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2 years ago

Please answer all questions COMPLETELY!! I NEED IT ASAP!!!

Physics

1 answer:

wel2 years ago

6 0

Answer:

bdbdbdbdbdbdbdbdbdbdbdbdbdbdbdbd

Explanation:

ddvrrbbrbrbddddbdbddbd

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What is the power through a device with a resistance of 100 ohms if a current of 8 A is running through it?

Kazeer [188]

Answer:

6400 W (or) 6.4 KW

Explanation:

Formula we use,

→ P = I²R

Let's solve for the power of device,

→ P = I²R

→ P = (8)² × 100

→ P = 64 × 100

→ [ P = 6400 W ]

Hence, the power is 6400 W.

5 0

2 years ago

Please help me answer these questions ASAP !

vodomira [7]

1. Layer A is the oldest. The law of superposition says that the oldest layer is at the bottom whilst the newest is at the top.
2. The law of original horizontality.
3. The law of original lateral continuity.

3 0

3 years ago

A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that

VMariaS [17]

Answer:

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

$F = qE$

so here the force is tension force on it

$F = 6.57 \times 10^{-2} N$

$Q = 6.80 \times 10^3 C$

now we have

$6.57 \times 10^{-2} = (6.80 \times 10^3)E$

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

4 0

4 years ago

Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a

suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

$F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N$