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2 years ago

PHYSICS, WORTH 10 POINTS!

Physics

1 answer:

11Alexandr11 [23.1K]2 years ago

5 0

Answer:

Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle. Joining the points together reveals the path of the magnetic field lines.

Explanation:

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how is the position of electrons involved in metallic bonding different from the position of electrons that form ionic and coval

Yuri [45]

While ionic bonds join metals to nonmetals, and covalent bonds join nonmetals to nonmetals, metallic bonds are responsible for the bondingbetween metal atoms. In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize.

I hope that this answer helps you out

7 0

3 years ago

What force must act on a 50.0-kg mass to give it an ancceleration of 0.30 m/s^2?

ratelena [41]

Answer:

15.0 N

Explanation:

see pic

3 0

2 years ago

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

NikAS [45]

Answer:

The gazelles top speed is 27.3 m/s.

Explanation:

Given that,

Acceleration = 4.2 m/s²

Time = 6.5 s

Suppose we need to find the gazelles top speed

The speed is equal to the product of acceleration and time.

We need to calculate the gazelles top speed

Using formula of speed

$v=at$

Where, v = speed

a = acceleration

t = time

Put the value into the formula

$v=4.2\times6.5$

$v=27.3\ m/s$

Hence, The gazelles top speed is 27.3 m/s.

6 0

2 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

2 years ago

Hydroelectric power relies on the transfer of energy from waves through what medium

Sedaia [141]

Through the medium of electromagnetism, energy from waves get transferred which a hydroelectric power plant heavily rely on. The turbine eventually converts the energy from the waves into mechanical energy which is transformed to electrical energy using a generator.

6 0

2 years ago