If only 1 option is correct then it is (D)
All the others can also make one component negative, all depends how u measured the angle.
all the best
Answer:
a = 3.61[m/s^2]
Explanation:
To find this acceleration we must remember newton's second law which tells us that the total sum of forces is equal to the product of mass by acceleration.
In this case we have:
![F = m*a\\\\m=mass = 3.6[kg]\\F = force = 13[N]\\13 = 3.6*a\\a = 3.61[m/s^2]](https://tex.z-dn.net/?f=F%20%3D%20m%2Aa%5C%5C%5C%5Cm%3Dmass%20%3D%203.6%5Bkg%5D%5C%5CF%20%3D%20force%20%3D%2013%5BN%5D%5C%5C13%20%3D%203.6%2Aa%5C%5Ca%20%3D%203.61%5Bm%2Fs%5E2%5D)
The linear speed of the pepperoni is 0.628 m/s. Its direction is tangential to the circle.
We know that;
v = rω
r = radius of the piece = 10 cm or 0.1 m
ω = angular velocity
We have to convert 60 revolutions per minute to radians per second
1 rev/min = 0.10472 rad/s
60 revolutions per minute = 60 rev/min × 0.10472 rad/s/1 rev/min
= 6.28 rad/s
v = 0.1 m × 6.28 rad/s
v = 0.628 m/s
The direction of this velocity is tangential to the circle.
Learn more: brainly.com/question/4612545
Answer:
The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher the photon's frequency, the higher its energy. Equivalently, the longer the photon's wavelength, the lower its energy.
Explanation:
Answer:
![\Delta S=1.69J/K](https://tex.z-dn.net/?f=%5CDelta%20S%3D1.69J%2FK)
Explanation:
We know,
..............(1)
where,
η = Efficiency of the engine
T₁ = Initial Temperature
T₂ = Final Temperature
Q₁ = Heat available initially
Q₂ = Heat after reaching the temperature T₂
Given:
η =0.280
T₁ = 3.50×10² °C = 350°C = 350+273 = 623K
Q₁ = 3.78 × 10³ J
Substituting the values in the equation (1) we get
![0.28=1-\frac{Q_2}{3.78\times 10^{3}}](https://tex.z-dn.net/?f=0.28%3D1-%5Cfrac%7BQ_2%7D%7B3.78%5Ctimes%2010%5E%7B3%7D%7D)
or
![\frac{Q_2}{3.78\times 10^3}=0.72](https://tex.z-dn.net/?f=%5Cfrac%7BQ_2%7D%7B3.78%5Ctimes%2010%5E3%7D%3D0.72)
or
![Q_2=3.78\times 10^3\times0.72](https://tex.z-dn.net/?f=Q_2%3D3.78%5Ctimes%2010%5E3%5Ctimes0.72)
⇒ ![Q_2 =2.721\times 10^3 J](https://tex.z-dn.net/?f=Q_2%20%3D2.721%5Ctimes%2010%5E3%20J)
Now,
The entropy change (
) is given as:
![\Delta S=\frac{\Delta Q}{T_1}](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cfrac%7B%5CDelta%20Q%7D%7BT_1%7D)
or
![\Delta S=\frac{Q_1-Q_2}{T_1}](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cfrac%7BQ_1-Q_2%7D%7BT_1%7D)
substituting the values in the above equation we get
![\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cfrac%7B3.78%5Ctimes%2010%5E%7B3%7D-2.721%5Ctimes%2010%5E3%20J%7D%7B623K%7D)
![\Delta S=1.69J/K](https://tex.z-dn.net/?f=%5CDelta%20S%3D1.69J%2FK)