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4 years ago

5

A 1.5m wire carries a 8 A current when a potential difference of 78 V is applied. What is the resistance of the wire?

1 answer:

Papessa [141]4 years ago

3 0

R=V/I
R=78/8
R=9.75 ohms

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__________ are combating competitive pressures by providing better value with private-label merchandise; adding new value-added

Neko [114]

Answer:

convinience store

Explanation:

convenience store are more like retail store providing everyday goods.

since this kind of store is usually small, maybe not as big as a department store, there is a growing number of such stores creating competitive pressures as such many are providing better value with private-label merchandise (more like re-branding or personalizing the products in the store)

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3 years ago

Sarah is sitting at the top of the slide in the school's playground. Which of the following changes will increase Sarah's kineti

Ray Of Light [21]

B; kinetic energy is anything in motion

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4 years ago

Three tiny charged metal balls are arranged on a straight line. The middle ball is positively charged and the two outside balls

Dmitrij [34]

Answer:

$(a) 189.23 N$ , $(b) 47.31 N$ and $(c) 141.92 N$ .

Explanation:

Three balls are shown in figure having charge $q=1.45 \mu C$ . The middle ball, $B$ , is positively charged having charge $+q$ , and the remaining two outside balls, $A$ and $C$ , are negatively charged having charged $-q$ as shown.

$AC=20 cm$ and $AB=BC=10$ cm as B is the mid-point of AC.

Let $d_1=AC=20\times 10^{-3}m$ and $d_2=AB=BC=10\times 10^{-3}m$

From Coulomb's law, the magnitude of the force, $F$ , between two point charges having magnitudes $q_1 \& q_2$ , separated by distance, $d$ , is

$F=\frac {1}{4\pi\epsilon_0}\frac {q_1q_2}{d^2}\;\cdots (i)$

where, $\epsilon_0$ is the permittivity of free space and

$\frac {1}{4\pi\epsilon_0}=9\times 10^9$ in SI units.

This force is repulsive for the same nature of charges and attractive for the different nature of charges.

Now, Using equations(i),

(a) The magnitude of attraction force between balls A and B is

$F_{AB}=F_{BC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_2)^2}$

$\Rightarrow F_{AB}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(10\times 10^{-3}\right)^2}$

$\Rightarrow F_{AB}=189.23 N$

(a) The magnitude of the repulsive force between balls A and C is

$F_{AC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_1)^2}$

$\Rightarrow F_{AC}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(20\times 10^{-3}\right)^2}$

$\Rightarrow F_{AC}=47.31 N$

(c) The magnitude of the net force, $F_{net}$ , on the outside of the ball is,

$F_{net}=189.23-47.31 N$

$\Rightarrow F_{net}=141.92 N$

4 0

3 years ago

What is characteristic of both sound waves andelectromagnetic waves?(1) They require a medium.(2) They transfer energy.(3) They

Amiraneli [1.4K]

The correct answer would be 2: they transfer energy.

3 0

3 years ago

Martha was leaning out of the window on the second floor of her house and speaking to Steve. Suddenly, her glasses slipped from

DiKsa [7]

Answer:

s = 23.72 m

v = 21.56 m/s²

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

u = 0 m/s

g = 9.8 m/s²

s = 0 + 0.5 × 9.8 × 2.2²

s = 23.72 m

b) impact velocity

v = √(2gh)

v = √(2× 9.8 × 23.72)

v = √464.912

v = 21.56 m/s²

6 0

4 years ago