What quantity measures the amount of space an object occupies?

A ball is thrown up into the air with an initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed

kaheart [24]

Answer:

B) t = 1.83 [s]

A) y = 16.51 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f} =v_{o} -g*t$

where:

Vf = final velocity = 0

Vo = initial velocity = 18 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.

A) The maximum height is reached when the final velocity of the ball is zero.

0 = 18 - (9.81*t)

9.81*t = 18

t = 18/9.81

t = 1.83 [s], we found the answer for B.

Now using the following equation.

$y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\$

where:

y = elevation [m]

Yo = initial elevation = 0

y = 18*(1.83) - 0.5*9.81*(1.83)²

y = 16.51 [m]

7 0

2 years ago

Shaping will happen if _____.

Pavlova-9 [17]

What is being done in shaping is that it directs and guides the change in one's behavior. Therefore, what will happen when one undergoes shaping is that you encourage a lot of minor behaviors which would add up to one large action. The answer to this would be the first option.

7 0

2 years ago

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

4 0

2 years ago

A spherical capacitor is formed from two concentric sphericalconducting shells separated by vacuum. The inner sphere has radius1

zubka84 [21]

Explanation:

(1). Formula to calculate the potential difference is as follows.

$\Delta V = -\int E dr$

= $-\int \frac{kq}{r^{2}} dr$

= $\frac{kq}{r_{f}} - \frac{kq}{r_{i}}$

= $\frac{kq(r_{f} - r_{i})}{r_{f}r_{i}}$

= $\frac{9 \times 10^{9} \times 3.30 \times 10^{-9}(0.1 - 0.015)}{0.1 \times 0.015}$

= 38.7 volts

Therefore, magnitude of the potential difference between the two spheres is 38.7 volts.

(2). Now, formula to calculate the energy stored in the capacitor is as follows.

E = $\frac{1}{2}QV$

= $\frac{1}{2} \times 3.30 \times 10^{-9} \times 3.87 V$

= $6.39 \times 10^{-8} J$

Thus, the electric-field energy stored in the capacitor is $6.39 \times 10^{-8} J$ .

7 0

2 years ago

Read 2 more answers

Part C Let’s start the analysis by looking at your “extreme usage” cases. Compare the two cases in detail—low usage period versu

Serga [27]

Answer:

Day 7 DataUsage notes (since last reading)day & datetimekWh readingkWh usedhours elapsedavg. kW usedb.Usage Extremes: Data CollectionFor this experiment, you’ll measure electrical usage during a time period when you expect to havevery light electrical usage (for instance, while you’re asleep at night or during the day when no oneis at home). Likewise you’ll measure electrical usage during a time period when you expect to have heavier than average electrical usage. This time period might be in the evening, when lights and other appliances are on. Both of these time periods should be at least 4 hours long, to increase the accuracy of your results. Record your results in the tables below for each situation. For each time period, you’ll need to takean initial and a final reading.Type your response here:Low Usage - Initial Readingday & datetimekWh readingLow Usage - Final ReadingEnergy Usage Notesday & datetimekWh readingkWh usedhours elapsedavg. kW usedHigh Usage - Initial Readingday & datetimekWh reading4

6 0

2 years ago