False. The force comes first as in tapping or hitting a balloon. Then comes the motion and that might be the balloon dating in the direction it was shoved.
Answer:
The constant torque required to stop the disk is 8.6 N-m in clockwise direction .
Explanation:
Let counterclockwise be positive direction and clockwise be negative direction .
Given
Radius of disk , r = 1.33 m
Mass of disc , m = 70.6 kg
Initial angular velocity , ![\omega_i =217 rpm](https://tex.z-dn.net/?f=%5Comega_i%20%3D217%20rpm)
Final angular velocity , ![\omega_f =0\, rpm](https://tex.z-dn.net/?f=%5Comega_f%20%3D0%5C%2C%20rpm)
Time taken to stop , t = 2.75 min
Let
be the angular acceleration
We know
![\omega _f=\omega _i+\alpha t](https://tex.z-dn.net/?f=%5Comega%20_f%3D%5Comega%20_i%2B%5Calpha%20t)
=>![0=217+2.75\alpha =>\alpha = -78.9\frac{rev}{min^{2}}](https://tex.z-dn.net/?f=0%3D217%2B2.75%5Calpha%20%3D%3E%5Calpha%20%3D%20-78.9%5Cfrac%7Brev%7D%7Bmin%5E%7B2%7D%7D)
=>![\alpha =-\frac{78.9\times 2\pi}{60\times 60}\frac{rad}{s^{2}}=-0.138 \frac{rad}{s^{2}}](https://tex.z-dn.net/?f=%5Calpha%20%3D-%5Cfrac%7B78.9%5Ctimes%202%5Cpi%7D%7B60%5Ctimes%2060%7D%5Cfrac%7Brad%7D%7Bs%5E%7B2%7D%7D%3D-0.138%20%5Cfrac%7Brad%7D%7Bs%5E%7B2%7D%7D)
Torque required to stop is given by
![\tau =I\alpha](https://tex.z-dn.net/?f=%5Ctau%20%3DI%5Calpha)
where moment of inertia ,
=>![\therefore \tau =-0.138\times 62.5\, N.m=-8.6\, N.m](https://tex.z-dn.net/?f=%5Ctherefore%20%5Ctau%20%3D-0.138%5Ctimes%2062.5%5C%2C%20N.m%3D-8.6%5C%2C%20N.m)
Thus the constant torque required to stop the disk is 8.6 N-m in clockwise direction .
M=F/A
Which means 30 divided by 5 m/s is 6kg(mass)
Answer:
2577 K
Explanation:
Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.
So, T = ⁴√(P/σεA)
Since P = 60 W, we substitute the vales of the variables into T. So,
T = ⁴√(P/σεA)
= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)
= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)
= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)
= ⁴√(0.00441 × 10¹⁶K⁴)
= 0.2577 × 10⁴ K
= 2577 K
The mass of this bag of cement in S.I. units (kg) is equal to 0.062 kilograms.
<u>Given the following data:</u>
- Mass of cement = 62 grams.
To calculate the mass of this bag of cement in S.I. units (kg):
<h3>How to convert to
S.I. units.</h3>
In Science, kilograms (kg) is the standard unit of measurement or S.I. units of the mass of a physical object. Thus, we would convert the value of the mass of this bag of cement in grams to kilograms (kg) as follows:
<u>Conversion:</u>
1000 grams = 1 kilograms.
62 grams = X kilograms.
Cross-multiplying, we have:
X = ![\frac{62}{1000}](https://tex.z-dn.net/?f=%5Cfrac%7B62%7D%7B1000%7D)
X = 0.062 kilograms.
Read more on mass here: brainly.com/question/13833323