Can life be found/ sustain on Mars or other planets? What conditions are required to support life on another planet?

E14. A ball rolls off a table with a horizontal velocity of 5 m/s. If

Shkiper50 [21]

a) Vertical velocity: 5.9 m/s

b) Horizontal velocity: 5 m/s

Explanation:

a)

The motion of the ball is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction)

- A uniformly accelerated motion (constant acceleration) along the vertical direction

Here we want to find the vertical component of the ball's velocity. This can be done by using the suvat equation for the vertical motion:

$v_y = u_y +gt$

where:

$v_y$ is the vertical velocity at time t

$u_y=0$ is the initial vertical velocity (zero because the ball has been thrown horizontally)

$g=10 m/s^2$ is the acceleration of gravity (here we take downward as positive direction)

Substituting t = 0.6 s, which is the total time of flight, we find the vertical velocity of the ball just before it hits the ground:

$v_y=0+(9.8)(0.6)=5.9 m/s$

b)

The motion along the vertical direction is an accelerated motion, because there is a force (the force of gravity) acting on the ball and that it causes an acceleration in the ball.

However, there are no forces acting in the horizontal direction on the ball (if we neglect the air resistance): this means that the acceleration of the ball in the horizontal direction is zero.

As a consequence, this also means that the horizontal component of the ball's velocity is constant during the motion.

Since the ball was thrown from the table with an initial horizontal velocity of 5 m/s, this means that the horizontal velocity of the ball just before it hits the floor is still

$v_x = 5 m/s$

8 0

3 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

3 years ago

_______ are different forms of a single element. A) Atoms B) Elements C) Ions D) Isotopes

ELEN [110]

The answer is D. Isotopes.
Hope that helped.

7 0

3 years ago

Read 2 more answers

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$