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3 years ago

Is this a scam/virus?

Physics

1 answer:

kaheart [24]3 years ago

6 0

Answer:

Yes, I'm pretty sure it is. That's why I don't click on it!

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Why mole is called fundamental unit.

gladu [14]

Explanation:

because it doesn't depend upon other unit like kg meter and second

4 0

3 years ago

A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 1.10 m. If her pisto

lapo4ka [179]

Answer:

The diameter of the piston of the players equals 55.136 cm.

Explanation:

from the principle of transmission of pressure in a hydraulic lift we have

$\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}$

Since the force in the question is the weight of the individuals thus upon putting the values in the above equation we get

$\frac{57.0\times 9.81}{\frac{\pi \times (19.0)^{2}}{4}}=\frac{4\times 120\times 9.81}{\frac{\pi \times D_{2}^{2}}{4}}$

Solving for $D_{2}$ we get

$D_{2}^{2}=\frac{4\times 120}{57}\times 19^{2}\\\\\therefore D_{2}=\sqrt{\frac{4\times 120}{57}}\times 19\\\\D_{2}=55.136cm$

5 0

3 years ago

An object carries a +15.5 µC charge. It is 0.525 m from a -7.25 µC charge. What is the magnitude of the electric force on the ob

lorasvet [3.4K]

Answer:

the force of attraction between the two charges is 3.55 N.

Explanation:

Given;

first charge carried by the object, q₁ = 15.5 µC

second charge carried by the q₂ = -7.25 µC

distance between the two charges, r = 0.525 m

The force of attraction between the two charges is calculated as;

$F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(15\times 10^{-6})(7.25\times 10^{-6})}{(0.525)^2} \\\\F = 3.55 \ N$

Therefore, the force of attraction between the two charges is 3.55 N.

3 0

3 years ago

Consider steady heat transfer between two large parallel plates at constant temperatures of T1 = 210 K and T2 = 150 K that are L

snow_lady [41]

Answer:

$Q=81.56\ W/m^2$

Explanation:

Given that

$T_1= 210 K$

$T_2= 150 K$

Emissivity of surfaces(∈) = 1

We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies

$Q=\sigma (T_1^4-T_2^4)\ W/m^2$

So now by putting the values

$Q=\sigma (T_1^4-T_2^4)\ W/m^2$

$Q=5.67\times 10^{-8}(210^4-150^4)\ W/m^2$

$Q=81.56\ W/m^2$

So rate of heat transfer per unit area

$Q=81.56\ W/m^2$

8 0

4 years ago

A hockey puck slides off the edge of a platform with an initial velocity of 20 m/s horizontally. The height of the platform abov

Rina8888 [55]

Answer:

20.96 m/s

Explanation:

Using the equations of motion

y = uᵧt + gt²/2

Since the puck slides off horizontally,

uᵧ = vertical component of the initial velocity of the puck = 0 m/s

y = vertical height of the platform = 2 m

g = 9.8 m/s²

t = time of flight of the puck = ?

2 = (0)(t) + 9.8 t²/2

4.9t² = 2

t = 0.639 s

For the horizontal component of the motion

x = uₓt + gt²/2

x = horizontal distance covered by the puck

uₓ = horizontal component of the initial velocity = 20 m/s

g = 0 m/s² as there's no acceleration component in the x-direction

t = 0.639 s

x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m

For the final velocity, we'll calculate the horizontal and vertical components

vₓ² = uₓ² + 2gx

g = 0 m/s²

vₓ = uₓ = 20 m/s

Vertical component

vᵧ² = uᵧ² + 2gy

vᵧ² = 0 + 2×9.8×2

vᵧ = 6.26 m/s

vₓ = 20 m/s, vᵧ = 6.26 m/s

Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s

4 0

3 years ago

Read 2 more answers