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3 years ago

Math the words in the box with the sentences below.

Physics

2 answers:

lilavasa [31]3 years ago

5 0

Answer:

1.friction

Non contract

Static force

Gravity

Tension

Spring force

Contact force

Magnetic force

Normal force

Applied force

Explanation:

Sana makatulong

Tanzania [10]3 years ago

4 0

In order of appearance on the table;

Magnetic force

Normal force

Gravity

String force

Friction

Contact force

Tension

Non Contact force

Static force

Applied force

please mark BRAINLIEST if I'm right.

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How is the phenomenon of reflection used in making a kaleidoscope

dusya [7]

Answer:

The kaleidoscope

Explanation:

gives a number of images formed by reflection from the mirrors inclined to one another. Designers and artists use kaleidoscope to get ideas for new patterns to design wallpapers, jewellery and fabrics.

8 0

3 years ago

a student pushes a 0.500 kg trolley along a frictionless surface and accelerates it from rest to 4m/s. how much kinetic energy d

aalyn [17]

mass=m=0.5kg
velocity=4m/s

$\\ \sf\longmapsto KE=\dfrac{1}{2}mv^2$

$\\ \sf\longmapsto KE=\dfrac{1}{2}(0.5)(4)^2$

$\\ \sf\longmapsto KE=0.25(16)$

$\\ \sf\longmapsto KE=4J$

7 0

2 years ago

An electron is moving in a circular orbit in a uniform magnetic field. Is the kinetic energy of the electron changing?

Lelechka [254]

Answer:

kinetic energy does not change

Explanation:

you can use the formula for the kinetic energy of the electron and for the radius of the trajectory of an electron in a uniform magnetic field:

$E_k=\frac{1}{2}m_ev^2\\\\r=\frac{m_e v}{qB}$

me: mass of the electron

B: magnetic field

q: charge of the electron

r: radius

By doing v the subject of the formula and replace it in the expression for the kinetic energy you obtain:

$v=\frac{rqB}{m_e}\\\\E_k=\frac{1}{2}m_e(\frac{rqB}{m_e})^2=\frac{r^2q^2B^2}{2m_e}$

as youcan see, all parameters r, q, B and me are constant.

hence, the kinetic energy does not change

3 0

3 years ago

A radio antenna broadcasts a 1.0 MHz radio wave with 30 kW of power. Assume that the radiation is emitted uniformly in all direc

kozerog [31]

Answer:

a) intensity $= 2.19 \times 10^{-6} W/m^2$

b) E = 0.0406 V/m

Explanation:

given data:

Power = 30 kW

R = 33×10^3 M

1) Intensity is given as

$Intensity I = \frac{P}{4\times \pi \times r^2}$

$= \frac{30 \times 10^3}{4\times 3.14 \times (33\times10^3)^2}$

$= 2.19 \times 10^{-6} W/m^2$

2) electric field is given as

$I = 1/2 \times \epsilon \times c \times E^2$

Solving for E

$E = \sqrt{\frac{2I}{\epsilon c}}$

$E =\sqrt{\frac{2\times 2.19 \times 10^{-6}}{8.85\times 10^{-12} 3\times 10^8}}$

E = 0.0406 V/m

6 0

3 years ago

Help with all !!!!!!!

k0ka [10]

Answer A: It will rotate to the right
Answer B: it will rotate to the left
Answer C: 17.5 N
Answer D: 2.25 m

8 0

2 years ago