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Elanso [62]
3 years ago
6

"How do changes in voltage or resistance affect current in an electric circuit?"

Physics
1 answer:
kicyunya [14]3 years ago
5 0
We can answer the question by looking at the Ohm's law, which gives us the relationship between voltage (V), current (I) and resistance (R) of a circuit:
V=IR
equivalently, we can rewrite it as
I= \frac{V}{R}

by looking at the equation, we can make the following observations:
1) The current is proportional to the voltage: therefore, if the voltage increases, the current increases as well; if the voltage decreases, the current decreases too.
2) The current is inversely proportional to the resistance: if the resistance increases, the current decreases, and if the resistance decreases, the current increases.
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If a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the wate
Ulleksa [173]
<span>s= 0.5 at^2
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0.5 x 9.81 x 3^2 = 44m
8 0
3 years ago
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A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origi
Liula [17]

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

<u>a) calculating work done by the force over the journey of the train</u>

F = mx + b  ------ ( 1 )

m = slope  = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8   ----- ( 2 )

hence to determine the work done by the force

W   = \int\limits^7_0 { ( -0.373) x + 2.8  )} \, dx     Note:  the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

<u>b) calculate the speed of the train at the end of its journey</u>

we will apply the work energy theorem

W = 1/2 m*v^2  -  1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 )  ( input values into equation )

 V^2 = 12.11

hence V = 3.48 m/s

6 0
3 years ago
Nuclear fusion occurs in stars.<br> a. True<br> b. False
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8 0
2 years ago
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A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho
fiasKO [112]

Answer:

v_f=8.17\frac{m}{s}

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}

The box is initially at rest, so v_i=0. Solving for v_f:

v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}

5 0
2 years ago
A stone is thrown, run a velocity of 15m/s is projected of an elevation of 30° to the horizontal calculate the time rate of flig
frozen [14]

Answer:

1.53 seconds

Explanation:

Applying,

T = 2usin∅/g................ Equation 1

Where, T = time of flight, u = initial velocity, ∅ = angle of projectile to the horizontal, g = acceleration due to gravity

From the question,

Given: u = 15 m/s, ∅ = 30°

Constant: g = 9.8 m/s²

Substitute these values in equation 1

T = 2(15)(sin30°)/9.8

T = 15/9.8

T = 1.53 seconds

Hence the time rate of flight is 1.53 seconds

3 0
2 years ago
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