At atmospheric pressure, what is the characteristic boiling point of water,in degrees Celsius

boyakko [2]

Explanation:

Pascal's principle, also called Pascal's law, in fluid (gas or liquid) mechanics, statement that, in a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container.

3 0

3 years ago

A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.

GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is $8.75 m/s^2$ in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, $F_f = 15 N$ , pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, $W=mg$ , where m is the mass of the book and $g=9.8 m/s^2$ is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

$W=(4)(9.8)=39.2 N$

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

$N-W=ma$

where a is the acceleration; however, since N = W, this becomes

$a=0$

And since the book is initially at rest on the desk, this means that there is no motion.

c)

We said there are two forces acting in the horizontal direction:

- The applied force, F = 50 N, forward

- The frictional force, $F_f = 15 N$ , backward

Since they act along the same line, we can calculate their resultant as

$\sum F = F - F_f = 50 - 15 = 35 N$

and therefore the net force is 35 N in the forward direction.

d)

The net force is obtained as the resultant of the net forces in the horizontal and vertical direction. However, we have:

- The net force in the horizontal direction is 35 N

- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction

Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.

e)

The acceleration of the book can be calculated by using Newton's second law:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

Here we have:

$\sum F = 35 N$ (in the forward direction)

m = 4 kg

Therefore, the acceleration is

$a=\frac{\sum F}{m}=\frac{35}{4}=8.75 m/s^2$ (forward)

Learn more about forces, weight and Newton's second law:

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#LearnwithBrainly

8 0

3 years ago

ver shines light up to the surface of a flat glass-bottomed boat at an angle of 30 relative to the normal. If the index of refr

Free_Kalibri [48]

Answer:

$\beta = 41.68°$

Explanation:

according to snell's law

$\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }$

refractive index of water n_w is 1.33

refractive index of glass n_g is 1.5

$sin\alpha = \frac{n_w}{n_g}* sin30$

$sin\alpha = 0.443$

now applying snell's law between air and glass, so we have

$\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}$

$sin\beta = \frac{n_g}{n_a} sin\alpha$

$\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha]$

we know that $sin\alpha = 0.443$

$\beta = 41.68°$

7 0

3 years ago

Energy changes in generator ?

Firdavs [7]

What do you mean? because yes the energy is converted into electricity but the question isn't specific

7 0

3 years ago

Read 2 more answers

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

IRINA_888 [86]

Answer:

Time : 7.96 s

Distance Traveled : 357.8 m

Explanation:

In order to solve this problem, we first consider the accelerated motion of rocket. We will be using the subscript 1 for accelerated motion.

So, for accelerated motion, we have:

Acceleration = a₁ = 14.5 m/s²

Time Period = t₁ = 3.1 s

Initial Velocity = Vi₁ = 0 m/s (Since, it starts from rest)

Final Velocity = Vf₁

Distance covered by sled during acceleration motion = s₁

Now, using 1st equation of motion:

Vf₁ = Vi₁ + (a₁)(t₁)

Vf₁ = 0 m/s + (14.5 m/s²)(3.1 s)

Vf₁ = 44.95 m/s

Now, using 2nd equation of motion:

s₁ = (Vi₁)(t) + (0.5)(a₁)(t₁)

s₁ = (0 m/s)(3.1 s) + (0.5)(14.5 m/s²)(3.1 s)

s₁ = 22.5 m

Now, we first consider the decelerated motion of rocket. We will be using the subscript 2 for decelerated motion.

So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

Time Period = t₂ = ?

Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

0 m/s = 44.95 m/s + (- 5.65 m/s²)(t₂)

t₂ = (44.95 m/s)/(5.65 m/s²)

t₂ = 7.96 s

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

Thus, the total distance covered by sled will be:

Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

S = 357.8 m

7 0

3 years ago