50 pounds is equal to 22,700 grams. How many pounds does this backpack full of gold weigh? Would you be able to carry out that backpack filled with gold?

Answer:

8.93*10^13 N.

Explanation:

       [tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]

       [tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]

Answer:

hello your question is incomplete attached below is the complete question

21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C

22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)

hence the voltage in the battery will be equal to the voltage across each bulb

Explanation:

The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C

The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)

hence the voltage in the battery will be equal to the voltage across each bulb

Answer:

1.680kN/m

Explanation:

Work done by the spring is expressed as shown:

[tex]W = \frac{1}{2}ke^2[/tex] where:

k is the spring constant

e is the extension

Given

W = 525Joules

extension = 25cm = 0.25m

Substitute into the formula:

[tex]525 = \frac{1}{2}k(0.25)^{2} \\525 = \frac{0.0625k}{2}\\ 525 = 0.03125k\\k = \frac{525}{0.3125}\\k = 1680N/m\\k = 1.680kN/m[/tex]

Hence the force constant of the spring is 1.680kN/m

Answer:

vₓ = 53.6 km/h

vy = 12.4 km/h

Explanation:

       [tex]v_{x} = v_{o} * cos \theta = 55 km/h * cos 13 = 53.6 km/h[/tex]

       [tex]v_{y} = v_{o} * sin\theta = 55 km/h * sin 13 = 12.4 km/h[/tex]

Answer:

(a) Wavelength = 3.21 m (b) Time = [tex]1.07\times 10^{-4}\ s[/tex]

Explanation:

Given that,

The frequency of FM radio station, f = 93.4 MHz

(a) We need to find the wavelength of the radio wave associated with this signal. The relation between wavelength and frequency is given by :

[tex]c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{93.4\times 10^6}\\\\\lambda=3.21\ m[/tex]

(b) It is given that, an FM radio station, 20 miles away. Let t is time taken for signal to reach your radio from the station. So,

[tex]t=\dfrac{d}{c}\\\\t=\dfrac{20\times 1609.34}{3\times 10^8}\\\\t=1.07\times 10^{-4}\ s[/tex]

Hence, this is the required solution.

Answer:

also choose D and E i belive those are also correct

Answer:

t= 0.68 s

Explanation:

        [tex]y = v_{o}* t -\frac{1}{2} * g * t^{2} (1)[/tex]

        [tex]y = y_{o} - \frac{1}2}*g*t^{2} (2)[/tex]

        [tex]y -y_{o} = -\frac{1}{2}*g* t^{2} (3)[/tex]

       [tex]y = v_{o}*t + (y- y_{o})[/tex]

       ⇒ [tex]t =\frac{y_{o} }{v_{o} } =\frac{14 m}{20.6 m/s} = 0.68 s[/tex]

       ⇒ t = 0.68s

Complete Question

The complete question is shown on the first uploaded image

Answer:

The equation is   [tex]c = \frac{1}{\sqrt{ \mu_o *  \epsilon_o} }[/tex]

The value of  c is [tex]c = 2.998 *10^{8} \  m/s  [/tex]

Explanation:

From the question we are told that

   Generally the equation that  relates the speed of light to other fundamental electromagnetic constants is

     [tex]c = \frac{1}{\sqrt{ \mu_o *  \epsilon_o} }[/tex]

Here  c is the speed of light

[tex]\mu_o[/tex] is the permeability of free space with value

    [tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]

and  [tex]\epsilon_o[/tex] is the permittivity of free space  with value  

      [tex]\epsilon_o  =  8.85*10^{-12} \ C/V \cdot m[/tex]

So

      [tex]c = \frac{1}{\sqrt{ 4\pi *10^{-7}  *  8.85*10^{-12}} }[/tex]

=>   [tex]c = 2.998 *10^{8} \  m/s  [/tex]

Answer:

Options B & E are correct

Explanation:

Looking at all the options, B & E are the correct ones.

Option B is correct because the thicker the wire per unit length, the lesser resistance it will posses and the lesser the energy that will be dissipated by the wire and in return more energy will be dissipated by the bulb.

Option E is also correct because the resistance of the copper wires is low enough to ensure that there's not much drop in voltage across the copper wires. Thus, there will not be any noticeable differences in the voltage across the bulb.

Option A is not correct because the current is not used up and thus the charge is conserved, and it will circulate just through the circuit.

Option C is not correct because although the Electric field along the wire is not zero, it is very small.

Option D is not correct because the wires and the light bulb are connected in series and as such, the current in both the wires and the light bulb will be identical.

The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is :

b. Very little energy is dissipated in the thick connecting wires.

e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.

The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is very little energy is dissipated in the thick connecting wires and the electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.

The thicker the wire per unit length, the lesser resistance it'll posses and the lesser the vitality that will be scattered by the wire and in return more vitality will be disseminated by the bulb.

The resistance of the copper wires is low sufficient to guarantee that there's not much drop in voltage over the copper wires. Hence, there will not be any noticeable contrasts within the voltage over the bulb.

Thus, the correct answer is B and E.

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Answer:

Philosophy is a study that involves the nature of knowledge and truth. It serves as a guide that helps an individual seek which things are valuable and essential in life. ... It gives you a sense of direction, knowing the weight of things, therefore making you more productive.

Answer:

5 blocks east. His total distance is 15 feet.

Explanation:

I hope that this helps! Have a good day!

Answer:

1.5m/s^2

Explanation:

Answer:

1.5 m/s2. accerelation =force ÷mass

Answer:

Its rotation will be 3.89x10⁴ rad/s.

Explanation:

We can find the rotation speed by conservation of the angular momentum:

[tex] L_{i} = L_{f} [/tex]

[tex] I_{i}\omega_{i} = I_{f}\omega_{f} [/tex]   (1)

The initial angular speed is:

[tex] \omega_{i} = \frac{1 rev}{7 d} = 0.14 \frac{rev}{d} [/tex]

The moment of inertia (I) of a sphere is:

[tex] I = \frac{2}{5}mr^{2} [/tex]    (2)

Where m is 9 times the sun's mass and r is the sun's radius

By entering equation (2) into (1) we have:

[tex] \frac{2}{5}m_{i}r_{i}^{2}\omega_{i} = \frac{2}{5}m_{f}r_{f}^{2}\omega_{f} [/tex]

[tex]9m_{sun}(696342 km)^{2}0.14\frac{rev}{d} = \frac{3}{4}9m_{sun}(13 km)^{2}\omega_{f}[/tex]

[tex]\omega_{f} = \frac{4}{3}*0.14 \frac{rev}{d}(\frac{696342 km}{13 km})^{2} = 5.36 \cdot 10^{8} \frac{rev}{d}*\frac{1 d}{24 h}*\frac{1 h}{3600 s}*\frac{2\pi rad}{1 rev} = 3.89 \cdot 10^{4} rad/s[/tex]          

Hence, its rotation will be 3.89x10⁴ rad/s.

I hope it helps you!                                                        

Answer:

7 Newton's East

Explanation:

when the force is going in the same direction in this case east, you add the forces.

Answer:

-9m/s²

Explanation:

Given parameters:

Initial velocity = 45m/s

Final velocity  = 0

duration  = 5s

Unknown:

acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time;

        Acceleration  = [tex]\frac{v- u}{t}[/tex]

v is the final velocity

u is the initial velocity

t is the time taken

 Input the parameters and solve;

     Acceleration  = [tex]\frac{0 - 45}{5}[/tex]   = -9m/s²

The car accelerates at a rate of -9m/s² which is a deceleration

Answer:

2.The centripetal force required to keep the pennies in place increases with the distance from the center. Eventually, as the turntable spins faster, the friction force between the turntable and the penny near the edge is not enough to supply the required centripetal force.

Explanation:

centripetal force = m ω² R

here m is mass , ω is angular velocity and R is distance of penny from centre

So this force depends upon R

penny on the outer edge will require  greater centripetal force to move in circular path .

The centripetal force will be provided by frictional force of table which is same for both the coin . Hence the penny on the outer edge will slip off first the moment , frictional force reach its maximum value for it . But it will be sufficient to keep in balance the penny nearer to the centre .

Answer:

W = 0.012 J

Explanation:

For this exercise let's use Hooke's law to find the spring constant

         F = K Δx

         K = F / Δx

         K = 3 / (0.16 - 0.11)

         K = 60 N / m

Work is defined by

         W = F. x = F x cos θ

in this case the force and the displacement go in the same direction therefore the angle is zero and the cosine is equal to 1

         W = ∫ F dx        

         W = k ∫ x dx

we integrate

         W = k x² / 2

          W = ½ k x²

let's calculate

         W = ½ 60 (0.19 -0.17)²

         W = 0.012 J

Answer:

Explanation:

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. FALSE. The specific lines are obseved because of the energy level transition of an electron in an specific level to another level of energy.

The energies of atoms are not quantized. FALSE. The energies of the atoms are in specific levels.

When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. FALSE. During absorption, a specific wavelength of light is absorbed, not emmited.

Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. TRUE. Again, you can observe just the transition due the change of energy of an electron in the quantized energy level

When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. TRUE. The electron decreases its energy releasing a specific wavelength of light.

The energies of atoms are quantized. TRUE. In fact, the energy of all subatomic, atomic, and molecular particles is quantized.

The reason why atoms emit only specific wavelengths is because the energy levels in atoms are quantized.

Max Plank introduced the idea of quantization of energy in the early 1900s. He introduced the idea that energy can only take on certain specific values. This idea was later extended to atoms by Neils Bohr.

The following statements explain why atoms only emit certain wavelengths of light when they are excited;

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Statements 1,2,3 and 4 match statements B, C, A, and D respectively.A wind turbine converts rotational energy into electrical energy.

According to the Law of conservation of energy. Energy can not be created nor be destroyed, it can transfer from one to another form.

1.A wind turbine converts rotational energy into electrical energy.

2.A roller coaster going downhill converts gravitational energy into mechanical energy

3. Toaster converts electrical energy into thermal energy

4.A car converts rotational energy into mechanical energy.

Hence,statements 1,2,3 and 4 match statements B, C, A, and D respectively.

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Answer:

he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

Explanation:

In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values ​​where some natural frequency of the system coincides with the excitation frequency.

In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.

From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

Answer:

F = 58.35 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that force is equal to the product of mass by acceleration. But first we must use the following equation of kinematics.

We have to convert speeds from kilometers per hour to meters per second

[tex]\frac{26km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=\frac{7.22m}{s} \\\frac{54km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex]

[tex]v_{f}=v_{o}+(a*t) \\[/tex]

where:

Vf = final velocity = 15 [m/s]

Vi = initial velocity = 7.22 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Note: the positive sign of the above equation is because the car increases its speed

15 = 7.22 + (a*4)

a = 1.945 [m/s^2]

Now we can use the Newton's second law:

F = m*a

F = 30*1.945

F = 58.35 [N]

(a) Differentiate the position vector to get the velocity vector:

r(t) = (3.00 m/s) t i - (4.00 m/s²) t² j + (2.00 m) k

v(t) = dr/dt = (3.00 m/s) i - (8.00 m/s²) t j

(b) The velocity at t = 2.00 s is

v (2.00 s) = (3.00 m/s) i - (16.0 m/s) j

(c) Compute the electron's position at t = 2.00 s:

r (2.00 s) = (6.00 m) i - (16.0 m) j + (2.00 m) k

The electron's distance from the origin at t = 2.00 is the magnitude of this vector:

||r (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the x-y plane, the velocity vector at t = 2.00 s makes an angle θ with the positive x-axis such that

tan(θ) = (-16.0 m/s) / (3.00 m/s)   ==>   θ ≈ -79.4º

or an angle of about 360º + θ ≈ 281º in the counter-clockwise direction.

Answer:

VB

Explanation:

Answer:

We must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).

Explanation:

LIFTING:

When an object is lifted, we first need to overcome the force exerted on it by the field of gravity. Due to this force, which is also called the weight of object, we must apply a force on the object to keep it moving at constant speed, otherwise the gravity force will cause the object to slow down and eventually fall back on ground.

PUSHING:

When pushing an object the person must apply the force to first overcome the frictional force. The frictional force acts in opposite direction of motion. Thus, to move the object at constant speed we must apply force to it.

Hence, we must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).

Answer:

147.27N

Explanation:

Power = workdone/time

Power = Force*distance/time

Given

Power = 90Watts

Distance = 1.1m

Time = 1.8secs

Force = ?

Substitute the given parameters into the formula:

[tex]90 = \frac{1.1d}{1.8}\\cross \ multiply\\ 90 \times 1.8 = 1.1F\\162 = 1.1F\\1.1F = 162\\F = \frac{162}{1.1} \\F = 147.27N[/tex]

Hence the magnitude of the force that you exert on the handle is 147.27N

Answer: The four basics of innate knowledge are as follows:

Explanation:

1. This knowledge is learned from birth and it is inborn knowledge. It is allows the organism to act naturally. For example, a dog is not taught to pant, but it pants to reduce heat from the body.

2. It is inherent.

3. It is essential for survival.

4. It arises from intellectual knowledge rather than being learned via experiences.

Answer:

Explanation:

If a particle move with time and expressed according to the formula:

f(t) = 0.01t⁴ − 0.03t³

a) Velocity is the change in motion of the particle with respect to time and it is expressed as;

[tex]v(t) =\frac{d(f(t))}{dt}[/tex]

[tex]v(t) = 4(0.01)t^{4-1} - 3(0.03)t^{3-1}\\v(t) = 0.04t^3 - 0.09t^2[/tex]

Hence the velocity of the particle at time t is [tex]v(t) = 0.04t^3 - 0.09t^2[/tex]

b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:

[tex]v(t) = 0.04t^3 - 0.09t^2\\v(1) = 0.04(1)^3 - 0.09(1)^2\\v(t) = 0.04 - 0.09\\v(t) = -0.05[/tex]

Hence the velocity after 1second is -0.05

c) The particle is at rest when when the time is zero.

Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second