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3 years ago

Mary cycled at an average speed of 8 km/h. How far has she traveled if she rides for 4 hours?

Physics

2 answers:

Sati [7]3 years ago

4 0

Answer:

Which sentence from the passage shows that the function of the river depicted here has carried through to modern times?

Explanation:

Which sentence from the passage shows that the function of the river depicted here has carried through to modern times?

Lana71 [14]3 years ago

3 0

If it’s 8km/h =8x4km/1x4h= 32km/4h

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What is the distance between two spheres, each with a charge of 2.5 x 10-6 c, when the force between them is 0.50 n?

lianna [129]

In this case, Coulomb's Law applies:

F = 1/(4πε₀) · (Q₁Q₂/r²)

You can solve it for r:

r = √[1/(4πε₀) · (Q₁Q₂/F<span>)]

Plugging in numbers:

r = </span>√[1/(4π·8.85×10⁻¹²) · (2.5×10⁻⁶)²/0.50]
= 0.335m

The correct answer is: the two charges are 0.335m apart.

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3 years ago

What does a star color tell you about the amount of energy it emits?

Rus_ich [418]

The whiter the color the more young it is and the more energy it emits

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3 years ago

Read 2 more answers

A 500kg elevator is raised to a height of 10 m in 10 seconds. Calculate the power of the motor.

Hunter-Best [27]

Answer:

Correct answer: Third statement P = 4900 W

Explanation:

Given:

m = 500 kg the mass of the elevator

h = 10 m reached height after t = 10 seconds

P = ? power of the motor

The formula for the calculating power of the motor is:

P = W / t

since work is a measure of change in this case of potential energy then it is:

W = ΔEp = Ep - 0 = Ep

In this case we must take g = 9.81 m/s²

Ep = m g h = 500 · 9.81 · 10 = 49,050 W ≈ 49,000 W

Ep ≈ 49,000 W

P = Ep / t = 49,000 / 10 = 4,900 W

P =4,900 W

God is with you!!!

3 0

3 years ago

A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and

alexira [117]

Answer:

A) $\omega_f=17.503\ rad.s^{-1}$

B) $t=55.6822\ s$

C) $\theta=1312\ rad$

Explanation:

Given:

mass of flywheel, $m=40\ kg$

diameter of flywheel, $d=0.72\ m$

rotational speed of flywheel, $N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}$

duration for which the power is off, $t_0=35\ s$

no. of revolutions made during the power is off, $\theta=180\times 2\pi=360\pi\ rad$

<u>Using equation of motion:</u>

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2$

$\alpha=-0.8463\ rad.s^{-2}$

Negative sign denotes deceleration.

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$\omega_f=15\pi-0.8463\times 35$

$\omega_f=17.503\ rad.s^{-1}$ is the angular velocity of the flywheel when the power comes back.

Here:

$\omega_f=0\ rad.s^{-1}$

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$0=15\pi-0.8463\times t$

$t=55.6822\ s$ is the time after which the flywheel stops.

Using the equation of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2$

$\theta=1312\ rad$ revolutions are made before stopping.

3 0

3 years ago

Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle

marusya05 [52]

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

$n_1sin\theta_1 = n_2sin\theta_2$

Where,

$\theta =$ Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

$n_1 sin\theta_1 = n_2sin\theta_2$

$(1.54) sin\theta_1 = (1.33)sin(90)$

$sin\theta_1 = \frac{1.33}{1.54}$

$\theta = sin^{-1}(\frac{1.33}{1.54})$

$\theta = 59.72\°$

Therefore the $\alpha_{max}$ would be equal to

$\alpha = 90\°-\theta$

$\alpha = 90-59.72$

$\alpha = 30.27\°$

Therefore the largest value of the angle α is 30.27°

3 0

3 years ago