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3 years ago

A baseball player throws 4 balls every 20 seconds. what is his throw frequency

Physics

1 answer:

frosja888 [35]3 years ago

3 0

Answer:

Frequency, f = 0.2 Hz

Explanation:

We have,

A baseball player throws 4 balls every 20 seconds.

It is required to find the frequency of the baseball.

Frequency of an object is defined as the number of times an event occurs. It is given by number of throws per unit time. It can be given by :

$f=\dfrac{4}{20}\\\\f=0.2\ Hz$

So, the frequency of his throw is 0.2 Hz.

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A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago

A machine which has an energy loss of 10% will have efficiency of

larisa [96]

Answer:

90%

Explanation:

if you lose 10% of a 100 you get 90

6 0

2 years ago

a ball is thrown with a speed of 17.7 m/s at an angle of 49.8° above the horizontal. how much time does the ball need to reach a

julsineya [31]

Answer:

The ball needs approximately $0.41\; \rm s$ to reach that height for the first time.

Explanation:

The initial speed of the ball $17.7\; \rm m \cdot s^{-1}$ . However, what would be the initial vertical speed of this ball?

The angle of elevation is $\theta = 49.8^\circ$ . Consider the initial speed of this ball as the length of the hypotenuse of a right triangle. If the angle of elevation is one of the two acute angles of this triangle, the initial vertical speed of this ball would be the leg opposite to that angle.

$\displaystyle \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ .

$\displaystyle \sin \theta = \frac{v(\text{vertical, initial})}{v(\text{initial})}$ .

Therefore:

$\begin{aligned}&v(\text{vertical, initial})\\ &= v(\text{initial}) \cdot \sin\theta \\ &= 17.7\; \rm m \cdot s^{-1} \times \sin \left(49.8^\circ\right) \approx 13.5\; \rm m \cdot s^{-1}\end{aligned}$ .

Let $t$ denote the time (in seconds) required for the ball to reach a height of $4.7\; \rm m$ .

Let $g$ denote the acceleration because of gravity (typically $g \approx 9.81\; \rm m \cdot s^{-2}$ near the surface of the earth.) The height of the ball $t$ seconds after it was thrown would be: $\displaystyle \frac{1}{2}\, g \cdot t^{2} + v(\text{vertical, initial}) \cdot t$ .

Assume that $g \approx 9.81\; \rm m \cdot s^{-2}$ . Set the value of this expression for height to $4.7\; \rm m$ and solve for $t$ :

$\displaystyle \frac{1}{2} \times 9.81 \, t^{2} + 13.5 \, t = 4.7$ .

Either $t \approx 0.41$ or $t \approx 2.3$ will satisfy this equation. Both of these two values are reasonable. The first value for $t$ ( $0.41\; \rm s$ ) is the time required for the ball to reach a height of $4.7\; \rm m$ for the first time. The second value ( $2.3\; \rm s$ ) is the time required for the ball to come under that height on its way back to the ground. The question seems to be asking only for the first (the smaller one) of these two times.