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VashaNatasha [74]
3 years ago
8

WILL GIVE YOU A MEDAL!! To increase the current in a circuit, the _____ can be increased

Physics
2 answers:
Maru [420]3 years ago
8 0

the answer is Voltage.

hope this helps you .

ser-zykov [4K]3 years ago
4 0

Answer:

the answer is voltage

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The car in the figure travels at a constant speed along the road shown. Draw vector showing its acceleration at the point C if t
yuradex [85]

Answer:

The vector form is as shown in the attachment

Explanation:

The figure as shown in the diagram, indicates that the car is moving along the road at a constant speed. Centripetal acceleration comes into play for an object moving in a circular motion at uniform speed. The centripetal acceleration is the acceleration experienced by an object while in uniform circular motion.

Mathematically from centripetal acceleration; a = v2/r

The equation shows that there is an inverse relationship between the acceleration and the radius of curvature as such the radius of curvature at the point A will be more than the radius of curvature at the point C, this shows that the centripetal acceleration at point C will be more than the centripetal acceleration at point A.

The attachment shows the figure and the representation in vectorial form.

6 0
3 years ago
A diffraction grating produces a first-order bright fringe that is 0.18 m away from the central bright fringe on a flat screen.
Georgia [21]

Answer:

The wavelength of the light is 562.5 nm

Solution:

As per the question:

Order, n = 1

Slit separation, d = 2.5\times 10^{- 6}\ m

Distance from the bright fringe, y = 0.18 m

Distance between the screen and the grating, D = 0.8 m

Now,

We know from the eqn for diffraction:

n\lambda = dsin\theta

n = 1

\lambda = dsin\theta            (1)

Also,

For very small angle, \theta:

sin\theta ≈ tan\theta = \frac{y}{D} = \frac{0.18}{0.8} = 0.225

Using the above value in eqn (1):

\lambda = 2.5\times 10^{- 6}\times 0.225 = 5.625\times 10^{- 7}\ m = 562.5\ nm

3 0
3 years ago
Hilda plans to display in a clear cube the needle betsy ross used to sew the first american flag. the needle will fit exactly as
inysia [295]

Based on the length of side of the cube given, the length of the needle is 2.28 inches.

<h3>What is the length of the needle?</h3>

The length of the side of a cube and its diagonal are related by the formula below:

Length \:  of \:  Diagonal = \sqrt{3s}

where s is length of side

Side \:  Length  \: of \:  the \:  Cube = \sqrt{3inches}

Diagonal \: of  \: a  \: face \:  of  \: the \: cube = \sqrt{6inches}

The \: length \: of \: the \: needle = \sqrt{3\sqrt{3}}  = 2.28 \: inches \\

Therefore, the length of the needle is 2.28 inches

Learn more about length of diagonal of cube at: brainly.com/question/800212

#SPJ4

8 0
2 years ago
Read 2 more answers
A(n)
lozanna [386]
A theory is an explanation that unites the findings of many experiments, and it can be changed when new experimental results need to be explained. 
6 0
3 years ago
Read 2 more answers
What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x
lyudmila [28]

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

(a) What is the change in potential energy of a +2.00nC test charge, U_{electric,b} - U_{electric,a} as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?

(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?

(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?

Answer: (a) ΔU = 3.2×10^{-6} J

(b) KE = 2×10^{-6} J

Explanation: <u>Potential</u> <u>Energy</u> (U) is the amount of work done due to its position or condition and its unit is Joule (J). <u>Kinetic</u> <u>Energy</u> (KE) is the ability to do work by virtue of velocity and the unit is also (J). <u>Mechanical</u> <u>Energy</u> is the sum of Potential and Kinetic Energies of a system.

(a) Related to electricity, Potential Energy can be calculated as:

ΔU = Eqd

where E is the electric field (in N/C);

q is the charge (in C);

d is the distance between plaques (in m);

For a at x = - 30cm and b at x = 50 cm:

E = 2×10^{3} N/C

q = 2×10^{-9} C

d = 50 - (-30) = 80×10^{-2} = 8×10^{-1}m

ΔU = U_{electric,b} - U_{electric,a} = Eqd

U_{electric,b} - U_{electric,a} = 2×10^{3} .  2×10^{-9} . 8×10^{-1}

ΔU = 3.2×10^{-6} J

(b) Mechanical Energy is constant, so:

KE_{i} + U_{i} = KE_{f} + U_{f}

Since the initial position is zero and there is no initial kinetic energy:

KE_{f} = - U{f}

KE_{f} = - (2×10^{3}. 2×10^{-9} . 5×10^{-1})

KE_{f} = - 2.10^{-6} J

(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.

8 0
3 years ago
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