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nlexa [21]
3 years ago
12

As the energy of a wave increases its ________ increase

Physics
1 answer:
weqwewe [10]3 years ago
6 0
First one is B, second one is A
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Sam lifted his backpack with 5 Newtons of force a total of 400
irga5000 [103]

2000joules

Explanation:

work done=force×meters

4 0
2 years ago
A sanding disk with rotational inertia 1.2 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of 10 N·m a
tensa zangetsu [6.8K]

Answer:

Angular momentum = 0.7 kg.m²/s

Angular velocity = 583.3 rad/s

Explanation:

1. The torque τ  is related to the angular momentum L by the relation

τ = ΔL/Δt

ΔL = τΔt

τ = 10 N. m

Δt = 70 ms = 70 × 10⁻³s

ΔL = (10 N. m) × (70 × 10⁻³s) = 700 × 10⁻³ kg.m²/s = 0.7 kg.m²/s

2. The rotational inertia I relates the angular momentum L to the angular velocity w

L = Iw

w = L/I

L =  0.7 kg.m²/s

I = 1.2 × 10⁻³ kg.m²

w = (0.7 kg.m²/s)/(1.2 × 10⁻³ kg.m²) = 583.3 rad/s

4 0
3 years ago
I need help plz!!! Please help...
RSB [31]

Answer: 60

Explanation:

4 0
2 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
2 years ago
A 30.0-μF capacitor is connected to a 49.0-Ω resistor and a generator whose rms output is 30.0 V at 60.0 Hz. (a) Find the rms
Natali5045456 [20]

Explanation:

Given that,

Capacitor = 30μC

Resistor = 49.0Ω

Voltage = 30.0 V

Frequency = 60.0 Hz

We need to calculate the impedance

Using formula of impedance

Z=\sqrt{R^2+X_{c}^2}.....(I)

We need to calculate the value of X_{c}

Using formula of X_{c}

X_{c}=\dfrac{1}{2\pi f c}

X_{c}=\dfrac{1}{2\times\pi\times60.0\times30\times10^{-6}}

X_{c}=88.42\ \Omega

Put the value of X_{c} into the formula of impedance

Z=\sqrt{(49.0)^2+(88.42)^2}

Z=101.08\ \Omega

(a). We need to calculate the rms current in the circuit

Using formula of rms current

I_{rms}=\dfrac{V}{Z}

I_{rms}=\dfrac{30.0}{101.08}

I_{rms}=0.30\ A

The rms current in the circuit is 0.30 A.

(b). We need to calculate the rms voltage drop across the resistor

Using formula of rms voltage

V_{rms}=I_{rms}\times R

Put the value into the formula

V_{rms}=0.30\times49.0

V_{rms}=14.7\ V

The rms voltage drop across the resistor is 14.7 V

(c). We need to calculate the rms voltage drop across the capacitor

Using formula of rms voltage

V_{rms}=I_{rms}\times X_{C}

V_{rms}=0.30\times88.42

V_{rms}=26.53\ V

The rms voltage drop across the capacitor is 26.53 V.

Hence, This is the required solution.

4 0
2 years ago
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