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jeyben [28]
3 years ago
10

Two 1.1 kg masses are 1 m apart (center to center) on a frictionless table. Each has +10 JC of charge. What is the initial accel

eration (in m/s2) of this massif it is released and allowed to move?
Physics
1 answer:
sweet-ann [11.9K]3 years ago
6 0

Answer:

acceleration = 0.8181 m/s²

Explanation:

given data

mass = 1.1 kg

apart d = 1 m

charge q = 10 μC

to find out

What is the initial acceleration

solution

we know that acceleration is

acceleration = \frac{force}{mass}   .................1

here force = k \frac{q1q2}{r^2}

here q1 q2 is charge and r is distance and Coulomb constant k = 9 × 10^{9} Nm²/C²

force = 9*10^{9} \frac{(10*10^{-6})^2}{1^2}

force = 0.9 N

so  from equation 1

acceleration = \frac{0.9}{1.1}

acceleration = 0.8181 m/s²

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Which example nest represents balanced forces ???
n200080 [17]

Answer:

an elevator stopped on the third floor a basketball shot into a hoop a sled sliding down a snowy hill a tow truck pulling a car out of a ditch

5 0
3 years ago
Most of the really bright stars in our sky are NOT among the stars that are very close to us. Why then do they look so bright to
Yanka [14]

Answer:

stars will emit more light due to their Luminosity, so they look very bright.

Explanation:

Luminous refers to..,

  • The total amount of energy radiated by a star or other celestial object per second.
  • Therefore it is the power output of a star.

Most of the really bright stars in our sky are not that very close to us yet they look bright because of the Luminosity of the star.

These stars are intrinsically so luminous.

A star's power output across all wavelengths is called its bolometric luminosity.

A star with large luminosity will have more measure of radiated electromagnetic power meaning.

so it will emit more light than a low luminosity star.

Hence,

those stars can easily be seen even across great distance.

learn more about  Luminosity of the star here:

<u>brainly.com/question/13912549</u>

<u />

#SPJ4

6 0
2 years ago
a cylindrical jar is 10cm long and has a cross sectional area of 36cm. if it is completely filled with a fluid of relative densi
ki77a [65]

Answer:

The mass of the fluid is 72 g.

Explanation:

The following data were obtained from the question:

Height (h) = 10 cm

Area of cross section (A) = 36cm²

Relative density = 0.2

Mass =..?

Next, we shall determine the volume of the cylinder. This can be achieved by doing the following:

Volume = Area x Height

Volume = 36 x 10

Volume = 360 cm³

Next, we shall determine the density of the liquid.

This can be obtained as follow:

Relative density = density of substance/density of water.

Relative density = 0.2

Density of water = 1 g/cm³

Density of fluid =...?

Relative density = density of substance/density of water.

0.2 = density of fluid / 1 g/cm³

Cross multiply

Density of fluid = 0.2 x 1 g/cm³

Density of fluid = 0.2 g/cm³

Finally, we shall determine the mass of fluid as follow:

Volume = 360 cm³

Density of fluid = 0.2 g/cm³

Mass of fluid =...?

Density = mass /volume.

0.2 g/cm³ = mass of fluid /360 cm³

Cross multiply

Mass of fluid = 0.2 g/cm³ x 360 cm³

Mass of fluid = 72 g

Therefore, the mass of the fluid in the jar is 72 g.

6 0
3 years ago
How many cubic feet are there in a gallon?
shtirl [24]
1 gallon has 0.133681 cubic feet
5 0
3 years ago
Read 2 more answers
Vector ????⃗ has a magnitude of 16.6 and is at an angle of 50.5∘ counterclockwise from the +x‑axis. Vector ????⃗ has a magnitude
natka813 [3]

Answer:

For vector u, x component = 10.558 and  y component =12.808

unit vector = 0.636 i+ 0.7716 j

For vector v, x component = 23.6316 and y component = -6.464

unit vector = 0.9645 i-0.2638 j

Explanation:

Let the vector u has magnitude 16.6

u makes an angle of 50.5° from x axis

So u_x=ucos\Theta =16.6\times cos50.5=10.558

Vertical component u_y=usin\Theta =16.6\times sin50.5=12.808

So vector u will be u = 10.558 i+12.808 j

Unit vector u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j

Now in second case let vector v has a magnitude of 24.5

Making an angle with -15.3° from x axis

So horizontal component v_x=vcos\Theta =24.5\times cos(-15.3)=23.6316

Vertical component v_y=vsin\Theta =24.5\times sin(-15.3)=-6.464

So vector v will be 23.6316 i - 6.464 j

Unit vector of v =\frac{23.6316i-6.464}{\sqrt{23.6316^2+6.464^2}}=0.9645i-0.2638j

8 0
3 years ago
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