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3 years ago

Read this excerpt from Through the Looking-Glass by Lewis Carroll.

Physics

2 answers:

balandron [24]3 years ago

7 0

Answer:

Explanation:

The answer is A. :)

saul85 [17]3 years ago

6 0

Answer:

The Correct Answer Is A

Explanation:

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a missile is moving 1810 m/s at a 20.0. It needs to hit a target 29,500m away in a 65.0 magnitude in 9.20s. What acceleration mu

Vesna [10]

Answer:

323 m/s²

Explanation:

Given:

x₀ = 0 m

y₀ = 0 m

x = 29500 cos 65°

y = 29500 sin 65°

v₀x = 1810 cos 20°

v₀y = 1810 sin 20°

t = 9.20

Find:

ax, ay, θ

First, in the x direction:

x = x₀ + v₀ t + ½ at²

29500 cos 32° = 0 + (1810 cos 20°) (9.20) + ½ ax (9.20)²

25017 = 15648 + 42.32 ax

ax ≈ 221.4

And in the y direction:

y = y₀ + v₀ t + ½ at²

29500 sin 32° = 0 + (1810 sin 20°) (9.20) + ½ ay (9.20)²

15633 = 5695 + 42.32 ay

ay ≈ 234.8

Therefore, the magnitude of the acceleration is:

a² = ax² + ay²

a² = (221.4)² + (234.8)²

a ≈ 322.7

Rounded to 3 significant figures, the magnitude of the acceleration is approximately 323 m/s².

7 0

3 years ago

A 60.watt light bulb runs for 5.0 seconds. How much energy does it use?

VikaD [51]

"watt" means "Joule of energy per second"

"60 watts" means "60 Joules per second"

(60 joules per second) x (5 seconds) = <em>300 Joules of energy</em>

5 0

4 years ago

Check the correctness of formula t=2π √m/k, dimensionally

pantera1 [17]

Hi there! Lets see!

m is mass, and its units are kg
k is the elastic constant measured in newtons per meter (N/m), or kilograms per second squared kg/s²

Therefore:

$\sqrt{\dfrac{m}{k}} =\sqrt{\dfrac{[kg]}{[\dfrac{kg}{s^2}]}} =\sqrt{\dfrac{[kg]}{[kg]}\cdot s^2} = \sqrt{[s]^2} = s$

The period is given in seconds so the formula is dimensionally correct.

4 0

3 years ago

Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determin

tia_tia [17]

Answer with Explanation:

We are given that

A=3i-3j m

B=i-4 j m

C=-2i+5j m

a. $D=A+B+C$

$D=3i-3j+i-4j-2i+5j$

$D=2i-2j$

Compare with the vector r=xi+yj

We get x=2 and y=-2

Magnitude= $\mid D\mid=\sqrt{x^2+y^2}=\sqrt{(2)^2+(-2)^2}=2\sqrt 2$ units

By using the formula $\mid r\mid=\sqrt{x^2+y^2}$

Direction: $\theta=tan^{-1}\frac{y}{x}$

By using the formula

Direction of D: $\theta=tan^{-1}(\frac{-2}{2})=tan^{-1}(-1)=tan^{-1}(-tan45^{\circ})=-45^{\circ}$

b.E=-A-B+C

$E=-3i+3j-i+4j-2i+5j$

$E=-6i+12j$

$\mid E\mid=\sqrt{(-6)^2+(12)^2}=13.4$ units

Direction of E= $\theta=tan^{-1}(\frac{12}{-6}=tan^{-1}(-2)=-63.4^{\circ}$

4 0

3 years ago

Three resistors having resistances of LaTeX: 4.0\;\Omega4.0 Ω, LaTeX: 6.0\;\Omega6.0 Ω, and LaTeX: 10.0\;\Omega10.0 Ω are connec

aalyn [17]

Answer:

Explanation:

Given that three resistor of resistance

R1 = 4 Ω

R2 = 6 Ω

R3 = 10 Ω

The resistor are connected in parallel, the equivalent resistance is

1 / Req = 1 / R1 + 1 / R2 + 1 / R3

Rearranging this we have

Req = R1•R2•R3 / (R2•R3 + R1•R3 + R1•R2)

Req = 4 × 6 × 10 / (6×10 + 4 × 10 + 4 × 6)

Req = 240 / (60 + 40 + 24)

Req = 240 / 124

Req = 1.94 Ω

The parallel connection is connected in series with a battery and a resistor.

Resistor resistance is

r = 2 Ω

Supply voltage of the battery is 12V

V = 12V

Let find the current flowing in the circuit.

V = I(Req + r)

I = V / (Req + r)

I = 12 / (1.94 + 2)

I = 12 / 3.94

I = 3.05A

So, this is the current flowing in the circuit and It is the same current that will be shared by the parallel resistance.

We can calculated the voltage across the parallel resistance

From ohms las

V = iR

V = I × Req

V = 3.05 × 1.94

V = 5.92 V.

Then, this the voltage across each parallel resistor.

Then, to know the current in the 10ohms resistance

V = iR

I = V / R3.

I = 5.92 / 10

I = 0.592 A.

The current in the 10 ohms resistor is 0.59A

5 0

3 years ago