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GenaCL600 [577]
3 years ago
11

What is the value of x in the equation 8+ 4 = 2 (x-1)

Physics
2 answers:
liraira [26]3 years ago
6 0

Answer: x = 7

Explanation: 12 = 14 -2 = 12.

mylen [45]3 years ago
5 0

Answer:

mgi

Explanation:

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Ethan made a diagram to compare examples of the first and second laws of thermodynamics. What belongs in the areas marked X and
bazaltina [42]

Answer:

The answer is X: Thermal energy is converted to light energy

Y: A cold spoon placed in hot liquid gets warmer

Explanation:

I took the quiz

4 0
3 years ago
A hockey ball accelerates from 0.m/s to 25m/s in 0.05 seconds what is the acceleration of the ball ?
S_A_V [24]

Answer:

500

Explanation:

25/0.05 .

I think this ?

3 0
3 years ago
The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give
Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

4 0
3 years ago
Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor
rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

e_{1}=\frac{E}{4\pi r^{2}}

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}

Hence we sense it as 4 times fainter than the nearer star.

5 0
3 years ago
11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0
Greeley [361]

Answer:

Explanation:

position of centre of mass of door from surface of water

= 10 + 1.1 / 2

= 10.55 m

Pressure on centre of mass

atmospheric pressure + pressure due to water column

10 ⁵ + hdg

= 10⁵ + 10.55 x 1000 x 9.8

= 2.0339 x 10⁵ Pa

the net force acting on the door (normal to its surface)

= pressure at the centre x area of the door

= .9 x 1.1 x 2.0339 x 10⁵

= 2.01356 x 10⁵ N

pressure centre will be at 10.55 m below the surface.

When the car is filled with air or  it is filled with water , in both the cases pressure centre will lie at the centre of the car .

7 0
3 years ago
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