what is the force when ur pulling heavy furniture that wont budge

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:

I think it's D

Explanation:

Most mobiles have touch screens

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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it is meters and ounces

Answer:

false

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

Answer:

The conversion factors you will use to multiply is 1/0.001 or simply 1000 to perform the unit conversion.

Explanation:

To convert from gram/centimeter³ to kilogram/meter³

First,

NOTE:

1 kilogram = 1000 gram, that is 1 gram = 0.001 kilogram; and

1 meter = 100 centimeter, that is 1 centimeter = 0.01 meter

then,

1 meter³ = 1000000 centimeter³, that is

1 centimeter³ = 0.000001 meter³

Now, we can write that

1 kilogram/meter³ = 1000 gram / 1000000 centimeter³

= 0.001 gram/centimeter³

If 1 kilogram/meter³ = 0.001 gram/centimeter³

Then, 1 gram/centimeter³ = 1/0.001 kilogram/meter³  = 1000 kilogram/meter³

Hence, 1 gram/centimeter³  = 1000 kilogram/meter³

∴The conversion factors you will use to multiply is 1/0.001 or simply 1000 to perform the unit conversion.

That is,

2.33 gram/centimeter³ will be 2.33 × 1000 kilogram/meter³

= 2330 kilogram/meter³

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:

F-ma

Explanation:

If you are speaking of objects like satellites, etc. then their mass is much less than that of the Earth. A good approximation is Newton's first law of motion:

Force = Mass ×  Acceleration

often written:

F = m a

The gravitational force is the same between the Earth and the object - only the mass differs. So the acceleration is inversely proportional to the mass.

Answer:

I NEED OPTIONS

Explanation:

Answer:

Monarchy is rule from kings and queens

Explanation:

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

Answer:

a)  vₓ = 100 + 7.07 = 107.07 knot     (East) ,   v_{y} = 7.07 knot            (North)

b) v_wind_ North = 0 ,   v_wind_West = 61.12 knot      (West)

Explanation:

a) This is a vector velocity addition problem, they tell us that the plane goes East at 100 Knots and the wind goes North East at 10 Knots

the components ask the total speed of the plane.

For this we decompose the wind speed

            cos 45 = v₂ₓ / V₂

            sin 45 = [tex]v_{2y}[/tex] / v₂

            v₂ₓ = v₂ cos 45

           v_{2y} = v₂ sin 45

            v₂ₓ = 10 cos 45 = 7.07 m / s

            v_{2y} = 10 sin45 = 7.07 m / s

the speed of the plane is

            vₓ = v_plane + v₂ₓ

            v_{y} = v_{2y}

            vₓ = 100 + 7.07 = 107.07 knot     (East)

            v_{y} = 7.07 knot            (North)

speed is

            v = (107.07 i ^ + 7.07 j ^) knot

b) Estimated time for Target VOR t = 10 min

for a time of t = 10 min and 15 s, it is at a distance of d = 300 m from the VOR in an easterly direction

find the average speed of time on the ride

in the North direction there is no deviation so the average wind speed is zero

            v_wind_ North = 0

In the east direction

as the estimated time was 10 min and the real time for the distance is 10 min 15 s

the time difference is t = 15 s to travel d = 300 m

with these data we can calculate the speed of the plane

              v = x / t

              v = 300/15

              v = 20 m / s

let's reduce this speed to knot

             v_real = 20 m / s (1knot / 0.5144 m / s) = 38.88 knots

therefore the wind speed is

              v_tral = v_plane + v_wind_Este

              v_wind_Este = v_real - v_avión

               v_wind_Este = 38.88 - 100

                v_wind_Este = -61.12 knot

the negative sign indicates that the wind is going west

                v_wind_West = 61.12 knot (West)

3)

* The compass is not calibrated, to correct the magnetic deviation, therefore it gives an appreciable error in the direction

* The wind speed is taken when leaving, but there is no constant monitoring, so a change in direction or wind speed in the trajectory can significantly affect the results.

* The aircraft may noir throughout the trajectory at level, due to pitch errors

Answer:

Acceleration and mass.

Explanation:

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

F=MA

Answer:

499 seconds or a little over 8 minutes

Explanation:

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

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]

Answer: 938

Explanation:

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

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.

Answer:

28 m/s

Explanation:

vf^2=vi^2+2a(delta y)

=sqrt(2* -9.8 m/s^2* -40 m)

=28 m/s

Answer:

Q / Q₀ = 3.4 ,      ΔQ = 31.14 10⁻⁶ C

Explanation:

For this exercise we will calculate the charge of the capacitor with and without dielectric

the capacitance is given by

          C = Q /ΔV

we also use the ratio of capacitances

          C = k C₀

           Q /ΔV = kQ₀ /ΔV

           Q = k Q₀

            Q / Q₀ = k

            Q / Q₀ = 3.4

If we want a numerical value

           C = k Q / DV

           Q = C ΔV / k

            Q = 2.5 10⁻⁶ 60 / 3.4

            Q = 44.12 10⁻⁶ C

whereby the charge without dielectric is

          Q₀ = Q / 3.4

          Q₀ = 44.12 / 3.4 10⁻⁶

          Q₀ = 12.98 10⁻⁶ C

therefore the charge when removing the dielectric is reduced, the amount of charge that flows is

          ΔQ = Q - Q₀

          ΔQ = (44.12 - 12.98) 10⁻⁶

          ΔQ = 31.14 10⁻⁶ C

The amount of charge should be Q / Q₀ = 3.4 ,    ΔQ = 31.14 10⁻⁶ C

For this given situation,  we will determine the charge of the capacitor with and without dielectric

We know that

the capacitance is given by

         C = Q /ΔV

Now here we can applied the ratio of capacitances

So,

         C = k C₀

          Q /ΔV = kQ₀ /ΔV

          Q = k Q₀

           Q / Q₀ = k

           Q / Q₀ = 3.4

In case of numerical value

          C = k Q / DV

          Q = C ΔV / k

           Q = 2.5 10⁻⁶ 60 / 3.4

           Q = 44.12 10⁻⁶ C

whereby the charge without dielectric is

         Q₀ = Q / 3.4

         Q₀ = 44.12 / 3.4 10⁻⁶

         Q₀ = 12.98 10⁻⁶ C

Thus the charge at the time when removing the dielectric is decreased , the amount of charge that flows is

         ΔQ = Q - Q₀

         ΔQ = (44.12 - 12.98) 10⁻⁶

         ΔQ = 31.14 10⁻⁶ C

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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]

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.

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Answer:

200000000000000000000000

Explanation:

Answer:

10000000000000000100

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: 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:

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:

Frame of reference

Answer:

Newton's first law says that if the net force on an object is zero ( Σ F = 0 \Sigma F=0 ΣF=0\Sigma, F, equals, 0), then that object will have zero acceleration. That doesn't necessarily mean the object is at rest, but it means that the velocity is constant.

Explanation:

According to Newton’s first law of motion, a moving object acted on by a net force of zero will continue in motion.

Newton's first law of motion states that an object in state of rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.

This first is also known as the law of inertia because it is the reluctance of an object in motion to stop moving or reluctance of an object at rest to start moving which depends on the mass of the object.

Thus, we can conclude that according to Newton’s first law of motion, a moving object acted on by a net force of zero will continue in motion.

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Answer:

They are perpendicular.

Explanation:

Explanation:

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

The lowest speed will lose