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3 years ago

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A ball tossed vertically upward rises, reaches its highest point, and then falls back to its starting point. During this time th

e acceleration of the ball is always
A) in the direction of motion.
B) opposite its velocity.
C) directed upward.
D) directed downward.

1 answer:

kozerog [31]3 years ago

3 0

Answer: D) directed downward.

Explanation:

We know that the constant acceleration of $9.87ms^{-2}$ of every-thing is always DIRECT DOWNWARDS because of the to the influence of the gravity of the Earth.

This motion is denoted as free fall motion , where the acceleration due to the gravity is constant $9.87ms^{-2}$ .

∴ A ball tossed vertically upward rises, reaches its highest point, and then falls back to its starting point. During this time the acceleration of the ball is always directed downward.

Hence, the correct answer is D) directed downward..

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A 40.0-mH inductor is connected to a North American electrical outlet (ΔVrms = 120 V, f = 60.0 Hz). Assuming the energy stored i

trasher [3.6K]

Explanation:

It is given that,

Inductance, $L=40\ mH=40\times 10^{-3}\ H$

RMS value of voltage, $v_{rms}=120\ V$

Frequency, f = 60 Hz

We need to find the energy stored at t = (1 /185) s. It is assumed that energy stored in the inductor is zero at t = 0. So,

The current flowing through the inductor is given by :

$I_t=\dfrac{V_o}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})$

$I_t=\dfrac{\sqrt{2} V_{rms}}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt{2}}{2\pi f L}\ sin(2\pi f t-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt{2}}{2\pi\times 60\times 40\times 10^{-3}}\ sin(2\pi \times 60\times \dfrac{1}{185})-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt2}{15.07}\ sin(2\pi \times 60\times \dfrac{1}{185}-\dfrac{\pi}{2})$

I = 0.091 A

Energy stored in the inductor is, $U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times 40\times 10^{-3}\times (0.091)^2$

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3 years ago

A racing car on a straight accelerates from 100mph to 316 mph in three seconds what is the acceleration?

omeli [17]

Answer:

72mph/sec

Explanation:

The car goes from 100mph to 316mph in three seconds. Meaning it increases its speed by (316 - 100)mph in three seconds. That is 216 mph increase in three seconds. So, we divide the speed increase by the amount of time the increase occurred over. We get:

216mph / 3sec = 72mph/sec, our final answer

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3 years ago

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3 years ago

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A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
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3 years ago

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