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3 years ago

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A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the g

round to rise 1.25 m above the floor in an attempt to get the ball? (Assume that the player gets the ball at his maximum height.)

1 answer:

maxonik [38]3 years ago

8 0

Answer:4.95 m/s

Explanation:

Given

Ball rises to a height of 1.25 m

Let u be the velocity of player while leaving ground .

Its final velocity will be zero as he reaches a height of 1.25 m

thus

$u^2=2gh$

$u=\sqrt{2\times 9.81\times 1.25}$

$u=\sqrt{24.525}$

u=4.952 m/s

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In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted?

agasfer [191]

Answer:

The frequency of the photon is $3.069\times10^{14}\ Hz$ .

Explanation:

Given that,

Energy $E=4.07\times10^{-19}\ J$

We need to calculate the energy

Using relation of energy

$E_{4}-E_{2}=\Delta E$

Where, $\Delta E$ = energy spacing

$4h\nu-2h\nu=4.07\times10^{-19}$

$\nu=\dfrac{4.07\times10^{-19}}{2h}$

Put the value of h into the formula

$\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}$

$\nu=3.069\times10^{14}\ Hz$

Hence, The frequency of the photon is $3.069\times10^{14}\ Hz$ .

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3 years ago

Recall that the blocks can only move along the x axis. the x components of their velocities at a certain moment are v1x and v2x.

Contact [7]

The center of mass is given with this formula:
$x_c=\frac{\sum_{n=1}^{n=i}m_ix_i}{M}$
Velocity is:
$v=\frac{dv}{dt}$
So, for the velocity of the center of mass we have:
$\frac{dx_c}{dt}=\frac{\sum_{n=1}^{n=i}d(m_ix_i)}{Mdt}\\
v_c=\frac{\sum_{n=1}^{n=i}p_i}{M}\\$
In our case it is:
$v_{xc}=\frac{m_1v_{x1}+m_2v_{x2}}{m_1+m_2}$

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3 years ago

A 878-kg (1940 lb) dragster, starting from rest, attains a speed of 25.9 m/s (57.9 mph) in 0.62 s. (a) Find the average accelera

salantis [7]

Answer:

41.8m/s^2

Explanation:

Since the dragster starts from rest, initial velocity (u) = 0m/s, final velocity (v) = 25.9m/s, time (t) = 0.62s

From the equations of motion, v = u + at

a = (v - u)/t = (25.9 - 0)/0.62 = 25.9/0.62 = 41.8m/s^2

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3 years ago

If the focal length of a concave mirror is 18cm, find its radius of curvature.

Galina-37 [17]

Given :

The focal length of a concave mirror is 18 cm.

To Find :

The radius of curvature of the concave mirror.

Solution :

We know,

$\text{Focal length}=\dfrac{\text{Radius of curvature}}{2}\\\\F=\dfrac{R}{2}\\\\R = 18\times 2\ cm\\\\R = 36 \ cm$

Therefore, the radius of curvature of concave mirror is 36 cm.

Hence, this is the required solution.

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3 years ago

A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a

Galina-37 [17]

Answer:

$E = (0.56 \times 10^8 ) r \ \ N/c$

Explanation:

Given that:

$\rho_o = (10^{-3} ) \ c/m^3$

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To find the electric field for r < R by using Gauss Law

${\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)$

For r < R

$Q_{enclosed}=(\rho) ( \pi r^2 ) l$

$E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}$

$E= \dfrac{\rho ( r)}{2 \varepsilon_o}$

where;

$\varepsilon_o = 8.85 \times 10^{-12}$

$E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}$

$E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}$

$E = (0.56 \times 10^8 ) r \ \ N/c$

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3 years ago