Answer:
A
Explanation:
Ke = 1/2 MV^2
An electric range has a constant current of 10 A entering the positive voltage terminal with a voltage of 110 V. The range is operated for two hours, (a) Find the charge in coulombs that passes through the range, (b) Find the power absorbed by the range, (c) If electric energy costs 12 cents per kilowatt-hour, determine the cost of operating the range for two hours.
Answer:
A. 72000 C
B. 1100 W
C. 26.4 cents.
Explanation:
From the question given above, the following data were obtained:
Current (I) = 10 A
Voltage (V) = 110 V
Time (t) = 2 h
A. Determination of the charge.
We'll begin by converting 2 h to seconds. This can be obtained as follow:
1 h = 3600 s
Therefore,
2 h = 2 h × 3600 s / 1 h
2 h = 7200 s
Thus, 2 h is equivalent to 7200 s.
Finally, we shall determine the charge. This can be obtained as follow:
Current (I) = 10 A
Time (t) = 7200 s
Charge (Q) =?
Q = It
Q = 10 × 7200
Q = 72000 C
B. Determination of the power.
Current (I) = 10 A
Voltage (V) = 110 V
Power (P) =?
P = IV
P = 10 × 110
P = 1100 W
C. Determination of the cost of operation.
We'll begin by converting 1100 W to KW. This can be obtained as follow:
1000 W = 1 KW
Therefore,
1100 W = 1100 W × 1 KW / 1000 W
1100 W = 1.1 KW
Thus, 1100 W is equivalent to 1.1 KW
Next, we shall determine the energy consumption of the range. This can be obtained as follow:
Power (P) = 1.1 KW
Time (t) = 2 h
Energy (E) =?
E = Pt
E = 1.1 × 2
E = 2.2 KWh
Finally, we shall determine the cost of operation. This can be obtained as follow:
1 KWh cost 12 cents.
Therefore, 2.2 KWh will cost = 2.2 × 12
= 26.4 cents.
Thus, the cost of operating the range for 2 h is 26.4 cents.
Consider a swimmer that swims a complete round-trip lap of a 50 m long pool in 100 seconds. The swimmer's... average speed is 0 m/s and average velocity is 0 m/s. average speed is 0.5 m/s and average velocity is 0.5 m/s. average speed is 1 m/s and average velocity is 0 m/s. average speed is 0 m/s and average velocity is 1 m/s.What is the swimmers average speed and average velocity?
Answer:
average speed is 1 m/s and average velocity is 0 m/s.
Explanation:
Given that :
Length of round trip = 50 m
Time taken = 100 seconds
The average speed :
Total distance / total time taken
Length of complete round trip :
(50 + 50) m, total. Distance = 100 m
100 / 100 = 1m/s
The average velocity :
Total Displacement / total time taken
Total Displacement of round trip = end point - start point = 0
0 / 100 = 0
Average speed is 1 m/s and average velocity is 0 m/s.
The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.
The speed is obtained from;
Speed = Distance/time = 2(50 m)/100 s = 1 m/s.
The velocity is 0 m/s since it is complete round-trip lap.
Learn more about speed: https://brainly.com/question/7359669
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
Answer:
it will option option A hope it helps
Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer distance
Please show working
Distance = (speed) x (time)
Car A: Distance = (8 m/s) x (43 s) = 344 meters
Car B: Distance = (7 m/s) x (50 s) = 350 meters
350 meters is a longer distance than 344 meters.
Car-B traveled a longer distance than Car-A did.
Answer:
[tex]\boxed {\boxed {\sf Car \ B : 350 \ meters }}[/tex]
Explanation:
Distance is equal to the product of speed and time.
[tex]d=s*t[/tex]
1. Car A
Car A has a speed of 8 meters per second and travels for 43 seconds.
[tex]s= 8 \ m/s \\t= 43 \ s[/tex]
Substitute the values into the formula.
[tex]d= 8 \ m/s *43 \ s[/tex]
Multiply and note that the seconds will cancel out.
[tex]d= 8 \ m*43= 344 \ m[/tex]
2. Car B
Car B has a speed of 7 meters per second and travels for 50 seconds.
[tex]s= 7 \ m/s \\t= 50 \ s[/tex]
Substitute the values in and multiply.
[tex]d= 7 \ m/s * 50 \ s[/tex]
[tex]d= 7 \ m * 50 = 350 \ m[/tex]
350 meters is a longer distance than 344 meters, so Car B traveled the longer distance.
take a picture of an object in your house, describe the
energy stores and transfers that happen with it. You can be as imaginative as you wish
with the object (choose something unusual), but the stores you identify and transfers
that happen must be real.
pls give me ideas of what to take a photo of for this I'm really stuck :(
A car enters a 105-m radius flat curve on a rainy day when the coefficient of static friction between its tires and the road is 0.4. What is
the maximum speed which the car can travel around the curve without sliding
Static friction (magnitude Fs) keeps the car on the road, and is the only force acting on it parallel to the road. By Newton's second law,
Fs = m a = W a / g
(a = centripetal acceleration, m = mass, g = acceleration due to gravity)
We have
a = v ² / R
(v = tangential speed, R = radius of the curve)
so that
Fs = W v ² / (g R)
Solving for v gives
v = √(Fs g R / W)
Perpendicular to the road, the car is in equilibrium, so Newton's second law gives
N - W = 0
(N = normal force, W = weight)
so that
N = W
We're given a coefficient of static friction µ = 0.4, so
Fs = µ N = 0.4 W
Substitute this into the equation for v. The factors of W cancel, so we get
v = √((0.4 W) g R / W) = √(0.4 g R) = √(0.4 (9.80 m/s²) (105 m)) ≈ 20.3 m/s
