A 50 g mass is freely hanging from a horizontal meter stick at a distance of 99 cm from the pivot. Calculate the weight force W that the mass exerts on the meter stick? ·Calculate the torque that the weight force exerts about the pivot? 4. What does the pivot exert on the meter stick?

When a substance undergoes fusion, it melts.

Fusion is a physical process. The process of fusion of substances takes place during the conversion of a solid substance into a liquid.

For this process to take place, heat is being applied to the solid substance, and the particles of the solid loose, then move freely thereby reducing the intermolecular forces in the solid substance before changing into liquid.

Learn more about physical changes here:

https://brainly.com/question/25014732?referrer=searchResults

Answer:

gravitational potential energy:

GPE = m g h

kinetic energy:

KE = 1/2 m v^2

Answer:

[tex]\boxed{\bold { \large { \boxed {KE=\frac{1}{2} mv^2 \ , \ PE=mgh}}}}[/tex]

Explanation:

Kinetic energy formula

[tex]\displaystyle KE=\frac{1}{2} mv^2[/tex]

Potential energy formula

[tex]\displaystyle PE=mgh[/tex]

[tex]\displaystyle KE \Rightarrow \sf kinetic \ energy \ (J)[/tex]

[tex]\displaystyle PE \Rightarrow \sf potential \ energy \ (J)[/tex]

[tex]\displaystyle m \Rightarrow \sf mass \ (kg)[/tex]

[tex]\displaystyle v \Rightarrow \sf velocity \ (m/s)[/tex]

[tex]\displaystyle g \Rightarrow \sf acceleration \ of \ gravity\ (m/s^2)[/tex]

[tex]\displaystyle h \Rightarrow \sf height \ (m)[/tex]

Answer:I need the answer pls

Explanation:

I don't have one

Answer:

The answer is D, what the cell parts are.

Explanation:

Answer:

the thickness of the glass divided by thickness of water is going to be 1.333 divided by 1.52, which is 0.877. So, the height of this glass, in order to have the same number of wavelengths as in water, the height of the glass will be 0.877 times the height of the water, and so it will be smaller.

Answer:

a)  P = 4.03 10⁵ Pa, b)  P = 4.03 10⁵ Pa

Explanation:

a) The pressure as a function of the depth of a fluid is

          P =[tex]P_{atm}[/tex] + ρ g y

where Patm is the atmospheric pressure, the sea density of about 1025 kg / m³

let's calculate

         P = 1.01825 10⁵ + 1025 9.8 30

         P = 4.03 10⁵ Pa

b) When the hood is submerged, the water exerts a perpendicular force on the entire surface, in the equilibrium position, the air is compressed by this force until the pressure it exerts is equal to the external pressure (open at the lower), therefore the air pressure is

        P = 4.03 105 Pa

"[tex]3.958\times 10^5 \ Pa[/tex]" would be the pressure outside the bell .as well as the air pressure inside it. A further explanation is below.

According to the question,

Depth,

Pressure,

             [tex]= 1.018\times 10^{5} \ Pa[/tex]

Now,

→ The water pressure outside the bell will be:

= [tex]P_o +p\times g\times d[/tex]

By putting the values, we get

= [tex]1.018\times 10^5+1000\times 9.8\times 30[/tex]

= [tex]3.958\times 10^5 \ Pa[/tex]

And inside the air pressure will be same as the water pressure.

Thus the response above is right.

Learn more:

https://brainly.com/question/22786261

Answer:

42

Explanation:

42

Answer:

energy

Explanation:

im sure

Answer:

The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.

Explanation:

We can determine the final velocity of the skier at the bottom of the ramp by Principle of Energy Conservation and Work-Energy Theorem, whose model is:

[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+W_{disp}[/tex] (1)

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.

[tex]W_{disp}[/tex] - Work dissipated by friction, measured in joules.

By definitions of gravitational potential and translational kinetic energy and work, we expand and simplify the model:

[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2}) =\mu_{k}\cdot N\cdot \Delta s[/tex] (2)

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the skier, measured in meters.

[tex]N[/tex] - Normal force from the incline on the skier, measured in newtons.

[tex]\Delta s[/tex] - Distance covered by the skier, measured in meters.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.

The normal force exerted on the skier and the covered distance are, respectively:

[tex]N = m\cdot g\cdot \cos \theta[/tex] (3)

[tex]\Delta s = \frac{z_{1}-z_{2}}{\sin \theta}[/tex] (4)

Where [tex]\theta[/tex] is the angle of the incline above the horizontal, measured in sexagesimal degrees.

By applying (3) and (4) in (2), we get that:

[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot m\cdot g \cdot \cos \theta \cdot \left(\frac{z_{1}-z_{2}}{\sin \theta} \right)[/tex]

[tex]g\cdot (z_{1}-z_{2}) +\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})= \mu_{k}\cdot g \cdot \left(\frac{z_{1}-z_{2}}{\tan \theta} \right)[/tex] (5)

Then, we clear the velocity of the skier at the bottom of the ramp is: ([tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.1[/tex], [tex]\theta = 40^{\circ}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}-z_{2} = 40\,m[/tex])

[tex]\left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = \frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex]

[tex]2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = v_{1}^{2}-v_{2}^{2}[/tex]

[tex]v_{2} = \sqrt{v_{1}^{2}-2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2})}[/tex] (6)

[tex]v_{2} = \sqrt{\left(0\,\frac{m}{s} \right)^{2}-2\cdot \left(\frac{0.1}{\tan 40^{\circ}} -1\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m)}[/tex]

[tex]v_{2} \approx 26.288\,\frac{m}{s}[/tex]

The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.

7.53 m

We are given:

Initial Horizontal Velocity of the Ball = 4.6 m/s

Initial Vertical Velocity of the Ball = 0 m/s

Height from which ball is kicked = 13.4 m

Time taken by the ball to reach the ground:

The ball has an initial vertical velocity of 0 m/s

it also has a downward acceleration of 10 m/s² due to gravity

Solving for the time taken:

s = ut + 1/2(at²)                 [second equation of motion]

replacing the values

13.4 = (0)(t) + 1/2 (10)(t²)

13.4 = 5t²

t² = 13.4/5                  [dividing both sides by 5]

t² = 2.68

t = 1.637 seconds     [taking the square root of both sides]

Horizontal distance covered by the ball:

Since there are no horizontal opposing forces on the ball,

the ball will more horizontally at a velocity of 4.6 m/s until it hits the ground

We calculated that the ball will hit the ground in 1.637 seconds

Distance covered:

s = ut + 1/2 (at²)                            [seconds equation of motion]

s = ut                                            [since a = 0m/s² in the horizontal plane]

replacing the values

s = 4.6 * 1.637

s = 7.53 m

Hence, the ball landed 7.53 m from the cliff

Answer:

Explanation:

Given

x component = +3m

y component = +4m

Required

The resultant vector has a magnitude

This can be gotten using the pythagras theorem as shown;

R = √x²+y²

R = √3²+4²

R = √9+16

R = √25

R = 5 m

Hence the resultant vector has a magnitude of 5m

Answer:

2.00 A

Explanation:

Total resistance = 10 + 20 + 30 = 60 ohms

Potential difference (V) = 120volts

Current (I) =?

from Ohms law V = IR

==> I = V/R = 120/60 = 2 A

Note: for resistors in series equal amount of electric current flows through the circuit

Answer:

10

Explanation:

10 is answer because velocity is time /position. So time is 50 and position is 5

Answer:

3.49m/s

Explanation:

We have mess of kit = 4kg

Then pieces are 2kg

3m/s due north

5m/s, 30⁰ due south

MiVi = m1v1 + m2v2

We break velocities to have x and y components

MiVi = m1v1(cosθ+sin θ1j) + m2v2(v2cos θ2i+sin2 θj)

θ1 = 90⁰north

θ2 = 30⁰ north east

Vi = 2/4x3m/s(cos90i + sin90j)+2/4x5m/s(cos30i + sin30j)

= 2.16i + 2.75j

Vi = √Vx²+Vy²

Vi = √ (2.16)²+(2.75)²

Vi = 3.49m/s

This is the original speed of the kit

Explanation:

Using the formula;

2x = vt

x is the distance up from the ocean floor the submarine is

v is the speed of sound in water

t is the time

Given

t = 2.3s

v = 1490m/s

Required

how high (approx) up from the ocean floor is the submarine x

From the formula;

x = vt/2

x = 1490(2.3)/2

x = 745(2.3)

x = 1,713.5m

Hence the submarine is 1713.5m high up from the ocean floor

Answer:

Explanation:

Step one:

given data

speed of bus= 35km/h forward

speed of ball= 4km/h backwards

`Step two:

Required

the magnitude of the velocity of the ball relative to the ground is the net velocity

that is

=35-4

=31km/h

The velocity of the ball relative to the ground is 31km/h

Answer:

Potential energy is the energy that is stored in an object. ... When you park your car at the top of a hill, your car has potential energy because the gravity is pulling your car to move downward; if your car's parking brake fails, your vehicle may roll down the hill because of the force of gravity.

Answer:

A

Explanation:

Hopefully this helps.

Answer:

1. A. 8.22. m/s

Explanation:

Answer:

960.4N

Explanation:

Given parameter:

Mass of the body  = 98kg

Unknown:

Force exerted by gravity  = ?

Solution:

Force exerted by gravity on the body is the weight

  Weight  = force x acceleration due to gravity

Acceleration due to gravity = 9.8m/s²

  Weight  = 98kg x 9.8m/s²

  Weight  = 960.4N

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, to determine the bicyclist that traveled the fastest at the end of the race, the speed of the bicyclists at the end of the race will determine this (not the bicyclist that came first nor there overall speed). The speed of the bicyclist at the end of the race can be determined by using the formula below

s = d ÷ t

Where s is the speed of each bicyclist at the end of the race

d is the specific distance covered by the bicyclist at the end of the race

t is the time taken for the bicyclist to complete that distance

It should be noted that to get an accurate result, the distance covered at the end of the race must be the same for all the bicyclists.

Answer:

0.1875 m/s leftward

Explanation:

Taking rightwards as positive

We are given:

Ball 1:

Mass (m1) = 1.5 kg

velocity (u1) = 2 m/s

Ball 2:

Mass (m2) = 2.5 kg

velocity (u2) = -1.5 m/s          [negative because it is in the opposite direction]

Speed and Direction of Ball 2:

We are told that the balls stick together after the collision

We can say that the balls have the same velocity since they are sticking together

So, Final velocity of Ball 1 (v1) = Final velocity of Ball 2 (v2) = V m/s

According to the law of conservation of momentum

m1u1 + m2u2 = m1v1 + m2v2

replacing the variables

1.5(2) + (2.5)(-1.5) = V (1.5 + 2.5)                     [v1 = v2 = V]

3 + (-3.75) = 4V

-0.75 = 4V

V = -0.75/4                                                    [dividing both sides by 4]

V = -0.1875 m/s

Hence, the balls will move at a velocity of 0.1875 m/s in the Leftward direction

Answer:

im just so focused on the fact that im going to mars :O

Answer:

2.9 cm

Explanation:

Assuming that the rear wheel has a radius of 0.330 m

Given that

r(a) = 12 cm -> 0.12 m

w(a) = 0.6 rev/s -> 3.77 rad/s

v = 5 m/s

r(w) = 0.330 m

The speed on any point on the rim at the sprocket in the front is

v(a) = w(a).r(a) = 3.77 * 0.12 = 0.4524 m/s

Also,

v(a) = speed at any point on the chain

v(b) = speed at any point on the rim of the rear sprocket

v(a) = v(b)

where v(b) = w(b).r(b)

Recall that the speed at any point on the rear wheel is v, where

v = w(b).r(w)

5 = w(b) * 0.330

w(b) = 5/0.330

w(b) = 15.15 rad/s

On substituting this in the equation, we have

v(b) = w(b).r(b).

Remember also, that v(a) = v(b), so

0.4524 = 15.15 * r(b)

r(b) = 0.4524 / 15.15

r(b) = 0.029 m -> 2.9 cm

Therefore, the radius of the rear sprocket needed is 2.9 cm

Answer: The Answer is attached to the image below

Answer: T = 692.82 and 346.4 N

Explanation:

Given that;

w = 600 N

∅ = 30°

ΣFy = ma

a = 0 m/s²

ΣF = T(cos30°) - W = 0

T(cos30°) = W

we Divide both sides by cos30°

T = W / cos30o

T= 600N / cos30°

T = 692.82

and ∑fx

F = T sin∅

F = 692.82 × (sin30°)

F = 346.4 N

The equilibrium condition allows finding the result for the force of the ball against the wall is:

Newton's second law gives a relationship between force, mass and acceleration of bodies. In the case where the acceleration is zero, it is called the equilibrium condition.

            ∑ F = 0

A free-body diagram is a diagram of the forces without the details of the bodies. In the attached we have a free-body diagram of the system.

Let's use trigonometry to break down stress.

          sin 30 = [tex]\frac{T_x}{T}[/tex]  

          cos 30 = [tex]\frac{T_y}{T}[/tex]  

          T_y = T cos 30

          Tₓ = T sin 30

Let's write the equilibrium condition for the system.

y-axis.

         T_y -W = 0

          T cos 30 = W

          [tex]T = \frac{W}{cos 30}[/tex]  

x-axis.

        R - Tₓ = 0

        R = T sin 30

We substitute

        [tex]R = \frac{W}{cos 30} \ sin 30 \\R = W \ tan 30[/tex]

Let's calculate.

        R = 600 tan 30

        R = 346.4 N

This force is directed from the wall towards the ball, by Newton's third law the force of the ball is of equal magnitude and opposite direction, that is, directed towards the wall.

In conclusion with the equilibrium condition we can find the result for the force of the ball against the wall is:

Learn more about the equilibrium condition here: brainly.com/question/18117041

Answer:

D Magnesia

Explanation:

Ore particles Fe3O4 (magnetite) were found in the region called Magnesia.

Magnesia was an antic city in Asia, named after the inhabitants of Greek Magnesia.

Magnetite is used as an ore, abrasive, in paint production, electrophotography, as a micronutrient fertilizer, and as a high-density concrete aggregate.

Answer:

The value is  [tex]n = 18.5 \ oscillations[/tex]

Explanation:

From the question we are told that

   The mass is  [tex]m = 350 \ g = 0.350 \ kg[/tex]

   The length is  [tex]L = 45 \ cm = 0.45 \ m[/tex]

    The angle is  [tex]\theta = 4.5^o[/tex]

    The damping constant is  [tex]b = 0.010 \ kg/s[/tex]

    The time taken is [tex]t = 25 \ s[/tex]

Generally the angular frequency of this damped oscillation is mathematically evaluated as

     [tex]w = \sqrt{ \frac{ g}{L} + \frac{b^2}{4m^2} }[/tex]  

=>   [tex]w = \sqrt{ \frac{9.80 }{ 0.45} + \frac{0.010 ^2}{4* 0.350^2} }[/tex]  

=>   [tex]w = 4.667 \ s^{-1}[/tex]

Generally the period of the oscillation is mathematically represented as

      [tex]T = \frac{2 \pi }{w}[/tex]  

=>  [tex]T = \frac{2 * 3.142 }{ 4.667 }[/tex]

=>  [tex]T = 1.35 \ s[/tex]

Generally the number of oscillation is mathematically represented as

        [tex]n = \frac{t}{T}[/tex]

=>     [tex]n = \frac{25}{1.35}[/tex]

=>     [tex]n = 18.5 \ oscillations[/tex]

Answer:

3.32 atm

__________________________________________________________

We are given:

side of the cubical container = 28 cm

number of molecules in the container = 3 * Nₐ  

[where Nₐ is the Avogadro's number]

Temperature = 24°C  OR  297 K

We need to find the pressure exerted by the gas on the walls of the container

__________________________________________________________

Some Calculations:

Volume of the container

we are given the side of the cubical container = 28 cm

Volume of the cubical container = side³

Volume = 28³

Volume = 21952 cm³

We know that 1 cm³ = 1 mL

So,

Volume = 21952 mL

We also know that 1 L = 1000 mL

Volume = 21.952 L

Number of moles of Gas

We know that:

number of moles = number of molecules / Avogadro's number

number of moles = 3 * Nₐ / Nₐ                   [number of molecules = 3 * Nₐ]

number of moles = 3 moles

__________________________________________________________

Pressure Exerted by the Gas:

Using the ideal gas equation:

PV = nRT

Since the volume is in L, and Temperature is in K. R is equal to

0.082 L atm /mol K and the pressure will be in atm

P(21.952) = 3*(0.082)*(297)        

P = 3.32 atm

Hence, the gas will exert a pressure of 3.32 atm on the walls of the container