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4 years ago

Would you measure the height of a building in meters? give reasons for your answer

1 answer:

5 0

No because, if we measure the height of a building in 'metres' it will be difficult for us as we will have a big number. Calculating the height in feet is easier.

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The electrical loads in _____ circuits each have the same voltage drop, with equals the total applied voltage of the circuit.

Vesna [10]

Answer:

The electrical loads in parallel circuits each have the same voltage drop, with equals the total applied voltage of the circuit.

Explanation:

I did some research and the voltage drop across any branch of a parallel circuit is the same as the applied voltage.

3 0

3 years ago

Describe a vibration that is not periodic. NO LINKS PLEASE

Paraphin [41]

Answer:

1)The position change of almost any manually operated room light switch.

2) Sunlight striking a point on the ground on a partly cloudy and windy day

Explanation:

4 0

3 years ago

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

$\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}$

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

6 0

4 years ago

A baseball is thrown by the center fielder (from shoulder level) to home plate where it is caught (on the fly at eye level) by t

marishachu [46]

At the highest point of the trajectory the vertical component will have its zero velocity, and the descent caused by the force of gravity will begin.

Since the ball is thrown with a certain speed, the vertical component reaches its highest point (upwards), until returning to the receiver who will receive the ball with the same vertical component but in the opposite direction (downwards).

Therefore the vertical component will have its highest value at launch.

8 0

3 years ago

Practice Problems

vodomira [7]

(a) The acceleration of the salt shaker is 1.18 m/s².

(b) The distance traveled by the baseball player before coming to rest is 204.1 m.

<h3>Acceleration of the salt shaker</h3>

The acceleration of the salt shaker at the given coefficient of kinetic friction is determined as follows;

a = μg

a = 0.12 x 9.8

a = 1.18 m/s²

Acceleration of the baseball player is calculated as follows;

a = μg

a = 0.4 x 9.8

a = 3.92 m/s²

<h3>Distance traveled by the baseball player</h3>

The distance traveled by the baseball player before coming to rest is calculated as follows;

v² = u² - 2as

0 = 40² - 2(3.92)s

0 = 1600 - 7.84s

7.84s = 1600

s = 204.1 m

The complete question is below:

A baseball player slides into third base with an initial speed of 40 m/s. If the coefficient of kinetic friction between the player and the ground is 0.40, how far does the player slide before coming to rest?

Learn more about coefficient of friction here: brainly.com/question/20241845

8 0

3 years ago