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Nadusha1986 [10]
3 years ago
8

A ball is thrown straight up at time t=0 with an initial speed of 19m/s. Take the point of release to be y0=0 and upwards to be

the positive direction. Calculate the displacement at the time of 0.50 s
Physics
1 answer:
Wittaler [7]3 years ago
4 0
First we write the corresponding kinematics equations:
 a = -g
 v = -g * t + vo
 y = -g * ((t ^ 2) / 2) + vo * t + yo
 Substituting the values:
 y = - (9.81) * (((0.50) ^ 2) / 2) + (19) * (0.50) + (0) = 8.27m
 answer:
 the displacement at the time of 0.50s is 8.27m
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find the average velocity of a bicycle that starts 100km south and is 120km south of town after 0.4 hours​
dusya [7]

Answer:

The average velocity is 50 km/h south

Explanation:

The average velocity of an object is its total displacement divided by

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That means it is the rate at which an object changes its position from

one place to another.

Average velocity is a vector quantity.

The SI unit is meters per second.

A bicycle that starts 100 km south and is 120 km south of town after

0.4 hour​.

The displacement = 120 - 100 = 20 km south

The time = 0.4 hour

The average velocity = \frac{D}{T}, where D is the displacement

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The average velocity of the bicycle = \frac{20}{0.4}=50 km/h

<em>The average velocity is 50 km/h south</em>

If you want it in meter per second, change the kilometer to meter

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1 km = 1000 m

1 hour = 60 × 60 = 3600 seconds

The average velocity of the bicycle = \frac{50(1000)}{3600}=13.89 m/s south

5 0
3 years ago
How is current related to voltage? ​
Tju [1.3M]

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8 0
3 years ago
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A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio
natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

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v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}

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Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0
1 year ago
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AnnyKZ [126]

Answer:

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c)  S = ΔQ/T

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dsp73

Answer:

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Explanation:

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