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3 years ago

Please I need help on this one

Physics

2 answers:

Lisa [10]3 years ago

5 0

lake effect snow is the correct answer

Vlad1618 [11]3 years ago

5 0

lake effect

plz mark as brainliest

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What is the formula for momentum?

olasank [31]

The momentum of any object is (the object's mass) times (its speed).

8 0

3 years ago

A 230 g air-track glider is attached to a spring. The glider is pushed in 8.2 cm and released. A student with a stopwatch finds

Natasha_Volkova [10]

Answer:

4.3 N/m

Explanation:

m = 230 g = 0.230 kg, x = 8.2 cm

in 13 oscillations, time taken = 19 s

In 1 oscillation, time taken = 19 / 13 = 1.46 s

By the use of formula of time period , Let k be the spring constant.

$T = 2\pi \sqrt{\frac{m}{k}}$

$1.46 = 2\pi \sqrt{\frac{0.230}{k}}$

0.054 = 0.230 / k

k = 4.26 N/m

k = 4.3 N/m

8 0

3 years ago

Which best defines transparent and provides an example of a material that is transparent to light?

Crank

Answer: option B. The ability to transmit light; for example water. The transparency is due to the pass of the light. The pass of light is what is called transmission of light. Light may be reflected, refracted or transmitted. Remember light is combination of several waves with different frequencies. When light falls upon an object the interaction may be some frequencies are reflected and other transmitted. Transparent objects are those that permit tha pass of (transmit) most light waves and so let the eyes and instruments to see through them.

6 0

4 years ago

A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby

8090 [49]

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, $O=(0\ nm,0\ nm)$

Position of desired magnetic field, $D\equiv(1\ nm,8\ nm)$

Magnitude of desired magnetic field, $E=0\ N.C^{-1}$

Let q be the positive charge magnitude placed at origin.

We know the distance between the two Cartesian points is given as:

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

For the electric field effect to be zero at point D we need equal and opposite field at the point.

$\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }$

$\therefore (1-0)^2+(8-0)^2=r^2$

$r^2=65\ nm$

$r=\sqrt{65}$

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

$x=2\times 1=2\ nm$