List and describe 3 conventional heating systems

According to the general theory of relativity, what are consequences of the curvature of space-time? Check all that apply. The d

Svetlanka [38]

Answer:

The dilation of time.

The falling of objects.

The changing of paths of light.

Explanation:

I have explained in the image attached below.

From the explanation, the correct ones are;

The dilation of time.

The falling of objects.

The changing of paths of light.

4 0

3 years ago

What is the change in internal energy if 30 J of thermal energy is released

alexira [117]

Answer:

ΔU = -70 J

Explanation:

ΔU = Q − W

where ΔU is the change in internal energy,

Q is the heat absorbed by the system,

and W is the work done by the system (on the surroundings).

30 J of thermal energy is released, so Q = -30 J.

40 J of work is done by the system, so W = 40 J.

Therefore, the change in internal energy is:

ΔU = -30 J − 40 J

ΔU = -70 J

6 0

3 years ago

4. Johnny exerts a 3.55 N rightward force on a 0.200-kg box to accelerate it across a low-friction track. If the total resistanc

muminat

a) $15.2 m/s^2$

b) 1.96 N

c) 1.96 N

Explanation:

a)

To find the acceleration of the box, we apply Newton's second law of motion:

$\sum F=ma$

where

$\sum F$ is the net force on the box

m is the mass of the box

a is its acceleration

Here we have to consider the horizontal direction, which is the one in which the box is moving. The net force is given by:

$\sum F=3.55 N - 0.52 N=3.03 N$

which is the difference between the forward force and the resistive force. Then we have

m = 0.200 kg (mass of the box)

Therefore, the acceleration of the box is:

$a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2$

b)

The gravitational force (also called weight) is the force with which an object is pulled by the Earth towards the Earth's centre.

It is given by

$F_g = mg$

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

m = 0.200 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration due to gravity

Therefore, the gravitational force on the box is:

$F_g=(0.200)(9.8)=1.96 N$

c)

The normal force is the force with which a surface pushes back on an object.

For an object lying on a flat surface, there are two forces acting along the vertical direction:

- The gravitational force, $F_g$ , which pushes the object downward

- The normal force, N, which pushes the object upward

As the object is at rest, the vertical acceleration of the object is zero, therefore according to Newton's second law of motion, the net force must be zero:

$\sum F=F_g-N=0$

Which means that the normal force is equal to the gravitational force:

$N=F_g$

And so, for the box in this problem, the normal force is

$N=1.96 N$

6 0

4 years ago

Junior or sophomore guys you can hit me up if ur bored or need anything

rodikova [14]

Does Physics involve science? Because I think there's a chemical bond between us. :3

4 0

3 years ago

An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens

Pavlova-9 [17]

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

$\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

Putting the values

$\frac{1}{v} +\frac{1}{15} =\frac{1}{14}$

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

$\frac{210}{15} \times \frac{D}{f_e}$

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

7 0

3 years ago