The rotational analogue of force is moment of force, also called torque.

a) The turning effect of force is maximum when the angle between R and F is.--------

b) A wheel starting from rest acquires an angular velocity of 10 rad/s in two seconds. The moment

of inertia of the wheel is 0.4 kg m 2 . Calculate the torque acting on it.

c) The possibility of falling backward with the ladder is more when you are high up on the ladder

than when you just begin to climb. Explain why.

Answer:

I think it's D

Explanation:

Most mobiles have touch screens

Answer: To isolate the unknown substances, of which only tiny amounts were present, the Curies were the first to use a new method of chemical analysis. They employed various standard (but sometimes demanding) chemical procedures to separate the different substances in pitchblende.

Explanation:

Answer:

Monarchy is rule from kings and queens

Explanation:

Answer: 938

Explanation:

1.67x10^-27kg (938V/c2)

Answer:

i got you

Explanation:

Unfortunately, it is also an incomplete chemical equation. ... But if we count the number of oxygen atoms in the reactants and products, we find that there are two oxygen atoms in the reactants but only one oxygen atom in the products.

Given parameters:

First velocity  = 2.50m/s

Time of travel = 3s

Second velocity  = 1.50m/s

Unknown:

The displacement during the first interval = ?

Velocity is the displacement of a body with time. Displacement is a distance move in a specific direction by a body.

    Velocity  = [tex]\frac{Displacement}{Time taken}[/tex]

So;

      Displacement  = Velocity x Time taken

Now input the parameter for the first velocity and time of travel;

      Displacement  = 2.5 x 3  = 7.5m

The displacement id 7.5m

it is meters and ounces

Answer:

Its Reversing the polarity of the magnet :)

Explanation:

The action that cause the induced current for reduce or be same is when the polarity is reverse with respect to the magnet.

The following actions should not induced the current:

Therefore we can conclude that the action that cause the induced current for reduce or be same is when the polarity is reverse with respect to the magnet.

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Answer:

Acceleration and mass.

Explanation:

Depends on how heavy the car is and how fast it is going.

F=MA

Answer:

The unit energy losses due to nonconservative forces is 881.40 joules per kilogram.

Explanation:

We can estimate the unit energy losses of gas eruption by Principle of Energy Conservation and Work-Energy Theorem:

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

Where:

[tex]U_{g,1}[/tex] - Gravitational potential energy of gas eruptions at surface, measured in joules.

[tex]U_{g,2}[/tex] - Gravitational potential energy of gas eruptions at highest height, measured in joules.

[tex]K_{1}[/tex] - Translational kinetic energy of gas eruptions at surface, measured in joules.

[tex]K_{2}[/tex] - Translational kinetic energy of gas eruptions at highest height, measured in joules.

[tex]W_{loss}[/tex] - Energy losses due to nonconservative forces, measured in joules.

We clear the component associated with energy losses in (Eq. 1):

[tex]W_{loss} = U_{g,1}-U_{g,2}+ K_{1}-K_{2}[/tex]

And we expand it afterwards:

[tex]W_{loss} = m\cdot g\cdot (z_{1}-z_{2}) + \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})[/tex] (Eq. 2a)

[tex]w_{loss} = g\cdot (z_{1}-z_{2})+\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex] (Eq. 2b)

Where:

[tex]W_{loss}[/tex] - Energy losses due to nonconservative forces, measured in joules.

[tex]w_{loss}[/tex] - Unit energy losses due to nonconservative forces, measured in joules per kilogram.

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

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Bottom and top height, measured in meters.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Gas eruption speeds at surface and highest heights, measured in meters per second.

If we know that [tex]g = 3.7\,\frac{m}{s^{2}}[/tex], [tex]z_{1} = 0\,m[/tex]. [tex]z_{2} =62\,m[/tex]. [tex]v_{1} = 36.111\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], the unit energy losses are:

[tex]w_{loss} = \left(3.7\,\frac{m}{s^{2}} \right)\cdot (62\,m-0\,m)+\frac{1}{2} \cdot \left[\left(36.11\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right][/tex]

[tex]w_{loss} = 881.40\,\frac{J}{kg}[/tex]

The unit energy losses due to nonconservative forces is 881.40 joules per kilogram.

Answer:d

Explanation:

Answer:

D  is correct

hope it helps

Answer: The control group is the part where you see what happens when you change a variable you want to study/examine. Basically, you need the control group because you need something to see what happens when you change something.

Hope this Helps! :))))

Answer:

They are perpendicular.

Explanation:

Answer:

The sum of the three vectors are 63.18 km.

Explanation:

Given that,

One vector = 74 km at 30°

Second vector = 54 km at 135°

Third vector = 62 km

We need to calculate the sum of the three vectors

Using given data

[tex]R=\vec{A}\sin\theta+\vec{B}\sin\theta+\vec{C}[/tex]

Put the value into the formula

[tex]R=-74\sin30+54\sin135+62[/tex]

[tex]R=63.18\ km[/tex]

Hence, The sum of the three vectors are 63.18 km.

Answer:

decreases, but frequency increases.

Explanation:

Wavelength and frequency are inversely proportional, meaning the higher the frequency, the shorter the wavelength, and the lower the frequency, the longer the wavelength.

Answer:

In the given equation y=Asin(ωt−kx)

Since, trigonometric function is a dimensionless quantity. So, (ωt−kx)is also dimensionless quantity.

So,

(ωt−kx)=1

ωt=1

[ω]= 1=[T] ^−1

[T]

kx=1

[k]= 1=[L]^ −1

[L]

Explanation:

If you want to get speed, u have to divided distance over time

The lowest speed will lose

Answer:

There is matter in gas but due to the high intermolecular forces, all the molecules of gas are far apart

Since the size of a molecule is almost negligible to the human eye, we cannot see the gas molecules and hence, gas is invisible to the naked eye

If you compress a gas against the intermolecular forces, the gas will turn into a liquid

Answer:

Ek1 = 900000 [J]

Ek1 = 400000 [J]

Explanation:

In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:

[tex]E_{k1}=\frac{1}{2}*m*v1^{2}[/tex]

where:

m = mass = 500 [kg]

v1 = 60 [m/s]

So we have:

Ek1 = 0.5*500*(60^2)

Ek1 = 900000 [J]

and:

Ek2 = 0.5*500*(40^2)

Ek2 = 400000 [J]

Answer:

Estimate slope= acceleration of -18.75 m/s^2

Explanation:

In the velocity vs time graph, acceleration is the slope

Let's take two points from the line and find the slope

(0,0) (0.8, -15)

slope = -15-0/ 0.8-0 = -18.75 m/s^2

Answer:

false

Explanation:  because the other scientists could've missed something in there experiment

Answer:

Low mass stars are: hydrogen burning in the core while on the Main Sequence. As the hydrogen fuel runs out, extreme pressure raises the temperature to 100 million degrees, where helium burning becomes possible.

The speed of the train is 150 Km/h.

Speed can be defined as the distance travelled over a period of time. Mathematically, it can be expressed as:

Speed = Distance / Time

With the above formula, we can obtain the speed of the train as shown below:

Speed = Distance / Time

Speed = 600 / 4

Speed = 150 Km/h

Therefore, the speed of the train is 150 Km/h.

Learn more about speed:

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Answer:

7.13781 m/s

Explanation:

X-direction             | Y-direction

[tex]x=v_{xo}t+\frac{1}{2}a_{x}t^2[/tex]      | [tex]y=v_{yo}t+\frac{1}{2}a_{y}t^2[/tex]

[tex]3.41=v_{o}cos(28.4)t[/tex]  | [tex]0.397=v_{yo}sin(28.4)t+\frac{1}{2} (-9.81)t^2[/tex]

[tex]3.41=v_{o}(0.87964)t[/tex]  | [tex]0.397=v_{yo}sin(28.4)(\frac{3.87658}{v_{o} })+\frac{1}{2} (-9.81)(\frac{3.87658}{v_{o} })^2[/tex]

[tex]\frac{3.41}{0.87964}=v_{o}t[/tex]             | [tex]0.397=1.84379-\frac{73.71171}{v^2}[/tex]

[tex]\frac{3.87658}{v_{o} } =t[/tex]                | [tex]7.13781=v[/tex]

Answer:

The answer is "[tex]q=0.0945\,C[/tex]".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

[tex]U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}[/tex]

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

[tex]U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}[/tex]

Potential energy shifts:

[tex]= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\[/tex]

[tex]=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J[/tex]

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

[tex]=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\[/tex]

[tex]=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\[/tex]

[tex]\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C[/tex]

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

Answer:

Frame of reference

Answer:

(a). The magnitude of the net force is [tex](2.1\times10^{-18}\ N)k[/tex]

(b). The magnitude of the net force is [tex](4.23\times10^{-19}\ N)k[/tex]

(c). The magnitude of the net force is [tex](8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]

Explanation:

Given that,

Magnetic field [tex]B=-3.54\times10^{-3}i\ T[/tex]

Velocity = 2230j m/s

We know that,

The net force acting on the proton is equal to the sum of electric and magnetic force.

[tex]F=F_{e}+F_{B}[/tex]

(a). If the electric field is in the positive z direction and has a magnitude of 5.25 V/m,

We need to calculate the magnitude of the net force acting on the proton

Using formula of net force

[tex]F_{net}=e(E+v\times B)[/tex]

Put the value into the formula

[tex]F_{net}=1.6\times10^{-19}(5.25k+2230\times-3.54\times10^{-3}(j\times i))[/tex]

[tex]F_{net}=1.6\times10^{-19}(5.25k+2230\times-3.54\times10^{-3}(-k))[/tex]

[tex]F_{net}=(2.1\times10^{-18}\ N)k[/tex]

(b). If the electric field is in the negative z direction and has a magnitude of 5.25 V/m,

We need to calculate the magnitude of the net force acting on the proton

Using formula of net force

[tex]F_{net}=e(E+v\times B)[/tex]

Put the value into the formula

[tex]F_{net}=1.6\times10^{-19}(-5.25k+2230\times-3.54\times10^{-3}(j\times i))[/tex]

[tex]F_{net}=1.6\times10^{-19}(-5.25k+2230\times-3.54\times10^{-3}(-k))[/tex]

[tex]F_{net}=(4.23\times10^{-19}\ N)k[/tex]

(c). If the electric field is in the positive x direction and has a magnitude of 5.25 V/m

We need to calculate the magnitude of the net force acting on the proton

Using formula of net force

[tex]F_{net}=e(E+v\times B)[/tex]

Put the value into the formula

[tex]F_{net}=1.6\times10^{-19}(5.25i+2230\times-3.54\times10^{-3}(j\times i))[/tex]

[tex]F_{net}=(8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]

Hence, (a). The magnitude of the net force is [tex](2.1\times10^{-18}\ N)k[/tex]

(b). The magnitude of the net force is [tex](4.23\times10^{-19}\ N)k[/tex]

(c). The magnitude of the net force is [tex](8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]

Answer:

499 seconds or a little over 8 minutes

Explanation:

~150 million km/ 300,000 km per second = 500 seconds