PLEASE HELP!!! PLEASE HELP!!!

640 nanometer setara dengan

lidiya [134]

Answer:

640 nanometer setara dengan 6.4e-7 meter

7 0

2 years ago

A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.

nikitadnepr [17]

The time spent in the air by the ball at the given momentum is 6.43 s.

The given parameters;

<em>momentum of the ball, P = 0.9 kgm/s</em>
<em>weight of the ball, W = 0.14 N</em>

The impulse experienced by the ball is calculated as follows;

$Ft = \Delta P$

where;

$Ft$ is impulse

$\Delta P$ is change in momentum

The time of motion of the ball is calculated as follows;

$t = \frac{\Delta P}{F} \\\\t = \frac{0.9 - 0}{0.14} \\\\t = 6.43 \ s$

Thus, the time spent in the air by the ball at the given momentum is 6.43 s.

Learn more here:brainly.com/question/13468390

7 0

2 years ago

She uses a voltmeter, that measures in volts, and an ammeter that measures in amps. Both were correctly placed in her circuit. I

Nesterboy [21]

Answer:

A) the ammeter is x

B)

voltage across R₁ (left resistor) = 0.75 V

voltage across the right one = 0.3 V

C) 1.05 V

Explanation:

From the diagram attached below;

A) Assuming the homes were wired in series, and one of the homes face short circuit then all the houses would face power cut but it doesn't happen. So they must be connected in parallel.

Therefore; The ammeter is connected in series, Hence, the ammeter is x and the voltmeter must be z.

B)

Given that:

x = 0.15 A

z = 0.3 V

Resistor (R) on the left = 5 ohms

Then, voltage across R₁ (left resistor) = 5×(x)

= 5×0.15

= 0.75 V

voltage across the right one = z = 0.3 V

C)

The total voltage of battery = 0.75+0.3 = 1.05 V

6 0

2 years ago

Based on Newton’s 3rd law if you were to push on the wall with a force of 100 N, how much force would the wall push back towards

rusak2 [61]

The answer is 100N. Look up the definition of Newton's third law.

3 0

3 years ago

Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be

avanturin [10]

Answer:

The mass rate of the cooling water required is: $1'072988.5\frac{kg}{h}$

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

${m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}$

Where w refers to the cooling water and s to the steam flow. Reorganizing,

$m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}$

Write the difference of enthalpy for water as Cp (Tout-Tin):

$m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}$

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, ${h_{s}}^{out}=251.40\frac{kJ}{kg}$ and ${h_{s}}^{in}$ can be calculated as:

${h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}$ .

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 $\frac{kJ}{kgK}$ . If the average temperature is actually different, it won't mean a considerable mistake. Also we know that ${T_{w}}^{out}-{T_{w}}^{in}\leq 10$ , so let's work with the limit case, which is ${T_{w}}^{out}-{T_{w}}^{in}=10$ to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

$m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}$

Download pdf

5 0

3 years ago