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2 years ago

What type of energy is at work when mini-blinds are being raised or lowered

1 answer:

4 0

The type of energy that is ar work when blinds are being raised or lowered is potential energy. because potential energy is the energy stored in an object due to its vertical position, the main cause of the potential energy is the acceleration due to gravity. so that is why its formual is PE = mgh
where m is the mass, g is the acceleration due to gravity and h is the height

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A capacitor is charged to 8.0×10−4 C , then discharged by connecting a wire between the two plates. 35us after the discharge beg

Anettt [7]

Answer:

The average current is 19.567 A

Solution:

As per the question:

Charge, Q = $8.0\times 10^{- 4}\ C$

Time, t = $35\times 10^{- 6}\ s$

Now,

We know that current is constituted by the rate of transfer of the charge per unit time. Thus we can write:

I = $\frac{Q}{t}$ (1)

Now, the charge that was transferred is 86 % of the original value.

Therefore,

We replace Q by 0.86Q in eqn (1):

I = $\frac{0.86\times 8.0\times 10^{- 4}}{35\times 10^{- 6}} = 19.657\ A$

7 0

2 years ago

A long solenoid that has 1 200 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00

Fantom [35]

Answer:

<h2>The current required winding is $2.65*10^-^2 mA$ </h2>

Explanation:

We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil and

n= N/L (number of turns per unit length)

Given data

The number of turns n= 1200 turns

length L= 0.42 m

magnetic field B= 1*10^-4 T

μ₀= $4\pi*10^-^7 T.m/A$

Applying the equation B=μ₀*n*I

I= B/μ₀*n

I= B*L/μ₀*n

$I= \frac{1*10^-^4*0.42}{4\pi*10^-^7*1.2*10^3 }$

$I= 2.65*10^-^2 mA$

8 0

3 years ago

An 18 month-old child is hypotonic and cannot fully extend the neck and trunk against gravity. Which position would be MOST EFFE

Flura [38]

Answer:

b. Prone on 45 degree wedge positioned on the floor.

Explanation:

According to the pediatric therapists the most effective position to place the child in to facilitate anti gravity movement and promote hand usage is to prone on 45 degree wedge positioned on the floor.

5 0

2 years ago

In need of some help here

Cloud [144]

Answer:

need more context!

Explanation:

7 0

2 years ago

In a thunderstorm, charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considered to be d

amm1812

Answer:

1) q = 83.4 C

2)52.13 × 10^(19) electrons

Explanation:

1) To calculate the total charge q on the cloud when the breakdown of the surrounding air begins, we will use the formula;

E = kq/r²

Making q the subject, we have;

q = Er²/k

Where k is a constant = 1/(4πε_o)

We are given;

ε_o = 8.85 × 10^(-12)) C²/N.m²

Thus;

k = 1/(4 × π × 8.85 × 10^(-12)) N.m²/C²

k = 8.99 × 10^(9)

Also,we are given E = 3 × 10^(6) N/C

Diameter = 1km = 1000 m

Radius(r) = diameter/2 = 1000/2 = 500 m

Thus;

q = (3 × 10^(6) × 500²)/(8.99 × 10^(9))

q = 83.4 C

2) To get the number of Excess electrons, we will divide the charge gotten in Part 1 above by the charge of a single electron.

Now, charge of a single electron = 1.6 × 10^(-19) C

Thus, number of Excess elecrons = 83.4/(1.6 × 10^(-19)) = 52.13 × 10^(19) electrons

5 0

2 years ago