The temperature of 5 pounds of water is 40 degrees. Btus are added until the temp of

Sveta_85 [38]

Answer: The first one

Explanation:

8 0

3 years ago

Two small identical conducting spheres are placed with their centers 0.41 m apart. One is given a charge of 12 ✕ 10−9 C, the oth

nataly862011 [7]

(a) $-1.48\cdot 10^{-5}N$

The electrostatic force exerted between the two sphere is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the charges on the two spheres

r is the separation between the centres of the two spheres

In this problem,

$q_1 = 12\cdot 10^{-9} C\\q_2 = -23\cdot 10^{-9} C\\r = 0.41 m$

Substituting these values into the equation, we find the force

$F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(12\cdot 10^{-9}C)(-23\cdot 10^{-9} C)}{(0.41 m)^2}=-1.48\cdot 10^{-5}N$

And the negative sign means the force is attractive, since the two spheres have charges of opposite sign.

(b) $+1.62\cdot 10^{-6}N$

The total net charge over the two sphere is:

$Q=q_1 +q_2 = 12\cdot 10^{-9}C+(-23\cdot 10^{-9}C)=-11\cdot 10^{-9} C$

When the two spheres are connected, the charge distribute equally over the two spheres (since they are identical, they have same capacitance), so each sphere will have a charge of

$q=\frac{Q}{2}=\frac{-11\cdot 10^{-9}C}{2}=-5.5\cdot 10^{-9}C$

So the electrostatic force between the two spheres will now be

$F=k\frac{q^2}{r^2}$

And substituting numbers, we find

$F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(-5.5\cdot 10^{-9} C)^2}{(0.41 m)^2}=+1.62\cdot 10^{-6}N$

and the positive sign means the force is repulsive, since the two spheres have same sign charges.

7 0

3 years ago

. Draw a Cartesian coordinate system on a sheet of paper. On this Cartesian coordinate system, draw each vector to scale, starti

Triss [41]

Answer:

a) the first vector has magnitude 58 cm and the angle is 15 measured clockwise from the positive side of the x-axis

b) the second vector, the magnitude is 55.7 cm and the angle is 35 half from the negative side of the x-axis in a clockwise direction

c) the magnitude is 54.2 cm with an angle of 18 measured counterclockwise from the x-axis

Explanation:

For this exercise we draw a Cartesian coordinate system in this system: East coincides with the positive part of the x-axis and North with the positive part of the y-axis.

a) the first vector has magnitude 58 cm and the angle is 15 measured clockwise from the positive side of the x-axis

b) the second vector, the magnitude is 55.7 cm and the angle is 35 half from the negative side of the x-axis in a clockwise direction

c) the magnitude is 54.2 cm with an angle of 18 measured counterclockwise from the x-axis

In the attachment we can see the representation of the three vectors

8 0

3 years ago

A force of 30 N is applied tangentially to the rim of a solid disk of radius 0.10 m. The disk rotates about an axis through its

den301095 [7]

Answer:

m = 4.0 Kg

Explanation:

Using an analogy with the Newton's 2nd law for point masses, for rigid bodies, the external net torque on a rigid body, is equal to its rotational inertia (I), times the angular acceleration (α) of the body, as follows:

$\tau_{ext} = I* \alpha (1)$

Since the magnitude of the torque is the product of the value of the force times the perpendicular distance between the line of action of the force and the axis of rotation, and the force is tangential to the rim of the disk, we can write the following expression:

$\tau = F*r*sin 90 = F*r (2)$

For a solid disk, the rotational inertia regarding an axis through its center, and perpendicular to its face is as follows:

$I = \frac{m*r^{2}}{2} (3)$

Replacing (3) in (1), and (2) in the left side of (1) also, we can solve for m, as follows:

$m = \frac{2*30.0N}{0.1m*150(1/s2)} = 4.0 Kg (4)$

So, the mass of the disk is 4.0 Kg.

7 0

3 years ago

5) Why does the first hill/drop on a roller coaster have to be the highest? (think about

prohojiy [21]

5) Due to the conservation of energy / due to the presence of friction

6) The work done is 4000 J

7) The distance travelled by the object is 4 m

8) The potential energy of the bird is 2450 J

Explanation:

5)

For a roller coaster in motion along the track, in absence of friction, the mechanical energy remains constant:

$E=U+K = const.$

where

U is the potential energy

K is the kinetic energy

At the beginning, the roller coaster is at rest, therefore its kinetic energy is zero and its mechanical energy is just equal to the potential energy:

$E=U=mgh$

where m is the mass of the roller coaster, g the acceleration of gravity, and h the height of the roller coaster above the ground at the first hill.

Since this amount of energy remains constant, the roller coaster cannot go to a higher hill: in fact, in that case it would reach a height $h'>h$ , but this is not possible, because it would mean that its mechanical energy would be $mgh' > E$ , larger than its mechanical energy, and since energy cannot be created (law of conservation of energy), the height of the next hills cannot be higher than the first one. Moreover, if we consider friction, then friction does work on the roller coaster in motion, and as a result, part of its mechanical energy is wasted (converted into thermal energy): therefore, the height of the following hills must be lower than the height of the first hill.

6)

The work done when lifting an object is equal to the gravitational potential energy gained by the object:

$W=Fh$

where

F is the weight of the object

h is the change in height of the object

For the object in this problem, we have

F = 200 N (weight)

h = 20 m (change in height)

Substituting, we find the work done:

$W=(200)(20)=4000 J$

7)

The work done when pushing an object in a direction parallel to the motion of the object is given by:

$W=Fd$

where

F is the magnitude of the force

d is the distance through which the object has travelled through

In this problem, we have:

W = 300 J (work done)

F = 75 N (magnitude of the force)

Solving the equation for d, we find the distance travelled:

$d=\frac{W}{F}=\frac{300}{75}=4 m$

8)

The potential energy of an object is the energy possessed by an object due to its position in a gravitational field. Near the Earth's surface, it is given by

$PE=mgh$