1. Convert the following measurements to the units specified.<br> 2.5 days to seconds

A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near sid

igor_vitrenko [27]

Answer:

a) attractiva, b) dF = $k \frac{Q_1 \ dQ_2}{dx}$ , c) F = $k Q_1 \frac{Q_2}{d \ (d+L)}$ , d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

dF = $k \frac{Q_1 \ dQ_2}{dx}$

where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

∫ dF = $k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2$

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

λ = dQ₂ / dx

DQ₂ = λ dx

we substitute

F = $k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}$

F = k Q1 λ ( $-\frac{1}{x}$ )

we evaluate the integral

F = k Q₁ λ $(- \frac{1}{d+L} + \frac{1}{d} )$

F = k Q₁ λ $( \frac{L}{d \ (d+L)})$

we change the linear density by its value

λ = Q2 / L

F = $k Q_1 \frac{Q_2}{d \ (d+L)}$

d) we calculate the magnitude of F

F =9 10⁹ (-4.2 10⁻⁶) $\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}$

F = -1.09 N

the sign indicates that the force is attractive

3 0

3 years ago

34. (Double points) You help your friend construct a soap box derby car for the All-American Soap Box derby's Stock Division. He

Elden [556K]

To accelerate a 34.01 kg-car at 0.55 m/s², a force of 19 N will be required, according to Newton's Second Law of Motion.

<h3>What does Newton's Second Law of Motion state?</h3>

Newton's Second Law of Motion states that acceleration (a) happens when a force (F) acts on a mass (m).

We want a car of mass 34.01 kg to have an acceleration of 0.55 m/s². We can calculate the required force using Newton's Second Law of Motion.

F = m × a = 34.01 kg × 0.55 m/s² = 19 N

To accelerate a 34.01 kg-car at 0.55 m/s², a force of 19 N will be required, according to Newton's Second Law of Motion.

Learn more about Newton's Second Law of Motion here: brainly.com/question/25545050

#SPJ1

4 0

2 years ago

A 100 g aluminum calorimeter contains 250 g of water. The two substances are in thermal equilibrium at 10°C. Two metallic blocks

ipn [44]

Answer:

A. 1,950 J/kgºC

Explanation:

Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the copper and the unknown metal.

The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:

Q = c * m* Δt

where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.

In our case, we can write the following equality:

(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (ccu*mcu*Δtcu) + (cₓ*mₓ*Δtₓ)

Replacing by the givens , and taking ccu = 0.385 J/gºC and cAl = 0.9 J/gºC, we have:

Qg= 0.9 J/gºC*100g*10ºC + 4.186 J/gºC*250g*10ºC = 11,365 J(1)

Ql = 0.385 J/gºC*50g*55ºC + cₓ*66g*80ºC = 1,058.75 J + cx*66g*80ºC (2)

Based on all the previous assumptions, we have:

Qg = Ql

So, we can solve for cx, as follows:

cx = (11,365 J - 1,058.75 J) / 66g*80ºC = 1.95 J/gºC (3)

Expressing (3) in J/kgºC:

1.95 J/gºC * (1,000g/1 kg) = 1,950 J/kgºC

3 0

3 years ago

The box is pushed to the right with a force of 40 Newtons and it just begins to move what is the maximum static frictional force

Vesnalui [34]

Answer:

The maximum static frictional force is 40N.

Explanation:

When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).

This coefficient defines the maximum static friction force that we can have.

So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.

In this case, we know that we apply a force of 40N and the object just starts to move.

Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.

3 0

3 years ago

A body is at equilibrium under the action of three forces. One force is 10N acting due east and one is 5N in the direction 60° n

olchik [2.2K]

Answer:

If the body is in equilibrium the two forces add up and the third is the opposite of the resultant.

F(1x)=F(1)=10 N

F(2x)=F(2)cos60=5•0.5=2.5 N

F(2y) =F(2)sin60 = 5•0.866= 4.33 N

F(3x) =- F(x)=- (10+2.5 )= -12.5 N

F(3y) =- F(2y)= - 4.33 N

F(3) = sqrt{ F(3x)²+F(3y)²} =13.23 N

tan φ = F(3y)/F(3x) =4.33/12.5=0.364

φ = 19.1⁰ (south-west)

Explanation:

7 0

3 years ago

1. Convert the following measurements to the units specified. 2.5 days to seconds