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3 years ago

What is a medum and why do waves travel through them?

1 answer:

6 0

A medium is a substance or material that carries the wave. Sounds waves need a medium in order to travel through a wave to help it travel as it cannot travel through a vacuum. Mechanical waves need a medium to transport energy.

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A 0.005 kg bullet is travelling horizontally at a velocity of 250 m/s when it strikes a wooden block of mass 12 kg on a table at

Verdich [7]

Use the conservation of momentum formula: m1v1 + m2v2 = v(m1 + m2)

(0.005kg)(250m/s) + (12kg)(0m/s) = v(0.005kg + 12kg)

v = ((0.005kg)(250m/s)) / (0.005kg + 12kg)

v = 0.104 m/s

6 0

4 years ago

Name 5 machine principles

lidiya [134]

The Wheel & Axle. The wheel has always been considered a major invention in the history of mankind

The Inclined Plane

The Wedge

The Pulley

The Screw

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4 years ago

Homeboy Joe is riding his skateboard and playing Among Us. Because he's distracted he doesn't notice he's about to skate right o

Aleks [24]

Answer: He wasn’t skating he was on Ice ;)

3 0

4 years ago

You are working in cooperation with the Public Health department to design an electrostatic trap for particles from auto emissio

Mademuasel [1]

Answer:

Explanation:

Given that:

Charge (q) on the particle = 3 × 10⁻⁸ C

mass (m) of the particle = 6 × 10⁻⁹ kg

at a distance x = 15 cm , the velocity in the plate = 900 m/s²

For the square plate, the surface charged density σ = -8 × 10⁻⁶ C/m²

To start with calculating the electric field as a result of the square plate; we use the formula;

$E = \dfrac{\sigma }{2 \varepsilon_o}$

$E = \dfrac{8 \times 10^{-6} }{2 \times 8.85 \times 10^{-12}}$

$E = 4.51977 \times 10^5 \ V/m$

On the square plate; The electric force F = Eq

$F = (4.51977 \times 10^5 \ V/m )(3\times 10^{-8} \ C)$

$F = 1.3559 \times 10^{-2} \ N$

The acceleration $a =\dfrac{ F}{m}{$

$a = \dfrac{1.3559\times 10^{-2} \ N}{6 \times 10^{-9} \ Kg}$

$a = 2.25988 \times 10^6 \ m/s^2$

For the particle, the velocity at distance x = 7 m can be calculated by using the formula:

$(\dfrac{1}{2}) mv^2 = \Delta Vq$

$v^2 = \dfrac{2 Eq}{dm}$

$v^2 = \dfrac{2 * 4.51977 \times 10^5 \times 3 \times 10^{-8} }{0.07 \times 6\times 10^{-9} }$

$v^2 = 64568142.86 \ m/s$

$v =\sqrt{ 64568142.86 \ m/s}$

$\mathbf{v = 8.035 \times 10^3 \ m/s}$

From the calculation, we realize that the charge acting between the particle and the plate is said to be "opposite".

Hence, the force is an attractive force.

Similarly, there is a gradual increase exhibited by the velocity of the particle.

Therefore, the particles get to the detector, but the detector failed to get detect due to the velocity which is greater than 1000 m/s.

6 0

3 years ago

The wheel of a car has a radius of 20.0 cm. It initially rotates at 120 rpm. In the next minute it makes 90.0 revolutions. (a) W

just olya [345]

Answer:

Explanation:

Given that,

Radius of the wheel, r = 20 cm = 0.2 m

Initial speed of the wheel, $\omega_i=120\ rpm=753.98\ rad/s$

Displacement, $\theta=90\ rev=565.48\ rad$

To find,

The angular acceleration and the distance covered by the car.

Solution,

Let $\alpha$ is the angular acceleration of the car. Using equation of rotational kinematics as :

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

$565.48=753.98\times 60+\dfrac{1}{2}\alpha (60)^2$

$\alpha =-24.81\ rad/s^2$

Let t is the time taken by the car before coming to rest.

$t=\dfrac{\omega_f-\omega_i}{\alpha }$

$t=\dfrac{0-753.98}{-24.81}$

t = 30.39 seconds

Let v is the linear velocity of the car. So,

$v=r\times \omega_i$

$v=0.2\times 753.98$

v = 150.79 m/s

Let d is the distance covered by the car. It can be calculated as :

$d=v\times t$

$d=150.79\ m/s\times 30.39\ s$

d = 4582.5 meters

d = 4.58 km

5 0

3 years ago