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4 years ago

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A 10kg object is near a planet’s surface such that the gravitational field strength is 4Nkg. With what force is the planet attra

cted to the 10kg object?
A-4 Nkg

b-10N

c- 40N

d-100N

1 answer:

dsp734 years ago

7 0

Answer:

C. 40 N

Explanation:

Gravitational field strength = force/mass.

Force = mass* gravitational field strength.

F = 10*4 = 40 N.

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At time t=0t=0 a proton is a distance of 0.360 mm from a very large insulating sheet of charge and is moving parallel to the she

insens350 [35]

Answer:

$1.34 * 10^{3}$ m/s

Explanation:

Parameters given:

distance of the proton form the insulating sheet = 0.360mm

speed of the proton, $v_{x}$ = 990m/s

Surface charge density, σ = 2.34 x $10^{-9}$ C/ $m^{2}$

We need to calculate the speed at time, t = 7.0 * $10^{-8}$ s.

We know that the proton is moving parallel to the sheet, hence, we can say it is moving in the x direction, with a speed $v_{x}$ on the axis.

The electric force acting on the proton moves in the y direction, so this means it is moving with velocity $v_{y}$ in the y axis.

Hence, the resultant velocity of the proton is given by:

$v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}$

$v_{x}$ = 990m/s from the question. We need to find $v_{y}$ and then the resultant velocity v.

Electric field is given in terms of surface charge density, σ as:

E = σ/ε0

where ε0 = permittivity of free space

=> $E = \frac{2.34 * 10^{-9} } {2 * 8.85418782 * 10^{-12} }$

E = 132 N/C

Electric Force, F is given in terms of Electric field:

F = eE

where e = electronic charge

=> F = ma = eE

∴ a = eE/m

where

a = acceleration of the proton

m = mass of proton

$a = \frac{1.60 * 10^{-19} * 132}{1.672 * 10^{-27} }$

a = 1.3 * $10^{10}$ m/ $s^{2}$

Therefore, at time, t = 7.0 * $10^{-8}$ , we can use one of the equations of linear motion to find the velocity in the y axis:

$a = \frac{v_{y} - v_{0}}{t} \\\\=> v_{y} = v_{0} + at$

$v_{y}$ = 0 + (1.3 * $10^{10}$ * 7.0 * $10^{-8}$ )

$v_{y}$ = 910 m/s

∴ $v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}$

$v = \sqrt{990^{2} + 910^{2} }$

$v = \sqrt{1808200}$

v = 1344.69 m/s = $1.34 * 10^{3}$ m/s

8 0

4 years ago

An ideal parallel plate capacitor with a cross-sectional area of 0.4 cm2 contains a dielectric with a dielectric constant of 4 a

Mrrafil [7]

Answer: $283.2\times 10^{-9}\ nC$

Explanation:

Given

Cross-sectional area $A=0.4\ cm^2$

Dielectric constant $k=4$

Dielectric strength $E=2\times 10^8\ V/m$

Distance between capacitors $d=5\ mm$

Maximum charge that can be stored before dielectric breakdown is given by

$\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC$

4 0

3 years ago

Read 2 more answers

Which does not represent a change in velocity?

alina1380 [7]

Answer:

A balloon is floating up into the air at 1 m/s.

Explanation:

Velocity is the displacement divided by the time;

Velocity = $\frac{displacement}{time}$

It is a vector quantity that has magnitude and direction.

The choice that shows a change in direction is a correct specification of velocity.

A balloon floating up into the air at 1m/s shows no directional change.

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3 years ago

If an object has a net positive charge of 4.0 coulombs, the object possesses (1) 6.3 × 10^18 more electrons than protons (2) 2.5

Readme [11.4K]

Answer:

(4) $2.5 \times 10^{19}$ more protons than electrons

Explanation:

Total positive change on the object is given as

$Q = 4 C$

now by the law of quantization of charge we know that the charge on any body is always in form of integral multiple of charge of an electron.

So we will have

$Q = N e$

here we know that

$4 = N(1.6 \times 10^{-19})$

$N = 2.5 \times 10^{19}$

Now we know that the object is positively charged so here electrons must be less in numbers than protons

so correct answer will be

(4) $2.5 \times 10^{19}$ more protons than electrons

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4 years ago

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frozen [14]

The answer would be A transverse

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3 years ago