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2 years ago

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A charge of 2 c is at the origin. when charge q is placed at 2 m along the positive x axis, the electric field at 2 m along the

negative x axis becomes zero. what is the value of q?

1 answer:

docker41 [41]2 years ago

4 0

The value of Q will be -8 C.

In the presence of an electric or magnetic field, matter experiences a force due to its electric charge.

A moving electric charge generates a magnetic field, and an electric charge has an accompanying electric field.

The information provided in the issue is;

The separation between and is 2m.

The separation between and is 2m.

An origin charge equals +2 C

The electric fields are identical in magnitude but are facing in different directions. As a result, the following relationship can be used

Q/16=1/2

The value of Q will be -8 C.

Learn more about electric charge here

brainly.com/question/8163163

#4174

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A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

so

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$

The change in momentum of the astronaut is

$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

And the duration of the push is

$\Delta t = 0.600 s$

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

3 0

3 years ago

Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co

patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of $^{3890}Sr\, t_{\frac{1}{2}}$ = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

$\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where

N = No. of nuclei left after time t

$N_{o}$ = No. of nuclei initially started with

$\frac{N}{N_{o}} = 1\times 10^{- 4}$

(Since, 100% - 99.99% = 0.01%)

Thus

$1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}$

Taking log on both the sides:

$- 4 = \frac{t}{28.5}log\frac{1}{2}$

$t = \frac{-4\times 28.5}{log\frac{1}{2}}$

t = 387.69 yrs

5 0

3 years ago

A pizza delivery driver must make three stops on her route. She will first leave the restaurant and travel 4 km due north to the

topjm [15]

<h2>5.3 km</h2>

Explanation:

This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.

Let us denote each of the individual displacements by a vector. Consider the unit vectors $\vec{i}\textrm{ and }\vec{j}$ as the unit vectors in the direction of East and North respectively.

By simple calculations, we can derive the unit vectors $\vec{j},\frac{-\vec{i}-\vec{j}}{2}\textrm{ and }\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2}$ in the directions North, $45^{o}$ South of West and $60^{o}$ North of West respectively.

So Total displacement vector = Sum of individual displacement vectors.

Displacement vector = $4(\vec{j})+6(\frac{-\vec{i}-\vec{j}}{2})+5(\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2})=-4.25\vec{i}+3.165\vec{j}$

Magnitude of Displacement = $|-4.25\vec{i}+3.165\vec{j}|=5.3km$

∴ Total displacement = $5.3km$

4 0

3 years ago

Needddd helppppppp!!!

yulyashka [42]

Answer:

2/9 times as strong.

Explanation:

From the question given above, the following assumptions were made:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

Next, we shall obtain an expression to determine the new force. This can be obtained as follow:

F = GMm / r²

Cross multiply

Fr² = GMm

Divide both side by Mn

G = Fr² / Mm

Since G is constant, then we have

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Finally, we shall determine the new force as follow:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Fr² / Mm = F₂ × (3r)² / 2M × m

Fr² / Mm = F₂ × 9r² / 2Mm

Cross multiply

Fr² × 2Mm = F₂ × 9r² × Mm

Divide both side by 9r² × Mm

F₂ = Fr² × 2Mm / 9r² × Mm

F₂ = F × 2 / 9

F₂ = 2/9 F

Thus, the new force is 2/9 times the original force i.e 2/9 times as strong.

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3 years ago

Tell the order of vessels a blood Cell travels through starting from the heart.

Natalka [10]

Arterie
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phloprant
veins
piscas

7 0

3 years ago