If an object is not moving, what’s the net force?

How much heat is required to convert 2.55g of water at 28 degrees c to steam?

pantera1 [17]

There are two different processes here:
1) we must add heat in order to bring the temperature of the water from $28^{\circ}C$ to $100^{\circ}C$ (the temperature at which the water evaporates)
2) other heat must be added to make the water evaporates

1) The heat needed for process 1) is
$Q_1=m C_s \Delta T$
where
$m=2.55 g$ is the water mass
$C_s = 4.18 g/J^{\circ}C$ is the water specific heat
$\Delta T=100^{\circ}C-28^{\circ}C=72^{\circ}C$ is the variation of temperature of the water
If we plug the numbers into the equation, we find
$Q_1 = (2.55 g)(4.18 J/g^{\circ}C)(72^{\circ}C)=767.4 J$

2) The heat needed for process 2) is
$Q=m L_e$
where
$m=2.55 g$ is the water mass
$L_e = 2264.7 J/g$ is the latent heat of evaporation of water
If we plug the numbers into the equation, we find
$Q_2=(2.55 g)(2264.7 J/g)=5775.0 J$

So, the total heat needed for the whole process is
$Q=Q_1+Q_2=767.4 J+5775.0 J=6542.4 J$

4 0

3 years ago

A 20.0-kg uniform plank (10.0 m long) is supported by the floor at one end and by a vertical rope at the other as shown in the f

Nataly [62]

Answer:

Tension= 475N

Force= 225N

Explanation:

The question is not complete, here is the complete question

Also, see attached a free body diagram for your reference

<em>"A 20.0-kg uniform plank is supported by the floor at one end by a vertical rope at the other as shown in the figure. A 50.0-kg mass person stands on the plank a distance three-fourths of the length plank from the end on the floor. </em>

<em>a. What is the tension in the rope? </em>

<em>b. What is the magnitude of the force that the floor exerts on the plank?"</em>

<em />

given data

mass of man=50kg

mass of plank=20kg

length of plank=10m

<em />

let us make the lenght of the rope be d

The torque about the floor

That is taking moment about the floor

$N*0+T*d=20*10*d/2 + 50*10*3d/4\\\\T=100+375=475N\\\\$

Force will be also zero

$N+T=20*10+50*10\\\\N+T=700 \\\\N=700-475=225 newtons\\\\N+T=20*10+50*10\\\\N+T=700\\\\N=700-475=225newtons$

<em />

7 0

4 years ago

A rock is thrown upwards and has a max altitude of 40 m. What was its initial velocity?

Natali [406]

Recall the formula,

${v_f}^2-{v_i}^2=2a\Delta x$

where $v_i$ and $v_f$ are the rock's initial and final velocities, respectively; $a$ is its acceleration; and $\Delta x$ is the displacement it undergoes.

At any point during its motion, the rock is subject to gravity, so $a=-g$ , where $g=9.80\frac{\rm m}{\mathrm s^2}$ . At its maximum height, the rock has zero vertical velocity, and if we take its starting height to be the origin, we have $\Delta x=x_{\rm max}$ .

So,

$0^2-{v_i}^2=-2gx_{\rm max}\implies-{v_i}^2=-2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(40\,\mathrm m\implies\boxed{v_i=28\dfrac{\rm m}{\rm s}}$

5 0

3 years ago

Jonny is pushing a 50-kg large package along the dorm room floor with 150 N force. Both the direction of push and the floor are

eduard

a. The push is 150 N and acts to the right while the frictional force is 150 N and acts to the left.

b. The weight is 490 N and acts downwards while the normal force is 490 N and acts upwards.

c. The package moves 1.8 m

d. The package's acceleration is 0 m/s²

<h3>a. The horizontal forces</h3>

The push is 150 N and acts to the right while the frictional force is 150 N and acts to the left.a

Since the direction of push and the floor is horizontal, and first horizontal force acting on the package is the push and its magnitude is 150 N.

Also, a frictional force also acts to oppose the motion of the package.

Since the packge moves at a constant velocity of 0.6 m/s, its acceleration is zero and thus the net force on the package is zero.

Let

F = push force and
f = frictional force

So, F - f = 0

F = f

= 150 N

So, the frictional force is 150 N and opposite to the push.

So, the push is 150 N and acts to the right while the frictional force is 150 N and acts to the left.

<h3>b Vertical forces on package</h3>

The weight is 490 N and acts downwards while the normal force is 490 N and acts upwards.

Since the floor is horizontal, the vertical forces that act on the package are its weight and the normal force due to the ground.

The direction of the weight is downwards while the direction of the normal force is upwards.

Since the floor is horizontal and the package does not move in the vertical direction, the net vertical force is zero.

Let W = weight of package = mg where
m = mass of package = 50 kg and
g = acceleration due to gravity = 9.8 m/s² and
N = normal force

So, the net force N - W = 0

N = W

= mg

= 50 kg × 9.8 m/s²

= 490 N

So, the weight is 490 N and acts downwards while the normal force is 490 N and acts upwards.

<h3>c. Distance package moves</h3>

The package moves 1.8 m

Since distance, d = vt where

v = velocity = 0.6 m/s and
t = time = 3 s

So, d = vt

= 0.6 m/s × 3 s

= 1.8 m

So, the package moves 1.8 m

<h3>d. The package's acceleration</h3>

The package's acceleration is 0 m/s²

Since the net force on the package is zero, its acceleration is also zero. Since force, F = ma where

m = mass of package and
a = acceleration of package

Since F = 0,

ma = 0

a = 0

So, the package's acceleration is 0 m/s²

Learn more about force here:

brainly.com/question/25239010

6 0

3 years ago

Can you put this in your own words plz.

strojnjashka [21]

Xenon can be used for things like high pressure arc lamps for motion picture projection, in high pressure arc lamps to produce ultraviolet light and photographic flashes. It is also used to help the radiation detection, for example X-ray counters.

hope it helps

6 0

4 years ago

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