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snow_tiger [21]
3 years ago
6

What do I say when commenting on my result?

Physics
1 answer:
nevsk [136]3 years ago
8 0
Result for what exactly?
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2 cycleists, 90 miles apart starting riding toward each other at some times one cycles twice as fast as the other if they meet 2
svetlana [45]

Average speed of the first car= 15 miles/hr

Average speed of the second car= 30 miles/hr

Explanations:

distance= d= 90 miles

Time= 2 hrs

let speed of first cyclist= v

speed of second cyclist= 2v

relative speed = v + 2v= 3v

using relative speed= Total distance/Total time

3v= 90/2

3 v=45

v=15 miles/hr

so speed of first cyclist= 15 miles/hr

speed of second cyclist= 2 (15)= 30 miles/hr

6 0
3 years ago
Everyone i need help so much
Veronika [31]

Answer:

b

Explanation:

3 0
2 years ago
Read 2 more answers
The drill used by most dentists today is powered by a small air-turbine that can operate at angular speeds of 350000 rpm. These
alexdok [17]

Answer:

θ  = 6.3 *10³ revolutions

Explanation:

Angular acceleration of the drill

We apply the equations of circular motion uniformly accelerated

ωf= ω₀ + α*t  Formula (1)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t :  time interval (s)

Data

ω₀ = 0

ωf = 350000 rpm = 350000 rev/min

1 rev = 2π rad

1 min= 60 s

ωf = 350000 rev/min =350000*(2π rad/60 s)

ωf = 36651.9 rad/s

t = 2.2 s

We replace data in the formula (2) :

ωf= ω₀ + α*t

36651.9 = 0 + α* (2.2)

α = 36651.9 / (2.2)

α = 17000 rad/s²

Revolutions made by the drill

We apply the equations of circular motion uniformly accelerated

ωf²= ω₀ ²+ 2α*θ Formula (2)

Where:  

θ : Angle that the body has rotated in a given time interval (rad)

We replace data in the formula (2):  

(ωf)²= ω₀²+ 2α*θ

(36651.9)²= (0)²+ 2( 17000 )*θ

θ = (36651.9)²/ (34000 )

θ  = 39510.64 rad = 39510.64 rad* (1 rev/2πrad)

θ  = 6288.31 revolutions

θ  = 6.3 *10³ revolutions

3 0
3 years ago
After hitting the spring, the block is bounced back up the ramp. The maximum compression of the spring is Δx=0.03m, and the spri
Lyrx [107]

Answer:

v = 1.28 m/s

Explanation:

Given that,

Maximum compression of the spring, \Delta x=0.03\ m

Spring constant, k = 800 N/m

Mass of the block, m = 0.2 kg

To find,

The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.

Solution,

When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,

Initial energy = Final energy

\dfrac{1}{2}kx^2=mgh+\dfrac{1}{2}mv^2

Substituting all the values in above equation,

\dfrac{1}{2}\times 800\times 0.03^2=0.2\times 9.8\times 0.1+\dfrac{1}{2}\times 0.2\times v^2

v = 1.28 m/s

Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.

5 0
2 years ago
An ideal gas is enclosed in a piston, and 1600 J of work is done on the gas. As this happens, the internal energy of the gas inc
Phantasy [73]

Answer:

- 1100 J heat flows out

Explanation:

dW = - 1600 J (as work is done on the gas)

dU = 500 J

dQ = ?

According to the first law of thermodynamics

dQ = dU + dW

dQ = 500 - 1600

dQ = - 1100 J

As heat is negative so it flows out.

3 0
3 years ago
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