What is the effect of changing the wire in a circuit from a straight thick wire to a straight thin wire?

As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -

postnew [5]

Answer:

-2.3 × 10^-9 Coulombs(C).

Explanation:

So, we are given the following data or information or parameters that is going to help us to solve the problem effectively and efficiently;

=> " the shuttle's potential is typically changed by -1.4 V during one revolution. "

=> " Assuming the shuttle is a conducting sphere of radius 15 m".

So, in order to estimate the value for the charge we will be making use of the equation below:

Charge, C =( radius × voltage or potential difference) ÷ Coulomb's law constant.

Note that the value of Coulomb's law constant = 9 x 10^9 Nm^2 / C^2.

So, charge = { 15 × (- 1.4)} / 9 x 10^9 Nm^2 / C^2.

= -2.3 × 10^-9 Coulombs(C).

4 0

2 years ago

ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 660.0 kg and was trav

Montano1993 [528]

Answer:

vₐ₀ = 29.56 m / s

Explanation:

In this exercise the initial velocity of car A is asked, to solve it we must work in parts

* The first with the conservation of the moment

* the second using energy conservation

let's start with the second part

we must use the relationship between work and kinetic energy

W = ΔK (1)

for this part the mass is

M = mₐ + m_b

the final velocity is zero, the initial velocity is v

friction force work is

W = - fr x

the negative sign e because the friction forces always oppose the movement

we write Newton's second law for the y-axis

N -W = 0

N = W = Mg

friction forces have the expression

fr =μ N

fr = μ M g

we substitute in 1

-μ M g x = 0 - ½ M v²

v² = 2 μ g x

let's calculate

v² = 2 0.750 9.8 6.00

v = ra 88.5

v = 9.39 m / s

Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.

Initial instant. Before the crash

p₀ = + mₐ vₐ₀ - m_b v_{bo}

instant fianl. Right after the crash, but the cars are still not moving

p_f = (mₐ + m_b) v

p₀ = p_f

+ mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v

mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}

let's reduce to the SI system

v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s

let's calculate

660 vₐ₀ = (660 +490) 9.39 + 490 17.778

vₐ₀ = 19509.72 / 660

vₐ₀ = 29.56 m / s

we can see that car A goes much faster than vehicle B

5 0

2 years ago

a package was determined to have a mass of 5.7kilograms. what's the force of gravity acting on the package on earth?(a)37.93N(b)

Nana76 [90]

The force of gravity on earth is 9.807 m/s^2 (or meters per second per second).
To determine the force applied, multiply the mass of the package (5.7 kg) by the force of gravity on Earth (9.807 m/s^2).
5.7 x 9.807 = 55.86 N The answer is D.

Note: the actual force is 55.89 Newtons.

8 0

2 years ago

Read 2 more answers

A man lifts a 120 kg barbell 2 m above the ground . What is the gain in gravitational PE of the barbell?

SpyIntel [72]

Answer:

2,352 Joules

Explanation:

At the ground, the barbell has a classical mechanical energy value of zero. There is no classical kinetic or potential energy for the barbell. The moment the man starts to lift the barbell, he does work on the barbell and transfers kinetic energy to it due to the motion. At its maximum height where the man lifts the barbell to a stop, the kinetic energy is zero because it transformed into gravitational potential energy stored in the gravitational field. Our reference point for potential was defined to be zero at the floor, therefore we can say that the gravitational potential energy at 2 meters is:

$U=mgh=(120kg)(9.8m/s^2)(2m)=2,352J$

7 0

1 year ago

You decide to clean the bathroom. You notice that the shower is covered in a strange green slime. You decide to try to get rid o

topjm [15]

The lemonade juice and water.

6 0

2 years ago