Answer:
1.8m/s^2
Explanation:
Since the two ropes are going up, their combined force is 105+115=220N. With a gravitational force of 186N, the force of the two ropes pulling up the will be 220-186=34N.
Now we need the mass of the bucket itself in order to find the acceleration of the bucket (remember that F=ma and m is needed to find a). Since gravitational acceleration is 9.8m/s^2 and F=186N, 186/9.8=18.97959184 kg for the mass of the bucket.
Now that we have the mass of the bucket, we can find the acceleration of the bucket. Since F=34N from earlier, 34N/18.97959184kg=1.791397849m/s^2=1.8m/s^2 is the acceleration of the bucket.
Therefore, 1.8m/s^2 is the correct answer.
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Full Question:
Consider an automobile with a mass of 5510 lbm braking to a stop from a speed of 60.0 mph.
How much energy (Btu) is dissipated as heat by the friction of the braking process?
______Entry field with incorrect answer Btu
Suppose that throughout the United States, 350.0 x 10^6 such braking processes occur in the course of a given day.
Calculate the average rate (megawatts) at which energy is being dissipated by the resulting friction.
______Entry field with incorrect answer MW
Answer:
Energy dissipated as heat = 852.10 Btu
Average rate of energy dissipation = 3.64 MW
Explanation:
1 lb = 0.453592 Kg
Mass = 5510 lb = 0.453592 * 5510 Kg = 2499.29 Kg
1 mph = 0.44704 m/s
Velocity = 60 mph = 60 * 0.44704 m/s = 26.82 m/s
Let E be equal to energy acquired
Energy dissipated as heat:
E = 852.10 Btu
Energy dissipated throughout the United States
Average power rate in MW, P
There are several ways they interact with each other.
Mechanically: Erosion, Mass movement, Sedimentation
Answer:
1124923453 electrons
Explanation:
The formula for charge in coulomb ,C =Current in amperes, A * Time in seconds, s
Given in question ;
Charge = 1.6 * 10⁻¹⁹ C
Current = 2.0 nA = 2*e⁻⁹ A
Calculate time in seconds as
1.6 * 10⁻¹⁹ = 2*e⁻⁹ * t
1.6 * 10⁻¹⁹ /2*e⁻⁹ = t
6.48e⁻¹⁶ s = t
So using t=6.48e⁻¹⁶ s and current =2*e⁻⁹ A , the charge will be;
C = 2*e⁻⁹ * 6.48e⁻¹⁶ =1.799e⁻¹⁰ C
But 1 coulomb = 6.25 x 10¹⁸ electrons
so 1.799e⁻¹⁰ C = ?,,,,{ 6.25 x 10¹⁸} *{1.799e⁻¹⁰} = 1124923453.06 electrons
