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Vera_Pavlovna [14]
3 years ago
12

What is a overly-simplified definition of Einstein's theory of general relativity?​

Physics
1 answer:
Phoenix [80]3 years ago
6 0

Answer:

the laws of physics are the same for all non-accelerating observers

Explanation:

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11) A man is on a 1/4 on a bridge. A train is coming the same direction he is going. The man can run across the bridge in the sa
lesya692 [45]

15mph

If the man turns and runs toward point A, he will cover

3/8 of the length of the bridge in the time that it takes

the train to reach A.

If the man runs forward toward point B, what part of the bridge

will he cover before the train reaches A? Well, he will cover

3/8 of the bridge, only heading forward toward B. This will put

him 3/8 + 3/8 = 6/8 = 3/4 of the way across the bridge by the

time the train reaches A.

since we know that the man and the train will meet at B, this

means that in the time it takes the man to run the remaining

1/4 of the bridge, the train will cover the entire length of

the bridge.

If it takes the man the same time to cover 1/4 of the bridge

that it takes the train to cover the whole bridge, then the train

must be going four times as fast as the man. Another way of saying

this is that the man runs at 1/4 the speed of the train.  

Since the train's speed is known to be 60 mph, this means that

the man runs at (1/4) 60 = 15 mph.

7 0
3 years ago
When a funnel shaped cloud considered as a tornado?
gladu [14]
It has to be one continuous column of cloud (air) connected to the ground and in constant rotation.
3 0
3 years ago
A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the
11Alexandr11 [23.1K]

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\

<u>v_f = 10.85 m/s</u>

4 0
3 years ago
A spaceship is traveling through deep space towards a space station and needs to make a course correction to go around a nebula.
expeople1 [14]

Answer:

Magnitude = 3.64 × 10^6  

စ = 43.9°

Explanation:

given data

ship to travel = 1.7 × 10^6    kilometers

turn = 70°

travel an additional = 2.7 × 10^6   kilometers

solution

we will consider here

Px = 1.7 × 10^6  

Py = 0

Qx =2.7 × 10^6  cos(70)

Qy= 2.7 × 10^6  sin(70)

so that

Hx = Px + Qx    ............1

Hx = 2.62 × 10^6  

and

Hy = Py + Qy      ..........2

Hy = 2.53 × 10^6  

so Magnitude = \sqrt{((2.62 \times 10^6  )^2+(2.53 \times 10^6)^2)}

Magnitude = 3.64 ×  

so direction  will be

tan စ = Hy ÷ Hx    ......................3

tan စ  = \frac{2.53}{2.62}

tan စ  = 0.9656

စ = 43.9°

3 0
3 years ago
A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o
Crank

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

\delta = \frac{PL}{AE}

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

P = 68*10^3 N

E = 200GPa  

A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2

Replacing we have,

\delta = \frac{PL}{AE}

\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}

\delta = 0.001932m

\delta = 1.93mm

Therefore the change in length is 1.93mm

7 0
3 years ago
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