Answer:
was there a reading that your class did on this
Explanation:
Answer:
Multiply the acceleration by time to obtain the velocity change:
Example:
Velocity change = 6.95 * 4 = 27.8 m/s . Since the initial velocity was zero, the final velocity is equal to the change of speed. You can convert units to km/h by multiplying the result by 3.6: 27.8 * 3.6 ≈ 100 km/h
Since the rocket’s acceleration is 3.00 m/s^3 * t, its acceleration is increasing at the rate of 3 m/s^3 each second. The equation for its velocity at a specific time is the integral of the acceleration equation.
<span>vf = vi + 1.5 * t^2, vi = 0 </span>
<span>vf = 1.5 * 10^2 = 150 m/s </span>
This is the rocket’s velocity at 10 seconds. The equation for its height at specific time is the integral velocity equation
<span>yf = yi + 0.5 * t^3, yi = 0 </span>
<span>yf = 0.5 * 10^3 = 500 meters </span>
<span>This is the rocket’s height at 10 seconds. </span>
<span>Part B </span>
<span>What is the speed of the rocket when it is 345 m above the surface of the earth? </span>
<span>Express your answer with the appropriate units. </span>
<span>Use the equation above to determine the time. </span>
<span>345 = 0.5 * t^3 </span>
<span>t^3 = 690 </span>
<span>t = 690^⅓ </span>
<span>This is approximately 8.837 seconds. Use the following equation to determine the velocity at this time. </span>
<span>v = 1.5 * t^2 = 1.5 * (690^⅓)^2 </span>
<span>This is approximately 117 m/s. </span>
<span>The graph of height versus time is the graph of a cubic function. The graph of velocity is a parabola. The graph of acceleration versus time is line. The slope of the line is the coefficient of t. This is a very different type of problem. For the acceleration to increase, the force must be increasing. To see what this feels like slowly push the accelerator pedal of a car to the floor. Just don’t do this so long that your car is speeding!!</span>
Temperature and pressure are directly proportional to each other. If this helps,please pick me the best. Also,please say thanks.
Answer:
A- series B- parallel
Explanation:
In order to measure current in a circuit, the multi-meter needs to be placed in series with the circuit while when measuring voltage, the multi-meter needs to be placed in parallel with the circuit.
It should be however noted that the same current flows in a series connected circuit and same voltage flows through loads connected in parallel. The ammeter is placed in series with the load to ensure that same value of currents is flowing in both the ammeter and loads(since same current flows in series connected circuit elements and all the amount of voltage must be made to appear on the load for the current to be measured accurately.
Voltmeter is connected in parallel to the load due to high value of current possessed by the voltmeter. The parallel connection will cause the current flowing through the voltmeter to reduce to zero so that it won't have effect (increase) on the amount of current initially on the resistor thereby measuring the exact amount of voltage on the load.
