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2 years ago

Why do the stars appear like pointed objects

Physics

1 answer:

saw5 [17]2 years ago

8 0

Answer:

because of where we are and our distance away from them, in reality they're giant balls of burning gas, like the sun. some stars are even bigger than the sun

Explanation:

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The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle

kozerog [31]

Answer:

0.191 s

Explanation:

The distance from the center of the cube to the upper corner is r = d/√2.

When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ. The new vertical distance from the center to the corner is r cos θ.

Sum of the torques:

∑τ = Iα

Fr cos θ = Iα

(k r sin θ) r cos θ = Iα

kr² sin θ cos θ = Iα

k (d²/2) sin θ cos θ = Iα

For a cube rotating about its center, I = ⅙ md².

k (d²/2) sin θ cos θ = ⅙ md² α

3k sin θ cos θ = mα

3/2 k sin(2θ) = mα

For small values of θ, sin θ ≈ θ.

3/2 k (2θ) = mα

α = (3k/m) θ

d²θ/dt² = (3k/m) θ

For this differential equation, the coefficient is the square of the angular frequency, ω².

ω² = 3k/m

ω = √(3k/m)

The period is:

T = 2π / ω

T = 2π √(m/(3k))

Given m = 2.50 kg and k = 900 N/m:

T = 2π √(2.50 kg / (3 × 900 N/m))

T = 0.191 s

The period is 0.191 seconds.

7 0

3 years ago

A student witnesses a flash of lightning and then t=2.5s later the student hears assiciated clap of thunder. (please show work)

frosja888 [35]

Answer:

857.5 m

2.8583×10⁻⁶ seconds

Explanation:

Time taken by the sound of the thunder to reach the student = 2.5 s

Speed of sound in air is 343 m/s

Speed of light is 3×10⁸ m/s

Distance travelled by the sound = Time taken by the sound × Speed of sound in air

⇒Distance travelled by the sound = 2.5×343 = 857.5 m

⇒Distance travelled by the sound = 857.5 m

Time taken by light = Distance the light travelled / Speed of light

$\text{Time taken by light}=\frac{857.5}{3\times 10^8}\\\Rightarrow \text{Time taken by light}=2.8583\times 10^{-6}$

Time taken by light = 2.8583×10⁻⁶ seconds

3 0

3 years ago

A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,

gavmur [86]

Answer:

a. ρ $_\beta=1.996J/m^3$

b. $U_E=9.445x10^{-15} J/m^3$

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

$i=14A$ , $d=2.5mm$ , $R=3.3$ Ω, $l=1 km$ , $E_o=8.85x10^{-12}F/m$ , $u_o=4*x10^{-7}H/m$

ρ $_\beta=\frac{\beta^2}{2*u_o}$

ρ $_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2$

ρ $_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}$

ρ $_\beta=1.996J/m^3$

b. The electric field can be find using the equation:

$U_E=\frac{1}{2}*E_o*E^2$

$E=(\frac{i*R}{l})^2$

$U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2$

$U_E=9.445x10^{-15} J/m^3$

4 0

3 years ago

Three resistors are wired in parallel with a battery. Two of the resistors have resistances of 38.7 Q/ and 89.5 Q. The current i

Lina20 [59]

Answer:

$214.9 \Omega$

Explanation:

The three resistors are connected in parallel: this means that the potential difference across each resistor is the same as the voltage of the battery. This can be calculated using the information about the $38.7 \Omega$ resistor: in fact, since we know its resistance and the current flowing through it (0.155 A), we can find the potential difference across this resistor, which is equal to the voltage of the battery:

$V=IR=(0.155 A)(38.7 \Omega)=6.0 V$

We also know the total current in the circuit, 0.250 A. This means that we can find the total resistance of the circuit, using Ohm's law:

$R_{eq}=\frac{V}{I}=\frac{6.0 V}{0.250 A}=24 \Omega$

So now we now the total resistance and the resistance of two of the 3 resistors; therefore, we can find the resistance of the 3rd resistor:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\frac{1}{R_3}=\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{24 \Omega}-\frac{1}{38.7\Omega}-\frac{1}{89.5\Omega}=0.00465 \Omega^{-1}\\R_3=\frac{1}{0.00465 \Omega^{-1}}=214.9 \Omega$

4 0

2 years ago

You are working in the finance department of Innotech Ltd (INT). The Company has spent $5 million

lozanna [386]

Answer: The Company has spent $5 million in research and development over the past 12 months developing cutting-edge battery technology which will be incorporated ...

Explanation: uhmmmmmm i dont know this one but it is pretty ez

5 0

3 years ago

Why do the stars appear like pointed objects​

Why do the stars appear like pointed objects