Given F1: a force of magnitude 6 N at an angle of 30°
F2: a force of magnitude 8 N at an angle of 50°C

a. Find F1+ F2 analytically (using equations instead of graphing) and write it in the form Fr1i + Fr2 j
b. Find the magnitude FR and θ_resultant

Answer: 117.6N

Explanation:

By the second Newton's law, we know that:

F = m*a

F = force

m = mass

a = acceleration

We know that in the surface of the Earth, the gravitational acceleration is g = 9.8m/s^2.

Then we just can input that acceleration in the above equation, and also replace m by 12kg, and find that the force due the gravity is:

F = 12kg*9.8m/s^2 = 117.6N

Answer:

The number of photons per second that strike the given area is 2.668 x 10⁸ photons/second

Explanation:

Given;

intensity of the sunlight, I = 2.00 kJ·s−1·m^−2

area of incident, A = 5.2 cm² = 5.2 x 10⁻⁴ m²

Energy of incident photons per second on the given area;

E = IA

E = (2000)( 5.2 x 10⁻⁴)

E = 1.04 J/s

Energy of a photon is given is by;

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{(6.626*10^{-34})(3*10^8)}{(510*10^{-9})}\\\\E = 3.898*10^{-19} \ J/photon[/tex]

The number of photons per second that strike the given area is;

[tex]n = \frac{1.04 \ J/s}{3.898*10^{-19} \ J/photon} \\\\n = 2.668*10^{18} \ photons/second[/tex]

Therefore, the number of photons per second that strike the given area is 2.668 x 10⁸ photons/second

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➡ 150hrs.

[tex]{\underline{\pink{\textsf{\textbf{Explanation : }}}}}[/tex]

➡ Time = distance × speed

➡ Time = 15*10

➡ Time = 150hrs ans.

Answer:

If the mass of one of the objects is doubled, then the force of gravity between them is doubled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces

Explanation:

Answer:

in the water cycle evaporation occurs when the sunlight warms the surface of the water the heat from the sun makes the water molecules move faster and faster until they move so fast they can escape as a gas once evaporated a molecule from water that vapor spends about 10 days in the air

Answer:

m₂ = 1.17 kg

Explanation:

Given that,

Force between two objects, [tex]F=1.01\times 10^{-8}\ N[/tex]

Mass of object 1, [tex]m_1=5.11\ kg[/tex]

Distance between masses, r = 19.9 cm = 0.199 m

The gravitational force between two masses is given by :

[tex]F=\dfrac{Gm_1m_2}{r^2}[/tex]

m₂ is the mass of object 2

[tex]m_2=\dfrac{Fr^2}{Gm_1}\\\\m_2=\dfrac{1.01\times 10^{-8}\times (0.199)^2}{6.67\times 10^{-11}\times 5.11}\\\\=1.17\ kg[/tex]

So, the mass of second object is 1.17 kg.

Answer:

B: lodestone

Explanation:

Each magnet has its magnetic poles, north (N) and south (S). Diversified ones are attracted and reptiles of the same name are repelled, similarly to charges, so it was considered possible to separate the magnet at the north and south poles.

Magnetic properties can be lost if the magnet is exposed to high temperatures if it falls or due to some mechanical shocks.

Answer:

Jerusalem

Explanation:

because now a days jerusalem is the royal capital of the kingdom of israel so i conclude that the answer is jerusalem

Answer:

The 300 Ohm resistor

Explanation:

They are in series, so the current has to be the same in both.  P = I^2R, so with the same current, the larger resistor dissipates more power.

Explanation:

the formula of speed is distance traveled by time it work

31.25 cm/sec

__________________________________________________________

We are given:

Distance travelled = 150 cm

Time taken = 4.8 seconds

Final Speed of the Marble:

Speed of the marble = Distance travelled / Time taken

Speed of the marble = 150 / 4.8

Speed of the marble = 31.25 cm/sec

Answer:

v = 0.823 m/s

Explanation:

A man walks south at a speed of 2.00 m/s for 60.0 minutes.

The distance covered in South = 60 × 60 × 2 = 7200 m

He then turns around and walks north a distance 3000 m in 25.0 minutes.

As they moved in opposite direction, net displacement will be : 7200 - 3000 = 4200 m

Average velocity of the man = net displacement/time

[tex]v=\dfrac{4200\ m}{(60+25)\times 60}\\\\=0.823\ m/s[/tex]

So, the average velocity of the man is 0.823 m/s.

Answer:

10J

Explanation:

In this question we have the following information

The charge of the particle is q = 5 C

The equipotenetial level is V1 = 5.5 v

and also the

equipotenetial level is V2 = 3.5 v

So we calculate the

work done W=q x (v1-v2)

workdone = 5 x (5.5-3.5)

= 5x2

=10 J

Workdone = 10 J

So we conclude that the workdone on a particle with these information is 10j

Answer:

e

Explanation:

cause you have any attachments are not sure if you can get it

When a block slides down an inclined plane, normal force does zero work. Hence, option (a) is correct.

The definition of force in physics is: The push or pull on a massed object changes its velocity.

An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.

The component of a contact force in mechanics known as the normal force is perpendicular to the surface that an item encounters.

As normal force always acts perpendicular to the motion, this force does zero work.

But friction force the weight acts in opposite direction of the motion, this force does work against the motion.

In an inclined plane, the parallel component of  weight does work along the motion.

Learn more about normal force here:

https://brainly.com/question/18799790

#SPJ2

Answer:

Evaporation from oceans = 460,000 km³

Explanation:

Given:

Precipitation on oceans = 420,000 km³

Precipitation on land = 130,000 km³

Evaporation from land = 90,000 km³

Find:

Evaporation from oceans

Computation:

Evaporation from oceans = Precipitation on oceans + Precipitation on land - Evaporation from land

Evaporation from oceans = 420,000 km³ + 130,000 km³ - 90,000 km³

Evaporation from oceans = 460,000 km³

Answer:ccccccC

Explanation:okay

Answer:

Gravity acts on all objects in the universe.

Explanation:

Answer:

750 nm

Explanation:

It is given that,

Third minimum is at an angle of 48.6º when it falls on a single slit of width 3.00 μm.

We need to find the wavelength of the light. For a single slit experiment, the relation between wavelength and slit width is given by :

[tex]d\sin\theta=n\lambda[/tex]

For third minimum, n = 3

[tex]\lambda=\dfrac{d\sin\theta}{n}\\\\\lambda=\dfrac{3\cdot10^{-6}\cdot\sin48.6}{3}\\\\\lambda=7.5\times 10^{-7}\ m\\\\\lambda=750\ nm[/tex]

So, the wavelength of light is 750 nm.

25/2=12.5s

12.5*9.8=122.5 m/s

Answer:

a) The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty in its perimeter is 0.8 centimeters.

Explanation:

a) From Geometry we remember that the perimeter of the rectangle ([tex]p[/tex]), measured in centimeters, is represented by the following formula:

[tex]p = 2\cdot (w+l)[/tex] (1)

Where:

[tex]w[/tex] - Width, measured in centimeters.

[tex]l[/tex] - Length, measured in centimeters.

If we know that [tex]w = 6.4\,cm[/tex] and [tex]l = 8.3\,cm[/tex], then the perimeter of the rectangle is:

[tex]p = 2\cdot (6.4\,cm+8.3\,cm)[/tex]

[tex]p = 29.4\,cm[/tex]

The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty of the perimeter ([tex]\Delta p[/tex]), measured in centimeters, is estimated by differences. That is:

[tex]\Delta p = 2\cdot (\Delta w + \Delta l)[/tex]  (2)

Where:

[tex]\Delta w[/tex] - Uncertainty in width, measured in centimeters.

[tex]\Delta l[/tex] - Uncertainty in length, measured in centimeters.

If we know that [tex]\Delta w = 0.2\,cm[/tex] and [tex]\Delta l = 0.2\,cm[/tex], then the uncertainty in perimeter is:

[tex]\Delta p = 2\cdot (0.2\,cm+0.2\,cm)[/tex]

[tex]\Delta p = 0.8\,cm[/tex]

The uncertainty in its perimeter is 0.8 centimeters.

Answer:

36.87°

Explanation:

Given

Maximum height = 60m

Horizontal distance (range) = 80,m

Required

Initial speed U

Angle of launch

To get the speed, we will use the range formula;

R = U √2H/g

80 = U√2(60)/9.8

80 = U√12.25

80 = 3.5U

U = 80/3.5

U = 22.86m/s

Get the angle of launch

Using the formula

Theta = tan^-1(y/x)

y is the vertical distance

x is the horizontal distance

Theta = tan^-1(60/80)

Theta = tan^-1(0.75)

Theta = 36.87°

Hence the angle of launch is 36.87°

Answer:1 because

Explanation: it’s pointing to the earth and gravity

Pulls things down to earth

Answer:

220v

Explanation:

Sorry, the question is incomplete

Answer:

on the potential difference applied and on the resistance of the wire.

Explanation:

Ohms law state that the current through a conductor between two points is directly proportional to the potential difference across the two points. Imtroducing the comstant of proportionality, the resistance, one arrives at the usual athematical equation that describes this relationship: I = V/R.

Answer:

distance i think

Explanation:

Answer:Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as  

K =-V dP/dV

Where  K, the bulk modulus of water = 2.15 x 10^9N/m^2

2.15 x 10^9N/m^2= - 4,300 x  4 × 106N/m2 / dV

dV = - 4,300 x  4 × 10^6N/m^2/ 2.15 x 10^9N/m^2

dV (change in volume)= -8.000cm^3

Final volume after pressure is applied,

V= V+ dV

V= 4300cm3 + (-8.000cm3)

=4300cm3 - 8.000cm3

Final Volume, V =4,292cm3

Answer:

θ = 122°

Explanation:

Components of a Vector

A vector in the plane can be defined by its rectangular components:

[tex]\vec A =<x,y>[/tex]

Or also can be given by its polar components:

[tex]\vec A =<r,\theta>[/tex]

Where r is the magnitude of the vector and θ is the angle it forms with the positive direction of x.

The relation between them is:

[tex]r=\sqrt{x^2+y^2}[/tex]

[tex]\displaystyle \theta=\arctan\frac{y}{x}[/tex]

It's given the x-component of vector A is x=-25 m and the y-component is y=40 m

(a)

The magnitude of the vector is:

[tex]r=\sqrt{(-25)^2+40^2}[/tex]

[tex]r=\sqrt{625+1600}[/tex]

[tex]r=\sqrt{2225}[/tex]

[tex]r\approx 47.2\ m[/tex]

(b)

[tex]\displaystyle \theta=\arctan\frac{40}{-25}[/tex]

[tex]\displaystyle \theta=\arctan (-1.6)[/tex]

The calculator gives us the value

θ = -58°

But the real angle lies on the second quadrant since x is negative and y is positive, thus:

θ = -58° + 180° = 122°

θ = 122°

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:

Explanation:

The weight of the object can be found by using the formula

[tex]f = \frac{w}{d} \\ [/tex]

w is the workdone

d is the distance

From the question we have

[tex]f = \frac{62500}{25} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

60.185 percent of the original kinetic energy is convertible to internal energy.

Explanation:

Let suppose that collision between both particles is entirely inellastic. If there is no external forces exerted on any of the particles, then we can apply the Principle of Linear Momentum Conservation. That is:

[tex]m_{A}\cdot v_{A,o} + m_{B}\cdot v_{B,o} = (m_{A}+m_{B})\cdot v[/tex]

[tex]v = \frac{m_{A}\cdot v_{A,o}+v_{B}\cdot v_{B,o}}{m_{A}+m_{B}}[/tex] (1)

Where:

[tex]m_{A}[/tex] - Mass of the 4.8-g particle, measured in kilograms.

[tex]m_{B}[/tex] - Mass of the 7.4-g particle, measured in kilograms.

[tex]v_{A,o}[/tex] - Initial speed of the 4.8-g particle, measured in meters per second.

[tex]v_{B,o}[/tex] - Initial speed of the 7.4-g particle, measured in meters per second.

[tex]v[/tex] - Final speed of the collided particles, measured in meters per second.

If we know that [tex]m_{A} = 4.8\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.4\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 3\,\frac{m}{s}[/tex] and [tex]v_{B,o} = 0\,\frac{m}{s}[/tex], then the final speed of the system is:

[tex]v = \frac{(4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s} \right)+(7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)}{4.8\times 10^{-3}\,kg+7.4\times 10^{-3}\,kg}[/tex]

[tex]v = 1.180\,\frac{m}{s}[/tex]

During the collision part of the initial energy is dissipated in the form of heat, which is related to the internal energy ([tex]\Delta U[/tex]), measured in joules. According to the Principle of Energy Conservation, we have the following model:

[tex]\Delta U = K_{A}+K_{B}-K[/tex] (2)

Where:

[tex]K_{A}[/tex], [tex]K_{B}[/tex] - Initial translational kinetic energies of each particle, measured in joules.

[tex]K[/tex] - Final translational kinetic energy of the collided particles, measured in joules.

By applying the definition of translational kinetic energy, we expand and simplify the equation above:

[tex]\Delta U = \frac{1}{2}\cdot m_{A}\cdot v_{A,o}^{2}+\frac{1}{2}\cdot m_{B}\cdot v_{B,o}^{2} -\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2}[/tex] (3)

If we get that  [tex]m_{A} = 4.8\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.4\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 3\,\frac{m}{s}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]v = 1.180\,\frac{m}{s}[/tex], the internal energy associated with the system is:

[tex]\Delta U = \frac{1}{2}\cdot (4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s} \right)^{2}+ \frac{1}{2}\cdot (7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-\frac{1}{2}\cdot (4.8\times 10^{-3}\,kg+7.4\times 10^{-3}\,kg)\cdot \left(1.180\,\frac{m}{s} \right)^{2}[/tex]

[tex]\Delta U = 0.013\,J[/tex]

And the initial energy of both particles is:

[tex]E_{o} = \frac{1}{2}\cdot (4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s}\right)^{2}+\frac{1}{2}\cdot (7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}[/tex]

[tex]E_{o} = 0.0216\,J[/tex]

Lastly, the percentage of the original kinetic energy that is convertible to internal energy is: ([tex]\Delta U = 0.013\,J[/tex], [tex]E_{o} = 0.0216\,J[/tex])

[tex]\%e = \frac{\Delta U}{E_{o}}\times 100\,\%[/tex] (4)

[tex]\%e = \frac{0.013\,J}{0.0216\,J}\times 100\,\%[/tex]

[tex]\%e = 60.185\,\%[/tex]

60.185 percent of the original kinetic energy is convertible to internal energy.

Answer:

When we love someone we experience the same positive thoughts and experiences as when we like a person. But we also experience a deep sense of care and commitment towards that person. Being “in love” includes all the above but also involves feelings of sexual arousal and attraction.

Explanation: