The change of state from a gas to a liquid?

a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its

lilavasa [31]

Answer:

D/H =15

Explanation:

We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
At this point, we can find the value of H, applying the following kinematic equation:

$v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)$

If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

$H = \frac{v_{0} ^{2} }{2*g} (2)$

We can use the same equation, to find the value of D, as follows:

$v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)$

In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
When we replace these values in (1), we find that v₁ = -v₀.
Replacing in (3), we have:

$(4*v_{0})^{2} - (-v_{0}) ^{2} = 2* g* D\\ \\ 15*v_{0}^{2} = 2*g*D$

Solving for D:

$D = \frac{15*v_{0} ^{2} }{2*g}$

From (2) we know that H can be expressed as follows:

$H = \frac{v_{0} ^{2} }{2*g}$

⇒ D = 15 * H

$\frac{D}{H} = 15$

3 0

3 years ago

You push very hard on a heavy desk, trying to move it. You do work on the desk:

frozen [14]

Answer:

(C) Only if it starts moving

Explanation:

We know that work done is given by

$W=F.d=Fdcos\Theta$

So there are two case in which work done is zero

First case is that when force and displacement are perpendicular to each other

And other case is that when there is no displacement

So for work to be done there must have displacement, if there is no displacement then there is no work done

So option (c) will be the correct option

3 0

3 years ago

A drunken sailor stumbles 550 meters north, 500 meters northeast, then 450 meters northwest. What is the total displacement and

cluponka [151]

Answer:

Resultant displacement = 1222.3 m

Angle is 88.3 degree from +X axis.

Explanation:

A = 550 m north

B = 500 m north east

C = 450 m north west

Write in the vector form

A = 550 j

B = 500 (cos 45 i + sin 45 j ) = 353.6 i + 353.6 j

C = 450 ( - cos 45 i + sin 45 j ) = - 318.2 i + 318.2 j

Net displacement is given by

R = (353.6 - 318.2) i + (550 + 353.6 + 318.2) j

R = 35.4 i + 1221.8 j

The magnitude is

$R = \sqrt{35.4^{2}+1221.8^{2}}R = 1222.3 m$

The direction is given by

$tan\theta =\frac{1221.8}{35.4}\\\\\theta = 88.3^{o}$

3 0

3 years ago

How much heat is needed to bring 12.0 g of water from 28.3 °C to 43.87 °C, if the specific heat capacity of water is 4.184 J/(g•

Vika [28.1K]

Answer:

Heat capacity, Q = 781.74 Joules

Explanation:

Given the following data;

Mass = 12g

Initial temperature = 28.3°C

Final temperature = 43.87°C

Specific heat capacity of water = 4.184J/g°C

To find the quantity of heat needed?

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 43.87 - 28.3

dt = 15.57°C

Substituting into the equation, we have;

$Q = 12*4.184*15.57$

Q = 781.74 Joules

7 0

3 years ago

A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density

blsea [12.9K]

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

$P = \rho h$