All categories

3 years ago

A 25kg ball is thrown into the air. when thrown it is going 10 m/s. calculate how high it travels?

Physics

1 answer:

Darina [25.2K]3 years ago

3 0

Answer:

Explanation:

V^2 = U^2 - 2gh

V=0 , U=10m/s, g=10m/s^2

0^2 = 10^2 - 2×10×h

0 = 100 - 20h

100=20h

H = 100/20 = 5m

You might be interested in

Starting from a location with position vector 1,=−17.5 m and 1,=23.1 m , a rabbit hops around for 10.7 seconds with average velo

Andru [333]

By definition of average velocity,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t} = \dfrac{(x_2-x_1)\,\vec\imath + (y_2-y_1)\,\vec\jmath}{10.7\,\mathrm s}$

$\vec v_{\rm ave} = (-2.25\,\vec\imath + 1.79\,\vec\jmath)\dfrac{\rm m}{\rm s}$

and $x_1=-17.5\,\mathrm m$ and $y_1=23.1\,\mathrm m$ , then

$-2.25\dfrac{\rm m}{\rm s} = \dfrac{x_2 - (-17.5\,\mathrm m)}{10.7\,\mathrm s} \\\\ 1.79\dfrac{\rm m}{\rm s} = \dfrac{y_2 - 23.1\,\mathrm m}{10.7\,\mathrm s}$

Solve for $x_2$ and $y_2$ :

$x_2 = 17.5\,\mathrm m + \left(-2.25\dfrac{\rm m}{\rm s}\right)(10.7\,\mathrm s) \approx \boxed{-6.58\,\mathrm m}$

$y_2 = 23.1\,\mathrm m + \left(1.79\dfrac{\rm m}{\rm s}\right)(10.7\,\mathrm s) \approx \boxed{42.3\,\mathrm m}$

8 0

2 years ago

A tug-of-war game is played by five c/hildren: three on one team and two on the other. How much force will the two child team ha

dem82 [27]

Answer:

no, 3 porces is more tha 2 so the power between the 3 should be more than 2

Explanation:

4 0

3 years ago

Calculate the length of a simple pendulum that oscillates with a frequency of 0.4Hz g=10m/s2 , ^=3.142

earnstyle [38]

Answer:

Explanation:

For simple pendulum the formula is

$T=2\pi\sqrt{\frac{l}{g} }$

Where T is time period , l is length and g is acceleration due to gravity .

$\frac{1}{n} =2\pi\sqrt{\frac{l}{g} }$

n is frequency

Putting the values

$\frac{1}{.4} =2\pi\sqrt{\frac{l}{10} }$

$\frac{l}{10} = .1584$

l = 1.584 m

4 0

3 years ago

An automatic dryer spins wet clothes at an angular speed of 5.2 rad/s. Starting from rest, the dryer reaches its operating speed

Dvinal [7]

<h2>Time taken by dryer to come up to speed is 1.625 seconds.</h2>

Explanation:

We have equation of motion v = u + at

Initial velocity, u = 0 rad/s

Final velocity, v = 5.2 rad/s

Time, t = ?

Acceleration, a = 3.2 rad/s²

Substituting

v = u + at

5.2 = 0 + 3.2 x t

t = 1.625 s

Time taken by dryer to come up to speed is 1.625 seconds.

6 0

3 years ago

The temperature and time t given in hours from 0 to 24 after midnight in downtown mathville is given by t=10-5 sin(pi/12 t) degr

Dvinal [7]

Answer:

$T_A_v_g=9.918192559$^{\circ}C$

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

$T(t)=10-5*sin(\frac{\pi}{12t})$

The average temperature over a given interval can be calculated as:

$T_a_v_g=\frac{T_o+T_f}{2}$

Where:

$T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature$

So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

Therefore:

$T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C$

$T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C$

Hence, the average temperature between noon and midnight is:

$T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C$

3 0

3 years ago