A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 45. what is the torque on the bar?
a. What is the direction of the torque.
b. what is the sign of the direction of the figure.

Answer:

6bar

Explanation:

According to pressure law;

P1/T1 = P2/T2

Given the following;

P1 = 1bar

T1  = 20oC

P2 = ?

T2 = 120oC

Substitute

1/20 = P2/120

20P2 = 120

P2 = 120/20

P2 = 6bar

Hence the new pressure will be 6bar

Answer:

The answer is "[tex]2.352 \ \frac{m}{s}[/tex]"

Explanation:

[tex]\to mass(m_1)=102 \ kg\\\\\to mass(m_2)=10 \ kg \\\\\to v=2.10\ \frac{m}{s}\\\\[/tex]

momentum before:

[tex]\to p=(m_1+m_2)v[/tex]

       [tex]=(102+10)2.10\\\\=(102\times 2.10 +10 \times 2.10)\\\\=214.2+21\\\\=235.2[/tex]

momentum After:

[tex]\to p=(m_1+m_2)v[/tex]

       [tex]=(102\times 0 +10 \times v)\\\\ =(0 +10v)\\\\=10v\\[/tex]

Calculating the conservation of momentum:

[tex]\to \text{momentum before = momentum After}[/tex]

[tex]\to 235.2=10v\\\\\to v= \frac{235.2}{10}\\\\ \to v=2.352 \ \frac{m}{s}[/tex]

Answer:

3.71 m/s

Explanation:

Given data :

radius of the spherical ball = 0.400 m

Density of the ball material  = 10.65 Kg/m^3

Density of water = 1000 Kg/m^3

Drag coefficient C_d= 0.520

Constant upward force  F_up= 2630

Concept: The moment before it starts to rise Upward force must be equal to downward drag force

⇒ F_up = F_D

F_D =C_d× 0.5 ×ρ_w×V^2×A

where V= terminal velocity and A=  area πR^2

equating the forces we get

2630= 0.52× 0.5×1000×V^2×π(0.4)^2

V= 3.171 m/s

Therefore, the terminal velocity of the ball = 3.71 m/s

The ball while moving through the water in the pool, experiences a drag force. Then, the magnitude of terminal speed during the rise of the ball is 3.171 m/s.

When an object is made to fall through a liquid medium, then the maximum velocity achieved at that instant is known as terminal velocity or terminal speed.

Given data-

The radius of the spherical beach ball is, r = 0.4000 m.

The density of the ball is, [tex]\rho'=10.65 \;\rm kg/m^{3}[/tex].

The density of water in the pool is, [tex]\rho =1000.0 \;\rm kg/m^{3}[/tex].

The drag coefficient of the ball in water is, [tex]C_{d}=0.520[/tex].

The magnitude of the constant upward force is, F = 2630 N.

As per the concept of Buoyancy, "The moment before the ball starts to rise upward force must be equal to downward drag force".

F = Fd

Here,

F is the upward force.

Fd is the downward force.

[tex]F=\dfrac{1}{2} \times C_{d} \times \rho \times v^{2} \times A[/tex]

Here,

A is the area of the ball.

Solving as,

[tex]2630=\dfrac{1}{2} \times 0.520 \times 1000 \times v^{2} \times (\pi r^{2})\\\\2630=\dfrac{1}{2} \times 0.520 \times 1000 \times v^{2} \times (\pi \times 0.4000^{2})\\\\v=3.171 \;\rm m/s[/tex]

Thus, we can conclude that the magnitude of terminal speed during the rise of the ball is 3.171 m/s.

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Answer:

To calculate the voltage, current, power and energy for an inductor and also to examine the behavior of an inductor under dc conditions.

an inductor is a passive circuit element that stores energy in its magnetic field.

Explanation:

An inductor has been described as a passive electronic component that stores electrical energy in a magnetic form.  It uses a conductor that is wound into a coil to produce DC current, acting as a short circuit when direct current is applied.  When electricity flows into the coil from the left to the right, a current is generated through a magnetic field in the clockwise direction.  It is also called a choke because of its frequency regulation.

Answer:

270 Kg e 130 kg

Explanation:

how many times is the number put together

Answer:

1km west

Explanation:

Because if the cyclist is going back 2km then its 3-2=1km

For the three-point charges fixed in a right triangle, we have:

1. The magnitude of the electric force on the +1.00 μC charge is 1.79 N.

2. The force does an angle of 21.68° with the x-axis.

First, let's denote the charges as is shown in the picture below:

The magnitude of the electric force on the charge 2 (+1.00 μC) is given by:  

[tex]|F_{net}| = \sqrt{(\Sigma\vec{F}_{x})^{2} + (\Sigma\vec{F}_{y})^{2}}[/tex]  (1)    

Where:  

We can calculate the electrical forces with Coulomb's law:

[tex]\vec{F} = \frac{Kq_{1}q_{2}}{d^{2}}[/tex]

Where:

Forces in the x-axis ([tex]\Sigma\vec{F}_{x}[/tex])

The forces in the x-component are given by:

[tex]\Sigma\vec{F}_{x} = \vec{F}_{21}_{x} + \vec{F}_{23}_{x}[/tex]

[tex]\Sigma\vec{F}_{x} = \vec{F}_{21}_{x}cos(\theta) + \vec{F}_{23}_{x}[/tex]

[tex]\Sigma\vec{F}_{x} = \frac{Kq_{1}q_{2}}{d_{12}^{2}}cos(\theta) + \frac{Kq_{2}q_{3}}{d_{23}^{2}}[/tex]   (2)

Where:              

We can calculate the angle θ with the following trigonometric function:

[tex]sin(\theta) = \frac{y}{z}[/tex]

[tex]\theta = sin^{-1}(\frac{y}{z}) = sin^{-1}(\frac{8.10 cm}{9.60 cm}) = 57.5 ^\circ[/tex]

To find the distance x (d₂₃), we need to use Pythagoras:

[tex]x = \sqrt{z^{2} - y^{2}} = \sqrt{(0.0960 m)^{2} - (0.0810 m)^{2}} = 0.051 m[/tex]

After entering θ and x (d₂₃) into equation 2, we have:

[tex]\Sigma\vec{F}_{x} = \frac{9.00\cdot 10^{9} Nm^{2}C^{-2}(0.800 \cdot 10^{-6} C)(1.00 \cdot 10^{-6} C)}{(0.0960 m)^{2}}cos(57.5) + \frac{9.00 \cdot 10^{9}Nm^{2}C^{-2}(1.00 \cdot 10^{-6} C)(-0.600 \cdot 10^{-6} C)}{(0.051 m)^{2}}[/tex]

[tex]\Sigma\vec{F}_{x} = [0.78*cos(57.5) + (-2.08) N] = -1.66 N[/tex]

Hence, the x-component of the force is -1.66 N.

Forces in the y-axis ([tex]\Sigma\vec{F}_{y}[/tex])

The only force acting on the y-axis is the y-component of the force F₂₁, so:

[tex]\Sigma\vec{F}_{y} = -\vec{F}_{21}_{y} = -\vec{F}_{21}sin(\theta)[/tex]

The minus sign is because the vector is pointing in the negative y-direction (see the picture below).

[tex]\Sigma \vec{F}_{y} = -\frac{9.00\cdot 10^{9} Nm^{2}C^{-2}(0.800 \cdot 10^{-6} C)(1.00 \cdot 10^{-6} C)}{(0.0960 m)^{2}}sin(57.5) = -0.66 N[/tex]

Hence, the y-component of the force is -0.66 N.

Finally, the magnitude of the electric force on the charge +1.00 μC is (eq 1):  

[tex]|F_{net}| = \sqrt{(-1.66 N)^{2} + (-0.66 N)^{2}} = 1.79 N[/tex]                                  

Therefore, the magnitude of the electric force on the +1.00 μC charge is 1.79 N.

According to the picture below, the angle of the force with respect to the x-axis is given by:  

[tex]tan(\beta) = \frac{\Sigma\vec{F}_{y}}{\Sigma\vec{F}_{x}}[/tex]

[tex]\beta = tan^{-1}(\frac{-0.66 N}{-1.66 N}) = 21.68 ^\circ[/tex]

Therefore, the force does an angle of 21.68° with the x-axis.

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Answer:

7976 Pascals significant figure= 7.9*10^3

Explanation:

formula of hpg = height*density*gravitational energy

.80*10*997=7976 pascals

Answer:

4

Explanation:

The momentum of the object is 100 kgm/s, therefore the magnitude of the velocity of the object is 4 meters per second.

Momentum of an object is the product of its mass and velocity. Momentum is the quantity of motion which is made up of the amount of matter which is being moved and the velocity at which the object moves. When we walk, run, etc., then we have momentum.

p = m × v

where, p is the momentum of the object,

m is the mass of the object,

v is the velocity of the object

p = 100 kgm/s

m = 25 kg

p = m × v

v = p/m

v = 100/ 25

v = 4 m/s

Therefore, the magnitude of velocity of the object is 4 meters per second.

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Answer:

Ti and O

Explanation:

It is what it is.

Answer:

B. Ti and O

Explanation:

A.P.E.X

Answer:

there are 2188 people at the game

Explanation:

Given the data in the question;

we know that decibel is defined as; dB = 10log(l/l₀)

so if one produces an intensity l₁, it results in 69.6 dB

69.6 = 10log(l₁/l₀) ---- equ1

also. if x number of people produces this intensity, it result to 103 dB

103 = 10log(xl₁/l₀)

103 = 10( log(l₁/l₀) + log(x) )

103 = 10log(l₁/l₀) + 10log(x) ----- equ2

input equation 1 into to equation 2

103 = 69.6 + 10log(x)

10log(x) = 103 - 69.6

10log(x) = 33.4

divide both side by 10

log(x) = 3.34

x = [tex]10^{3.34}[/tex]

x = 2187.76 ≈ 2188

Therefore, there are 2188 people at the game

Answer:

This link was diagram

Explanation:

https://doubtnut.app.link/FnsNC80Dccb

Answer:

plasma

Explanation:

Cause in the universe many exist in either in gaseous or plasma form but plasma is greater

Answer:

A negative slope results when an individual is moving away

Explanation:

Answer:

Parallel circuit

Explanation:

A parallel circuit has a direct current allowing it to go back and forth making it have more pathways and loops. if one light bulb goes out the pathways will allow the electricity to still flow inti the other light bulbs.

Answer:

B

Explanation:

A tractor is the heaviest

Answer:

25 (kg) and 40.5 (N).

Explanation:

1) the value of 'g' on the Earth is ~9.81 (N/kg), the value of 'g' on the Moon is ~1.62 (N/kg).

It means, the mass is not depended on the value of 'g', but the weight is depended on it.

2) the mass on the Moon is the same, 25 kg.

the weight is 25*1.62=40.5 N

Answer:

Its radius is thought to be around 1400 times than of our sun, and its luminosity 270,000 greater than our sun. Also, if a star has the same radius as the sun but a higher surface temperature, the hotter star exceeds the sun in luminosity

Explanation:

As the solar radius of the star increases, the brightness of the star also increases.

Luminosity is defined as the measure of brightness of a celestial object or the relative quantity of light.

Here,

Solar radius is a unit of distance, which is used to determine the size of the stars relative to the sun.

The size of the star and the luminous intensity are directly proportional. As the size of the star increases, the luminosity of the star also increases. Therefore, larger stars shine more brighter than the smaller ones.

The brightness of a star is directly proportional to the square of its solar radius. Therefore, the luminosity increases with increase in solar radius.

Hence,

As the solar radius of the star increases, the brightness of the star also increases.

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Answer:

The number of O₂ molecules that are left in the cylinder is 1.70x10²⁴.

Explanation:

The number of oxygen molecules can be found using the Ideal Gas law:

[tex] PV = nRT [/tex]        

Where:

P: is the pressure = 100 psi

V: is the volume = 10 L

n: is the number of moles =?

T: is the temperature = 20 °C = 293 K

R: is the gas constant = 0.082 L*atm/(K*mol)

Hence, the number of moles is:

[tex]n = \frac{PV}{RT} = \frac{100 psi*\frac{1 atm}{14.7 psi}*10 L}{0.082 L*atm/(K*mol)*293 K} = 2.83 moles[/tex]

Now, the number of molecules can be found with Avogadro's number:

[tex]n_{m} = \frac{6.022 \cdot 10^{23}\: molecules}{1\: mol}*2.83 moles = 1.70 \cdot 10^{24} \: molecules[/tex]

Therefore, the number of O₂ molecules that are left in the cylinder is 1.70x10²⁴.

I hope it helps you!              

Answer:

208m

Explanation:

since the angle is with the ground which gives a horizontal component to the velocity = v(cos)30°

so the velocity equals 40(cos)30°= 34.64m/s

range(distance) = speed × time

34.64m/s × 6s = 207.8m = 208m

Answer:

11.06797181 ms^-1 only.......

Answer:

Fn=m(a + g) :

Explanation:

nf=a(m +g)

an=f(g+m)