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Norma-Jean [14]
1 year ago
13

How much charge must pass by a point in a wire in 10 s for the current inb the wire to be 0.50 a?

Physics
1 answer:
Ostrovityanka [42]1 year ago
3 0

Charge must pass by a point in a wire is 5 Coulomb

Given:

time = 10 s = t

current = 0.50 A = I

To Find:

charge = q

Solution: Current is defined as amount of charge passing through a region per unit time. It is given by the formula

I = q/t

q = I x t = 0.50 A x 10 s = 5 Coulomb

Hence, charge must pass by a point in a wire is 5 Coulomb

Learn more about Charge here:

brainly.com/question/25922783

#SPJ4

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A car dropped from a height of 44 meters fall to a height of zero meters. How fast will the car be traveling as it hits the grou
FinnZ [79.3K]
Vi=0m/s
Vf=?
A=9.81
D=44
T=not needed

Vf^2=Vi^2+2ad
Vf=2ad square rooted
Vf=2(9.81)(44) square root it
Vf=29.3m/s
6 0
3 years ago
. A 2.0-kg block is on a perfectly smooth ramp that makes an angle of 30° with the horizontal. (a) What is the block’s accelerat
AfilCa [17]

Answer:

a) a = 4.9 m / s²,  N = 16.97 N   and b)   F = 9.8 N

Explanation:

a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry

        sin 30 = Wx / W

        cos 30 = Wy / W

        Wx = W sin30

        Wy = W cos 30

Let's write the equations on each axis

X axis

        Wx = ma

Y Axis  

       N- Wy = 0

       N = Wy = mg cos 30

       N = 2.0 9.8 cos 30

       N = 16.97 N

We calculate the acceleration

       a = Wx / m

       a = mg sin 30 / m

       a = g sin 30

       a =9.8 sin 30

       a = 4.9 m / s²

b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component

      F -Wx = 0

      F = Wx

      F = m g sin 30

      F = 2.0 9.8 sin 30

      F = 9.8 N

5 0
3 years ago
The speed of light in a transparent medium is 0.6 times that of its speed in vacuum. Find the refractive index of the medium.
grin007 [14]
<h2>Answer:</h2>

The refractive index is 1.66

<h2>Explanation:</h2>

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum .

Refractive index of medium = speed of light in vacuum / speed of light in medium  

So

RI = 1/0.6 = 5/3 or 1.66

3 0
3 years ago
Read 2 more answers
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}

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Answer D. The pitch is in the frequency.
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