Answer:
The officer's unit detects this 135-mile-per-hour speed and should subtract the patrol car's 70-mile -per-hour ground speed to get your true speed of 65 miles per hour. Instead, the officer's ground-speed beam fixes on the truck ahead and measures a false 50-mile-per-hour ground speed.
Explanation:
A speedometer or speed meter is a gauge that measures and displays the instantaneous speed of a vehicle. Now universally fitted to motor vehicles, they started to be available as options in the early 20th century, and as standard equipment from about 1910 onwards.
Answer:
1. 21.66 Ohms
2. 3.38 A
3. 6.7 V
Explanation:
1. Req = 6+2 = 8 Ohms (2 and 6 are in a series circuit)
Req = 1/8 +1/4 = 3/8 = 8/3 = 2.66 Ohms (8 and 4 are parallel, so we will add them using this equation)
Req = 2.66 + 1 + 9 + 3 + 6 = 21.66 Ohms
2. I = V/R = 9/2.66 = 3.38 A (In a series circuit, the current is the same across the resistors, so we will add them and divided them by 9 volts)
3. V = IR = 3.38 x 2 = 6.7 V (In a series circuit, the voltage is different, so each resistor will have a different voltage.)
I hope this helps. I am not an expert in physics but its ok :)
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Answer:
The time is 133.5 sec.
Explanation:
Given that,
One side of cube = 10 cm
Intensity of electric field = 11 kV/m
Suppose How long will it take to raise the water temperature by 41°C Assume that the water has no heat loss during this time.
We need to calculate the rate of energy transfer from the beam to the cube
Using formula of rate of energy
Put the value into the formula
We need to calculate the amount of heat
Using formula of heat
Put the value into the formula
We need to calculate the time
Using formula of time
Put the value into the formula
Hence, The time is 133.5 sec.
Answer:
The wheelbarrow's wheel and axle help the wheelbarrow to move without friction thus making it easier to push or pull. That's why it will be easier to lift a load in wheel barrow of the load is transferred towards the wheel.
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
