I The Circuit shows for Tamps Connected by Tovde supply

Choose one inner planet and one outer planet and compare their similarities and differences.

Katena32 [7]

They both have gas in there cloudy atmosphere which creates the color of the planets. They both have a weird and off centered magnetic field which gives them strength

7 0

2 years ago

Read 2 more answers

The the symptoms and treatments for these common nervous system

kykrilka [37]

Answer:

the symptoms are cough, cold, cattah and the treatment is antibiotics

7 0

2 years ago

An object of mass m 1 moving with speed v collides with another object of mass m 2 at rest and stick to it. Find the impulse to

garri49 [273]

Answer:

(m1+m2)vo=m1v+m2×0

⟹vo=m1+m2m1v

The impulse imparted to second object is equal to change is momentum is -

J=m2vo=m1+m2m2m1v

8 0

2 years ago

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike

zysi [14]

a) $t=\sqrt{\frac{2h}{g}}$

b) $v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c) $v=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d) $\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Explanation:

a)

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where here:

$s=h$ (the vertical displacement is the height of the counter)

$u=0$ (the initial vertical velocity of the cup is zero)

$a=g$ (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

$h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}$

b)

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

$v=\frac{d}{t}$

where

d is the distance covered

t is the time taken

Here, the distance covered is $d$ , the distance from the base of the counter, while the time taken is the time of flight:

$t=\sqrt{\frac{2h}{g}}$

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

$v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c)

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

$v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}$

The vertical speed instead is given by the suvat equation:

$v_y=u_y + at$

where:

$u_y=0$ is the initial vertical velocity

$a=g$ is the acceleration

$t=\sqrt{\frac{2h}{g}}$ is the time of flight

Substituting,

$v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}$

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d)

Just before hitting the floor, the velocity of the cup has two components:

$v_x=d\sqrt{\frac{g}{2h}}$ is the horizontal component (in the forward direction)

$v_y=\sqrt{2gh}$ is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

$tan \theta = \frac{v_y}{v_x}$

where here $\theta$ is measured as below the horizontal direction.

Substituting the expressions for $v_x,v_y$ , we find:

$tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}$

So

$\theta=tan^{-1}(\frac{2h}{d})$ (radians)

4 0

3 years ago

How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential d

Archy [21]

Answer:

The voltage is $V = 0.993V_b$

Explanation:

From the question we are told that

The time that has passed is $t = \frac{\tau}{2}$

Here $\tau$ is know as the time constant

The voltage of the power source is $V_b$

Generally the voltage equation for charging a capacitor is mathematically represented as

$V = V_b [1 - e^{- \frac{t}{\tau} }]$

=> $V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]$

=> $V = V_b [1 - e^{- \frac{\tau}{2\tau} }]$

=> $V = V_b [1 - e^{- \frac{1}{2} }]$

=> $V = 0.993V_b$

5 0

2 years ago