By definition of average acceleration,
<em>a</em> = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²
Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:
∑ <em>F</em> = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N
If you just want the magnitude, drop the negative sign.
For an ideal transformer power loss is assumed to be zero
i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage
this can be written in form of equation
here we know that
![i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1](https://tex.z-dn.net/?f=i_1%20%3D%2010%20A%7B%2Ftex%5D%3C%2Fp%3E%3Cp%3Enow%20we%20will%20use%20above%20equation%3C%2Fp%3E%3Cp%3E%5Btex%5D140%2A3.5%20%3D%2010%20%2A%20V_1)
So primary coil voltage is 49 Volts
Answer:
B. 6.6%
Explanation:
The percentage error of a measurement can be calculated using the formula;
Percent error = (experimental value - accepted value / accepted value) × 100
In this question, the calibrated 250.0 gram mass is the accepted value while the weighed mass of 266.5 g is the experimental or measured value.
Hence, the percentage error can be calculated thus;
Percent error = (266.5-250.0/250.0) × 100
Percent error = 16.5/250 × 100
Percent error = 0.066 × 100
Percent error = 6.6%
Answer:
8.94*10^22 kg
Explanation:
Given that
Mass of Lo, M = ?
Radius of Lo, r = 1.82*10^6 m
Acceleration on Lo, g = 1.80 m/s²
Gravitational constant, G = 6.67*10^-11
Using the formula
g = GM/r²
Solution is attached below
Answer is 8.94*10^22 kg
Explanation:
M₂ = Fr²/GM₁
M₂ = [(132N)(.243m)²]/[(6.67*10^-11N*m²/kg)(1.175*10^4kg)]
M₂ = (7.79N*m²)/(7.84*10^-7N*m²)
M₂ = 9.94*10^6 kg
