One person can hold mutiple types of authority at a time<br><br><br> T/F

two boxes sit on a frictionless surface and are in contact with one another. the first box has a mass of 7 kg and the second box

egoroff_w [7]

The acceleration of the boxes depends on the mass and weight.

we have a mass of 7 and 8 kilograms

if it took 25 N force to move box A, then you would take 25 and multiply by 8 then divide by 2.

It will leave you with 100 N.

finally take the sq rt of 100 to get 10

7 0

3 years ago

Renewable resources need to be conserved because

creativ13 [48]

Answer:

(A) We are using them faster than they are replenished by nature

4 0

3 years ago

The Displacement is 5m. We found that using the

Llana [10]

Answer:

A vector can be written as:

(R, θ)

Where R is the magnitude, in this case, we know that the magnitude of the displacement is 5m

Then:

R = 5m

and θ defines the direction, it's an angle measured from the positive x-axis.

(In the image, θ would be the angle located at the point A)

Now, if you look at the image, you can see a triangle rectangle.

Where the adjacent cathetus has a length of 4,

the opposite cathetus has a length of 3 units

the hypotenuse has a length of 5 units.

So we can use any trigonometric rule to find the value of θ, like:

sin(θ) = (opposite cathetus)/hypotenuse

Then:

sin(θ) = 3m/5m

Now we can use the inverse sin function, Asin(x), in both sides

Asin( sin(θ)) = θ = Asin( 3/5) = 36.87°

then the vector is:

(5m, 36.87°)

Now, if we define the positive y-axis as the North, and the positive x-axis as the East.

This vector would point at 36.87° North of East.

(or almost Northeast)

7 0

2 years ago

To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv

Lesechka [4]

Answer:

E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

λ = Q / L

If we derive from the length we have

λ = dq/dx ⇒ dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

dE = k dq / x²2

dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

E = k $\int\limits^{d+L}_d {\lambda/x^{2}} \, dx$

We take out the constant magnitudes and perform the integral

E = k λ (-1/x) ${(-1/x)}^{d+L} _{d}$

Evaluating

E = k λ [ 1/d - 1/ (d+L)]

Using λ = Q/L

E = k Q/L [ 1/d - 1/ (d+L)]

let's use a bit of arithmetic to simplify the expression

[ 1/d - 1/ (d+L)] = L /[d(d+L)]

The final result is

E = k Q / [d(d+L)]

3 0

3 years ago

On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo

ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

$\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i$

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

$\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i$

42 m/s, negative direction (to the west).

5 0

3 years ago

One person can hold mutiple types of authority at a time T/F