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777dan777 [17]
4 years ago
6

How are voltage (V), current (I), and resistance (R) related?

Physics
2 answers:
Studentka2010 [4]4 years ago
6 0

Voltage, current, and Resistance are put together into an equation such as

V = IR.

marin [14]4 years ago
3 0
The relationship between Voltage<span>, </span>Current<span> and </span>Resistance<span> forms the basis of Ohm’s law. In a linear circuit of fixed resistance, if we increase the voltage, the current goes up, and similarly, if we decrease the voltage, the current goes down. This means that if the voltage is high the current is high, and if the voltage is low the current is low.</span>
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Phoenix [80]

i think number 1 is D number 2 is B

4 0
3 years ago
A particle of mass 4.0 kg is constrained to move along the x-axis under a single force F(x) = −cx3 , where c = 8.0 N/m3 . The pa
Pie

Answer:4.58 m/s

Explanation:

Given

mass of Particle m=4 kg

F=-cx^3

a=\frac{F}{m}

a=-\frac{cx^3}{m}

a=-\frac{8x^3}{4}

a=-2x^3

v\frac{\mathrm{d} v}{\mathrm{d} x}=-2x^3

vdv=-2x^3dx

integrating

\int_{6}^{v_b}vdv=\int_{1}^{-2}-2x^3dx

\frac{v_b^2-6^2}{2}=-\frac{1}{2}\left [ \left ( -2\right )^4-\left ( 1\right )^4\right ]

\frac{v_b^2-36}{2}=-0.5\times 15

v_b^2=36-15

v_b=\sqrt{21}

v_b=4.58 m/s

6 0
3 years ago
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
4 years ago
A pair of opposite electric charges of equal magnitude is called a(n)
ladessa [460]

A pair of opposite electric charges of equal magnitude is called a(n) A) Dipole

8 0
3 years ago
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5.) How could we have improved this activity to be more precise?<br><br>ball on ramp ​​
AleksAgata [21]

Explanation:

I don't understand this question

could you please explain

7 0
3 years ago
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