Ask question

Ask question

All categories

3 years ago

12

A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules (with the poten

tial energy equal to zero at ground level) and is moving upward with a kinetic energy of 50 Joules. What is the maximum height h reached by the ball?

1 answer:

Mariana [72]3 years ago

8 0

Answer:

The maximum height reached by the ball is 20 meters

Explanation:

Given that,

Height, h = 10 m

Potential energy at the height of 10 m, $h_1=50\ J$

The ball is moving upward with a kinetic energy of 50 Joules

Let h is the maximum height reached by the ball. At a height of 10 meters above the ground, the energy is given by :

$mgh=50$

$mg\times 10=50$

mg = 5 N

At this point, total energy will be 100 J. We know that the velocity is zero at maximum height. So, the ball possess only potential energy. it is given by :

$100=mgh$

$100=5\times h$

h = 20 meters

So, the maximum height reached by the ball is 20 meters. Hence, this is the required solution.

You might be interested in

A projectile is fired vertically from Earth's surface with

scoray [572]

Answer:

$h=25.52\times10^6 m$

Explanation:

Initial speed, v = 10 x 10^3 m/s

Mass of the earth, M = 6 x 10^24 kg

Radius of the earth, R = 6.4 x 10^6 m

Maximum from the surface of earth, h = ?

Let m = Mass of the projectile

Solution:

Potential energy at maximum height = ( Potential + Kinetic energy ) at the surface

$-G M m / ( R + h )=- G M m / R + (1/2) m v^2$

$- G M / ( R + h ) = - G M / R + (1/2) v^2$

$-2\times G M / ( R + h ) = ( - 2 G M / R ) + v^2$

$-2\times6.67\times10^{-11}\times6\times10^{24}/ ( R + h )$

= $( (- 2\times 6.67\times10^{-11}\times6\times10^{24}) /(6.4\times10^6)} +10000^2$

= $-2.50625\times10^7 J$

$=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7$

$R+h=31.92\times10^{6}$

$h=31.92\times10^{6}-6.4\times10^6$

$h=25.52\times10^6 m$

5 0

2 years ago

Scientists want to place a telescope on the moon to improve their view of distant planets. The telescope weighs 200 pounds on Ea

LuckyWell [14K]

Answer:

Explanation:

the weight of the telescope decreases because the moon attract the body with less force as compared to earth due to less gravity as compared to earth

6 0

3 years ago

PROBLEMA DI FISICA:il volume di un attizzatoio da camino in ferro, a temperatura ambiente (20°C), è 12,2 cm cubi. A contatto con

lesantik [10]

Responder:

20.3 ° C

Explicación:

<u>Según la ley de Charles</u>: <em>cuando la presión sobre una muestra de gas seco se mantiene constante, la temperatura y el volumen estarán en proporción directa. </em>

Paso uno:

datos dados

Temperatura T1 = 20 ° C

Temperatura T2 =?

Volumen V1 = 12.2 cm ^ 3

Volumen V2 = 12.4 cm ^ 3

Aplicar la relación temperatura y volumen

$\frac {V_{1}}{T_{1}}=\frac {V_{2}}{T_{2}}$

sustituyendo tenemos

$\frac {12.2}{20}=\frac {12.4}{T_{2}}\\\\$

Cruz multiplicar tenemos

$T_2=\frac{20*12.4}{12.2} \\\\T_2=\frac{248}{12.2}\\\\T_2=20.3$

Temperatura delle braci 20.3°C

4 0

3 years ago

The pressure of a sample of gas is measured as 49 torrent. Convert this to atmosphere

spin [16.1K]

Answer:

P = 0.0644 atm

Explanation:

Given that,

The pressure of a sample of gas is measured as 49 torr.

We need to convert this temperature to atmosphere.

The relation between torr and atmosphere is as follow :

1 atm = 760 torr

1 torr = (1/760) atm

49 torr = (49/760) atm

= 0.0644 atm

Hence, the presssure of the sample of gas is equal to 0.0644 atm.

5 0

2 years ago

Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a

nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

$Q = \int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {r^3} \, dr * d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \, d\phi$

$Q = \frac{\pi^2 r^4}{2}}$

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

3 0

2 years ago