What might cause matter to change states

Maximum current problem. If the current on your power supply exceeds 500 mA it can damage the supply. Suppose the supply is set

VikaD [51]

To solve this problem we will apply Ohm's law. The law establishes that the potential difference V that we apply between the ends of a given conductor is proportional to the intensity of the current I flowing through the said conductor. Ohm completed the law by introducing the notion of electrical resistance R. Mathematically it can be described as

$V = IR \rightarrow R = \frac{V}{I}$

Our values are

$I = 500mA = 0.5A$

$V = 37V$

Replacing,

$R = \frac{V}{I}$

$R = \frac{37}{0.5}$

$R = 74 \Omega$

Therefore the smallest resistance you can measure is $74 \Omega$

8 0

3 years ago

A submersible pump is put under the water at the bottom of a well and is used to push water up through a pipe. What minimum outp

Maslowich

Answer:

695800 N/m^2 or Pa

Explanation:

Height of the water from the ground H = 71 m

Acceleration due to gravity g =9.8 m/s^2

density of water ρ= 1000 kg/m^3

The minimum output gauge pressure to make water reach height H

P= ρgH

= 1000×9.8×71= 695800 N/m^2 or Pa

5 0

3 years ago

Hi, Solve for λ<br> E=hc/λ

Paul [167]

Answer:

λ=hc/E

Explanation:

E=hc/λ

Eλ=hc

λ=hc/E

4 0

3 years ago

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago

Explain the process that causes dew to form on blades of grass?

luda_lava [24]

<span>Heat is radiated, atmospheric moisture condenses at a rate greater than that at which it can evaporate, resulting in the formation of water droplets.</span>

7 0

3 years ago

Read 2 more answers