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3 years ago

What factors determine the magnitude of the electric force between two particles ?

Physics

2 answers:

Harrizon [31]3 years ago

8 0

Answer: Charge and distance

Explanation:

svp [43]3 years ago

7 0

The electric force between the two particles are calculated through the equation,

F = kQ₁Q₂ / d²

where F is the force, k is a constant called Coulomb's law constant, Q₁ and Q₂ are the charges, and d is the distance. This equation is called the Coulomb's law.

It can be seen from the equation above that the electric forces between the objects are majorly affected by the substance's charges and distance.

The answer to this item is therefore letter A.

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A bakery measured the mass of a whole cake as 0.870 kg. A customer bought one slice of the cake and measured the mass of the sli

Brrunno [24]

Answer:

0.7549kg

Explanation:

The mass of the slice + mass of the remaining cake = total mass of cake.

mass of remaining cake = total mass of cake - the mass of the slice

total mass=0.870kg

mass of slice = 0.1151kg

mass of remaining cake = 0.870 - 0.1151

mass of remaining cake=0.7549kg

6 0

3 years ago

Kids are pelting a window with snowballs. On average, two snowballs of roughly 300-g mass hit the window each second, moving hor

hram777 [196]

Answer:

The average force exerted on the window due to two snowballs is 6 N

Explanation:

Given:

Mass of snowballs $m = 300 \times 10^{-3}$ Kg

Velocity of snowball $v = 10 \frac{m}{s}$

For finding the average force,

Force is equal to the change in momentum,

$F = \frac{dP}{dt}$

Here, final velocity is zero so we write,

$F = \frac{mv}{1}$

Where $dt = 1$ sec

$F = 300 \times 10^{-30} \times 10$

$F = 3$ N

Above value of force is due to one ball, but here given in question there are two ball,

$F = 3 \times 2$

$F = 6$ N

Therefore, the average force exerted on the window due to two snowballs is 6 N

5 0

3 years ago

Determine the magnitude of the effective value of g⃗ at a latitude of 60 ∘ on the earth. assume the earth is a rotating sphere.

dezoksy [38]

In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>

5 0

3 years ago

Water at 25°C and 1 atm is flowing over a long flat plate with a velocity of 14 m/s. Determine the distance from the leading edg

Gre4nikov [31]

Answer:

L=31.9 mm

δ = 0.22 mm

Explanation:

Given that

v= 14 m/s

ρ=997 kg/m³

μ= 0.891 × 10⁻3 kg/m·s

As we know that when Reynolds number grater than 5 x 10⁵ then flow will become turbulent.

$Re=\dfrac{\rho vL}{\mu}$

$L=\dfrac{Re\mu}{\rho v}$

$L=\dfrac{5\times 10^5\times 0.891\times 10^{-3}}{ 14 \times 997}\ m$

L=0.0319 m

L=31.9 mm

The thickness of the boundary layer at that location L given as

$\delta =\dfrac{5L}{\sqrt{Re}}$

$\delta =\dfrac{5\times0.0319}{\sqrt{5\times 10^5}}$

δ = 0.00022 m

δ = 0.22 mm

7 0

3 years ago

How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances

Tatiana [17]

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = $x+y=149.6\times 10^6\ m$

G = Gravitational constant

$M_s$ = Mass of Sun = $1.989\times 10^{30}$

$M_e$ = Mass of Earth = $5.972\times 10^{24}\ kg$

According to the question

$\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y$

$x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m$

The probe should be 258774.9441 m from Earth

7 0

3 years ago