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kondaur [170]
3 years ago
9

Find the instantaneous velocity at 1 s . can anyone help with c-h!!!

Physics
1 answer:
Lelu [443]3 years ago
7 0
Rise over run at 1 second
It’s the same slope from 0 to 2 seconds
10/2=5mps
As a note all time points between 0and 2 will have this instantaneous velocity

Instantaneous velocity at time 2 is 0
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A very powerful vacuum cleaner which has a hose of circular cross section can lift a brick of mass 12 kg when the hose is placed
valkas [14]

Answer:

A). 1.9 cm

Explanation:

m = Mass of brick = 12 kg

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of hose

A = Area = \pi r^2

F = Force = mg

Let us assume that the pressure required to lift the brick would be atmospheric pressure

P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{F}{\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{F}{\pi P}}\\\Rightarrow r=\sqrt{\dfrac{12\times 9.81}{\pi\times 101325}}\\\Rightarrow r=0.01923\ m=1.9\ cm

The radius of the hose should be 1.9 cm

6 0
3 years ago
The Sun is expected to undergo hydrogen fusion for a total of _____ years. a million 10 million a billion 10 billion 100 billion
Elanso [62]
The correct answer is 10 billion years. The Sun is expected to undergo hydrogen fusion for a total of 10 billion years. The Sun generates its energy by nuclear fusion of hydrogen and produces helium nucleus. It fuses 620 million metric tons every second.
3 0
3 years ago
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water flows horizontally from a hose that is 3m above the ground. the water lands 7.2m to the right of the hose . how long does
Sidana [21]

Answer:

0.782 s

Explanation:

The water flows horizontally from the hose, so its initial vertical velocity is 0.

Given:

y₀ = 3 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 m = 3 m + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 0.782 s

Round as needed.

8 0
3 years ago
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Which of the following must be true about the object labeled X in the circuit below?
Alex787 [66]

the answer is c and if I help you  thank me
4 0
3 years ago
A 0.5 kg block of aluminum (caluminum=900j/kg⋅∘c) is heated to 200∘c. the block is then quickly placed in an insulated tub of co
Alex_Xolod [135]

To solve this problem, we should recall the law of conservation of energy. That is, the heat lost by the aluminium must be equal to the heat gained by the cold water. This is expressed in change in enthalpies therefore:

- ΔH aluminium = ΔH water

where ΔH = m Cp (T2 – T1)

The negative sign simply means heat is lost. Therefore we calculate for the mass of water (m):

- 0.5 (900) (20 – 200) = m (4186) (20 – 0)

m = 0.9675 kg

 

Using same mass of water and initial temperature, the final temperature T of a 1.0 kg aluminium block is:

- 1 (900) (T – 200) = 0.9675 (4186) (T – 0)

- 900 T + 180,000 = 4050 T

4950 T = 180,000

T = 36.36°C

 

The final temperature of the water and block is 36.36°C

4 0
4 years ago
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