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2 years ago

BRAINLIST!!!

Physics

1 answer:

ss7ja [257]2 years ago

4 0

<em><u>QB=-8×10^-9C.</u></em>
<em><u>QB=-8×10^-9C.We fixed balls A and B 5cm apart. Given e=1.6×10^-9C and k=9×10^9 S.I unit.</u></em>

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A 4 kilogram box is at rest on a frictionless floor. A net force of 12 newtons acts on it. What is the acceleration of the box?

Rashid [163]

Answer:

Explanation:

Step one:

given

Mass m= 4kg

Force F= 12N

Required

Acceleration the relation between force, acceleration, and mass is Newton's first equation of motion, which says a body will continue to be at rest or uniform motion unless acted upon by an external force

F=ma

a=F/m

a=12/4

a=3 m/s^2

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2 years ago

An unknown material has a mass

atroni [7]

Answer: 1896.55J/kg°C

Explanation:

The quantity of Heat Energy (Q) required to heat a material depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = 1320 joules

Mass of material = 5.61kg

C = ? (let unknown value be Z)

Φ = 0.124°C

Then, Q = MCΦ

1320J = 5.61kg x Z x 0.124°C

1320J = 0.696kg°C x Z

Z = (1320J / 0.696kg°C)

Z = 1896.55 J/kg°C

Thus, the specific heat of the material is 1896.55J/kg°C

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3 years ago

Read 2 more answers

Hooke's law describes a certain light spring of unstretched length 34.8 cm. When one end is attached to the top of a doorframe a

DaniilM [7]

Answer:

a) $k = 1343.6\,\frac{N}{m}$ , b) $l = 0.501\,m\,(50.1\,cm)$

Explanation:

a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:

$F \propto \Delta l$

In this case, the force is equal to the weight of the object:

$F = m\cdot g$

$F = (8.22\,kg)\cdot (9.807\,\frac{m}{s^{2}} )$

$F = 80.614\,N$

The spring constant is:

$k = \frac{F}{\Delta l}$

$k = \frac{80.614\,N}{0.408\,m-0.348\,m}$

$k = 1343.6\,\frac{N}{m}$

b) The length of the spring is:

$F = k\cdot (l-l_{o})$

$l = l_{o} + \frac{F}{k}$

$l=0.348\,m+\frac{205\,N}{1343.6\,\frac{N}{m} }$

$l = 0.501\,m\,(50.1\,cm)$

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3 years ago

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garri49 [273]

Answer:

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Explanation:

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2 years ago

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$