V/m)

Answer: the maximum charge q is 716.85 μF

Explanation:

Given data;

with = 3.0 cm = 0.03

breathe = 5.0 m

Area = 0.03 × 5 = 0.15 m²

dielectric strength E = 1.00 × 10⁸

∈₀ = 8.85 × 10⁻¹²

constant K = 5.4

maximum charge = ?

the capacitor C = KA∈₀ / d

q = cv so c = q/v

now

q/v = KA∈₀ / d

q =  vKA∈₀/d = EKA∈₀

we substitute

q = (1.00 × 10⁸) × 5.4 × 0.15 ×  8.85 × 10⁻¹²    

q = 716.85 × 10⁻⁶ F

q = 716.85 μF

the maximum charge q is 716.85 μF

7 0
3 years ago
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