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nata0808 [166]
3 years ago
13

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa

l to 89.0N is applied to the rim of the wheel. The wheel has radius 0.100m . Starting from rest, the wheel has an angular speed of 12.2rev/s after 3.35s .
A
What is the moment of inertia of the wheel?
Express your answer with the appropriate units.
Physics
1 answer:
charle [14.2K]3 years ago
3 0
Find alpha from  ωf=ωi+αt
once you have it you can use: τ=Fr=89.0∗.1=Iα
to find I
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Write an Hypothesis whether land or water heats faster
Hatshy [7]

Answer:

Land heats faster

Explanation:

Land heats faster because absorbs much more solar radiation.

5 0
3 years ago
A person is standing on a scale in an unmoving elevator. The elevator starts to move upwards at 1 m/s squared. Is the scale read
timama [110]

Answer: GREATER

Explanation:when elevator does not move it reads weight of the person . when elevator moves up let apparent weight be F . W acts downwards so net force is F-W

HENCE

F-W =ma

F= ma+W

AS a= 1 m/s^2

F = m (1)+W

HENCE GREATER

7 0
3 years ago
A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun
Lera25 [3.4K]

Answer:

The distance is 0.53 m.

Explanation:

Given that,

Target distance = 100.0 m

Speed of bullet = 300 m/s

We need to calculate the total time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t=\dfrac{100.0}{300}

t=0.33\ sec

Now, consider vertical motion of bullet.

Initial velocity of bullet in vertical direction = 0 m/s

We need to calculate the vertically distance

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Put the value in the equation

s=0+\dfrac{1}{2}\times9.8\times(0.33)^2

s=0.53\ m

Hence, The distance is 0.53 m.

5 0
3 years ago
Please answer fast in hindi ​
Ipatiy [6.2K]

Answer:

1. Dheere Dheere (slowly slowly)

2. Har (every)

3. Kal (tomorrow)

4. Mat (don't)

5. Andar (inside)

sorry I wasn't able to write in hindi

8 0
2 years ago
A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00
gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:W=Q=Q_{in}-Q_{out} solving to Q_out we get:Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ) this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: n=1-\frac{T_{Low} }{T_{High}} where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:n=1-\frac{423.15}{723.15} =0.415=41.5%

5 0
3 years ago
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