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Alex777 [14]
3 years ago
14

After running for a while, the fan becomes hot to the touch, and ever so often releases ___________________ energy into the air.

Physics
1 answer:
kykrilka [37]3 years ago
4 0

Answer:

thermal

Explanation:

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A machinist turns on the power on to a grinding wheel at time t= 0 s. The wheel accelerates uniformly from rest for 10 s and rea
Ugo [173]

Answer:

θt = 514.3 revolutions

Explanation:

(1)The wheel accelerates uniformly from rest for 10 s and reaches the operating angular speed of 58rad/s.

The uniformly accelerated circular movement  a circular path movement in which the angular acceleration is constant.

We apply the equations of circular motion uniformly accelerated

ωf = ω₀ + α*t  Formula (1)

θ = ω₀*t + (1/2)*α*t² Formula (2)

ωf² = ω₀² +2*α*θ Formula (3)

Where:

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

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

Number of revolutions made by the wheel from t = 0 to t = 10 s

Data

ω₀ = 0

t = 10 s

ωf = 58 rad/s

We replace data in the formula (1) to calculate α

ωf = ω₀ + α*t

58 = 0 + α*(10)

α = 58 /10

α = 5.8 rad/s²

We replace data in the formula (2) to calculate θ

θ = ω₀*t + (1/2)*α*t²

θ = 0 + (1/2)*( 5.8)*(10)²

θ₁ = 290 rad

(2)The wheel is run at that angular velocity for 30 s, and then power is shut off.

The movement of the wheel is circular with constant angular speed and the formula to calculate θ is:

θ = ω*t

ω = 58 rad/s  , t= 30s

θ = (58 rad/s)*(30)

θ = (58 rad/s)*(30)

θ ₂= 1740 rad

(3)The wheel slows down uniformly at 1.4 rad/s² until the wheel stops.

ω₀ = 58 rad/s

α = -1.4 rad/s²

ωf = 0

We replace data in the formula (3) to calculate θ

(ωf)² = (ω₀)² + (2)*(α )*θ

0 = (58)² + (2)*(-1.4)*θ

(2)*(1.4)*θ = (58)²

θ = (58)² / (2.8)

θ = (58)² / (2.8)

θ₃ = 1201.42 rad

Total number of revolutions made by the wheel (θt)

θt =θ₁+θ₂+θ₃

θt  = 290 rad+ 1740 rad + 1201.42 rad

θt  = 3231.42 rad

1 revolution = 2π rad

θt = 3231.42 rad* ( 1revolution/2π rad)

θt = 514.3 revolutions

7 0
3 years ago
Does the distance a person kicks a soccer ball
3241004551 [841]

Answer:

Independent variable: how far the soccer ball is kicked.

Dependent variable: how well the person does on their math test

Explanation:

The distance the ball is kicked is what the scientist can change. On the other hand, in this experiment, how well the person does on their test supposedly relies on how far the soccer ball is kicked.

5 0
3 years ago
Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the f
gulaghasi [49]

Answer:

A)3.8196 * 10^{9} Newtons        B) 2.153 * 10^{9} Newtons

Explanation:

Force = Pressure * Area.

Pressure is given as  120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = πR^{2}.

Area Expansivity β =    \frac{Change in Area}{Original Area * Temperature Rise}

Area Expansivity β = 2α

α =  16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = \frac{0.2m}{2} = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = \frac{0.15}{2} = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π * 0.1^{2} =  0.0314m^{2}

Area of Rod BC = π * 0.075^{2} = 0.0177m^{2}.

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9*10^{-6} * 0.0314 * 30 = 3.183 * 10^{-5}m^{2}.

For Rod BC we have = 2 * 16.9*10^{-6} * 0.0177 * 30 = 1.794∈-5m^{2}.

The forces on each rods will be given by.

Force AB = Pressure * Area  of Rod AB

                 = 120GPa * 3.183 * 10^{-5} gives 3.8196 * 10^{9} Newtons.

Force BC = Pressure * Area of Rod BC

                = 120GPa * 1.794 * 10^{-5} gives 2.153 * 10^{9} Newtons

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3 years ago
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The magnetic north pole changes location but the geographic north pole does not true or false
a_sh-v [17]
The North Magnetic Pole is the point on the surface of Earth's Northern Hemisphere at which the planet's magnetic field points vertically downwards (in other words, if a magnetic compass needle is allowed to rotate about a horizontal axis, it will point straight down). There is only one location where this occurs, near (but distinct from) the Geographic North Pole and the Geomagnetic North Pole. So yes true
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3 years ago
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