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weeeeeb [17]
2 years ago
11

How many volts would it take to push 1 amp through a resistance of 1 ohm?

Physics
2 answers:
ELEN [110]2 years ago
5 0
V=I x R so V= 1 x 1 =1V
Diano4ka-milaya [45]2 years ago
4 0

Answer:

1\; \rm V.

Explanation:

Ohm's Law give the following relationship between the potential V across a conductor, the electrical resistance R of that conductor, and the current I through that conductor:

V = I \, R.

It is given that the current through this conductor is I = 1\; {\rm A}, and that the electrical resistance of this conductor is R = 1\; \rm \Omega.

Note, that the unit of electrical resistance Ohm is a composite unit; 1\; {\rm \Omega} = 1\; {\rm V \cdot A^{-1}} (volts per ampere.)

Substitute the value of I and R into the equation V = I \, R from Ohm's Law to find the potential across this conductor:

\begin{aligned}V &= I \, R \\ &= 1\; {\rm A} \times 1\; {\rm \Omega} \\ &= 1\; {\rm A} \times 1\; {\rm V \cdot A^{-1}} \\ &= 1\; {\rm V}\end{aligned}.

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An object has a mass of 7g and a volume of 14cm. What is the density
Andreyy89
Density = mass / volume = 7/14 = 0.5!g/cm^3
8 0
3 years ago
A bowling ball with a mass of 7.0 kg, generates 56.0 kgᐧm/s units of momentum. What
Misha Larkins [42]

Answer:

8 m/s

Explanation:

p = mv \\ 56 = (7.0)(v) \\ v = 8.0 \: m {s}^{ - 1}

7 0
2 years ago
A motorcycle begins at rest and accelerates uniformly S7.9 we want to find a time to take the motorcycle to reach a speed of 100
Len [333]

The motorbike reaches 100 km/h in 3.5 seconds

Explanation:

The motion of the motorbike is a uniformly accelerated motion (= constant acceleration), therefore we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the motorbike in this problem,

u = 0 (it starts from rest)

v = 100 km/h = 27.8 m/s is the final velocity

a=7.9 m/s^2 is the acceleration

Solving for t, we find the time it takes for the bike to reach that velocity:

t=\frac{v-u}{a}=\frac{27.8-0}{7.9}=3.5 s

Learn more about accelerated motion:

brainly.com/question/9527152

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6 0
3 years ago
One speed skater starts across a frozen lake at an average speed of 8 m/s. Ten seconds later, a second speed skater starts from
Luba_88 [7]

Answer:

40 s

Explanation:

After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start

Let t be the number of seconds after the second skater starts will the second skater overtake the first skater

The distance traveled by the first skater after t seconds is

s_1 = v_1t = 8t

Similarly the distance traveled by the 2nd skater after t seconds is

s_2 = v_2t = 10t

Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater

s_2 = s_1 + 80

We can substitute s_1 = 8t, s_2 = 10t

10t = 8t + 80

2t = 80

t = 80 / 2 = 40 s

7 0
3 years ago
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite
sladkih [1.3K]

Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

       The gravitational force is  F_g  = 6600 \ N

The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

     v  = \sqrt{\frac{G  M}{r^2} }

=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

7 0
3 years ago
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