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2 years ago

HELPPP IM DESPERATE

Physics

1 answer:

damaskus [11]2 years ago

5 0

the first one cuz I know

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Plz help me make a nice C-E-R for science I will give brainly for the best C-E-R

expeople1 [14]

C-E-R stands for Chemistry Eastern Right. Rights has five words so add five to your question that’ll be 45 and boom you got it

6 0

3 years ago

Calculate the weight of a 1 kg mass at earth's surface. The mass of the of the Earth's surface if the mass of the earth is 6 x 1

yuradex [85]

Answer:

9.8N

Explanation:

Here we can get gravitational acceleration according to the place where object is placed by bellow equation

g = GM/R²

g - Gravitational Acceleration

G - Gravitational constant (6.67×10-11)

R - Distance ( Radius )

g = 6.67 × 10-11 × 1024 /(6.37×106)²

g = 9.8 m/s²

There for

Weight = Mass × Gravitational acceleration

= 1×9.8

= 9.8 N

4 0

3 years ago

A net force of 15 N is applied to a cart with a mass of 2.1 kg. a. What is the acceleration of the cart? b. How long will it tak

Gennadij [26K]

The first question's answer is :
If, F=ma
Then, 15N= 2.1kg (a)
15/2.1=a
7.14=a
Therefore, acceleration = 7.14m/s^2

sorry I am not sure about the second question :-(

6 0

2 years ago

With Christmas lights wired in series, if one light goes out, what happens to the rest?

daser333 [38]

Answer:

The ones that are after the light that went out are also out.

Explanation:

3 0

2 years ago

Read 2 more answers

Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

5 0

3 years ago