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3 years ago

Dimensionally, which of the following could be a velocity? *

Physics

1 answer:

astra-53 [7]3 years ago

5 0

Answer:

5 meters per second

Explanation:

5m is the distance

5m west is the vector

5m per second is the velocity

5m per second west is unknown

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A. Obtain the following: microwave, ruler, something meltable (e.g. candy bar, marshmallows) B. A microwave works by setting up

Makovka662 [10]

Answer:

E = 124.7 N / C

Explanation:

Let's analyze the exercise: the microwave creates an electromagnetic wave of frequency F = 2.45 GHz, this wave is introduced into the microwave cavity and is reflected on the metal walls, which is why one or more standing waves are formed.

The electric field of the standing wave is

I = E²

E =√I

where I is the intensity of the radiation.

What is it

I = P / A

where P is the effective emission power, almost all the power of the microwave and A is the area of the cavity, in the most used microwaves

P = 700 W and the area is A = 25 x 18 cm² = 0.045 m²

I = 700 / 0.045

I = 15555.56 W/m²

let's calculate the electric field

E = √15555.56

E = 124.7 N / C

7 0

3 years ago

NEED ANSWER ASAP WILL MARK BRAINLIEST

lukranit [14]

Speed=30 m/s - 1.5 m/s = 28.5 m/s forward

7 0

3 years ago

Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could

Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

$\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}$

Where:

$d=17.91 mm =17.91(10)^{-3} m$ is the diameter of a dime

$D$ is the diameter of the Sun

$x_{sun-pinhole}=150,000,000 km=1.5(10)^{11} m$ is the distance between the Sun and the pinhole

$x_{sunball-pinhole}=100 d=1.791 m$ is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding $D$ :

$D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}$

$D=\frac{17.91(10)^{-3} m}{1.791 m} 1.5(10)^{11} m$

$D=1.5(10)^{9} m$ This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

$1 AU=149,597,870,700 m$

So, we have to divide this distance between $D$ in order to find how many suns could it fit in this distance:

$\frac{149,597,870,700 m}{1.5(10)^{9} m}=99.73 suns \approx 100 suns$

8 0

3 years ago

Using the data table to the right, determine the

andreyandreev [35.5K]

Silver sable is in spiderman lol

6 0

3 years ago

Read 2 more answers

A 1500 kg car skids to a halt on a wet road where μk = 0.47. You may want to review (Pages 141 - 145) . Part A How fast was the

shusha [124]

The car travels at a speed of 25m/s.

<u>Explanation:</u>

Given-

Mass, m = 1500kg

Coefficient of friction, μk = 0.47

Distance, x = 68m

Speed, s = ?

We know,

$Force, F = ma$

and

F = μ X m X g

Therefore,

μ * m * g = m * a

μ * g = a

Let, g = 9.8m/s²

So,

$a = 0.47 * 9.8 m/s^2$

$a = 4.606m/s^2$

We know,

$v^2 - u^2 = 2as$

where, v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

If the car comes to rest, the final velocity, v becomes 0.

So,

$u^2 = 2 * 4.606 * 68\\\\u^2 = 626.416m/s\\\\u = 25m/s$

The car travels at a speed of 25m/s.

6 0

3 years ago