Your friend's Frisbee has become stuck 19 m above the ground in a tree. You want to dislodge the Frisbee by throwing a rock at it. The Frisbee is stuck pretty tight, so you figure the rock needs to be traveling at least 4.1 m/s when it hits the Frisbee.

If you release the rock 1.8 m above the ground, with what minimum speed must you throw it?

Answer:

where are the options

it's not full question

Answer:

Explanation:capacitor

Answer:

Ammeter

Explanation:

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Explanation:

the car will be brought back

Answer:

The angle is 153.9 degree.

Explanation:

Let the magnetic field is B and the charge is q. Angle = 26.1 degree

The force is F.

Let the angle is A'.

Now equate the magnetic forces

[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]

Temperature in kelvin scale=300k

We know

[tex]\boxed{\sf 0°C=273K}[/tex]

[tex] \\ \Large\sf\longmapsto 300K[/tex]

[tex] \\ \Large\sf\longmapsto 300-273[/tex]

[tex] \\ \Large\sf\longmapsto 27°C[/tex]

Solution :

Let the positive [tex]x-axis[/tex] is along the East and the positive [tex]y[/tex] direction is along the north.

Given :

Mass of the Tesla car, [tex]m_1[/tex] = [tex]750 \ kg[/tex]

Mass of the Ford car, [tex]m_2 = 1250 \ kg[/tex]

Now let the initial velocity of Tesla car in the south direction be = [tex]-v_1j[/tex]

The initial momentum of Tesla car, [tex]p_1 = -750 \ v_1[/tex]

Let the initial velocity of Ford car in the east direction be = [tex]v_2 \ i[/tex]

So the initial momentum of the Ford car is [tex]p_2=1250\ v_2 \ i[/tex]

Therefore, the initial velocity of both the cars is [tex]p_i = p_1+p_2[/tex]

                                                                  [tex]=1250 \ v_2 \ i - 750\ v_1 \ j[/tex]

Now the final velocity of both the cars is [tex]v = 18 \ m/s[/tex]

So the vector form is :

[tex]v = 18\cos 32\ i-18 \sin 32 \ j[/tex]

  [tex]= 15.26 \ i - 9.54 \ j[/tex]

Therefore the momentum after the accident is

[tex]p_f=(m_1+m_2) \times v[/tex]

    [tex]=(750+1250) \times (15.26 \ i - 9.54 \ j)[/tex]

    [tex]= 30520\ i -19080\ j[/tex]

According to the law of conservation of momentum, we know

[tex]p_i = p_f[/tex]

[tex]1250 \ v_2 \ i - 750\ v_1 \ j[/tex]  [tex]= 30520\ i -19080\ j[/tex]

[tex]1250 \ v_2 = 30520[/tex]

[tex]v_2=24.4 \ m/s[/tex]

From, [tex]750\ v_1 = 19080[/tex]

We get, [tex]v_1=25.4 \ m/s[/tex]

Therefore the speed of Tesla car before collision = 25.4 m/s

The speed of ford car before collision = 24.4 m/s

Answer:

I = 3.6 kg•m²

Explanation:

Conservation of angular momentum

Let's assume CW is the positive direction

3.4(-6.1) + I(9.3) = 3.4(1.8) + I(1.8)

I(9.3 - 1.8) = 3.4(1.8 + 6.1)

I(7.5) = 3.4(7.9)

I = 3.4(7.9)/(7.5) = 3.5813333333...

The moment of inertia of the second disk will be  [tex]I=3.58\ kg-m^2[/tex]

The moment of inertia is defined as the product of mass of section and the square of the distance between the reference axis and the centroid of the section.

here it is given that

MOI of disk one   [tex]I_1=3.4\ kg-m^2[/tex]

Angular velocity  [tex]w_1=6.1\ \frac{rad}{s}[/tex]

Angular velocity of disk two  [tex]w=1.8\ \frac{rad}{s}[/tex]

MOI of the disk two [tex]I=?[/tex]

The final angular velocity [tex]w_f= 1.8\ \frac{rad}{sec}[/tex]

Now from the conservation of the momentum the angular momentum before collision will be equal to the angular momentum after collision.

[tex]I_1w_1+I_2w_2=(I_1+I_2)w_f[/tex]

Now put the values in the formula

[tex](3.4\times 6.10)+(I_2\times 9.3)=(3.4+I_2)\times 1.8[/tex]

[tex]I_2=3.58\ kg-m^2[/tex]

Thus the moment of inertia of the second disk will be  [tex]I=3.58\ kg-m^2[/tex]

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Answer:

English please?

Explanation:

Explain in english?

Answer:

   x = 9.32 cm

Explanation:

For this exercise we have an applied torque and the bar is in equilibrium, which is why we use the endowment equilibrium equation

Suppose the counterclockwise turn is positive, let's set our reference frame at the left end of the bar

          - W l / 2 - W_{child} x + N₂ l = 0

             x = [tex]\frac{-W l/2 + n_2 l}{W_{child}}[/tex]             1)

now let's use the expression for translational equilibrium

         N₁ - W - W_(child) + N₂ = 0

indicate that N₂ = 4 N₁

we substitute

           N₁ - W - W_child + 4 N₁ = 0

           5 N₁ -W - W_{child} = 0

           N₁ = ( W + W_{child}) / 5

we calculate

           N₁ = (450 + 250) / 5

          N₁ = 140 N

we calculate with equation 1

           x = -250 1.50 + 4 140 3) / 140

           x = 9.32 cm

Answer:

12

Explanation:

i agree the answer is 12 because the acceleration is given by the difference in the velocity divided by the time taken

a=v-u/t

60-0/5

=12m/s²

I hope this helps

Answer:

From far away, Saturn looks like it has seven large rings. Each large ring is named for a letter of the alphabet. The rings were named in the order they were discovered.

Answer:

Hence the battery's emf E is ε = 11.9 V.

The internal resistance is r = 0.739 ohms.

Explanation:

Now we know that

Voltage V = 11.9 V.

Current I = 16.1 A.

Hence this is an ideal voltmeter there are no current flows when the Voltmeter is applied.

ε = V + I r

∵ I = 0

ε = V

ε = 11.9 V

Then the ammeter is applied.

Let's take ( r ) to be the total resistance which is equal to internal resistance.

V = I r

r = [tex]\frac{V}{I}[/tex]

 [tex]= \frac{11.9}{16.1}[/tex]

r = 0.739 ohms

The battery's emf (E) and internal resistance (r) are 11.9 Volts and 0.739 Ampere respectively.

Given the following data:

To determine the battery's emf (E) and internal resistance (r):

For an ideal voltmeter, there isn't a flow of current and as such the current is equal to 0.

Mathematically, emf (E) is given by this formula:

[tex]E = V + IR[/tex]

Substituting the given parameters into the formula, we have;

[tex]E = 11.9 + 0R\\\\E = 11.9 + 0[/tex]

E = 11.9 Volts.

For the internal resistance (r):

Note: The total resistance is equal to internal resistance.

Applying Ohm's law, we have:

[tex]R = \frac{V}{I} \\\\R = \frac{11.9}{16.1}[/tex]

R = r = 0.739 Ampere.

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Answer:

a)   # _photon = 2.5 10¹⁸ photons / s,   b) E = 10⁻² N / C,  c)     B = 3 10⁻¹¹ T

d)  r=  2 10⁹ m

Explanation:

a) Let's solve this exercise in part, let's start by finding the energy of each photon using the Planck relation

          E₀ = h f

          c = λ f

          E₀ = h c /λ

          E₀ = 6.63 10⁻³³⁴   3 10⁸/500 10⁻⁹

          E₀ = 3.978 10⁻⁻¹⁹ J

Let's use a direct ratio rule to find the number of photons

         #_foton = E / Eo

         #_fototn = 1 / 3.978 10⁻¹⁹

         # _photon = 2.5 10¹⁸ photons / s

b) The intensity received by the detector is related to the electric field

          I = E²

Let's look for the intensity that the detector receives, suppose that the emission is shapeless throughout the space

          I = P / A

          P = I A

Let's use index 1 for the point on the bulb and index 2 for the point on the detector.

The area of ​​a sphere is

          A = 4π r²

         P = I₁ A₁ = I₂ A₂

         I₁ r₁² = I₂ r₂²

         I₂ = I₁  r₁²/r₂²

         I₂ = I₁    1 / 100²

         I₂ = I₁ 10⁻⁴

we must know the intensity at the output of the bulb suppose that I₁ = 1 J

          I₂ = 10⁻⁴ J

let's look for the electric field

         E =√I

         E = √10⁻⁴

         E = 10⁻² N / C

c) for the calculation of the magnetic field we use that the field is in phase

               E / B = c

               B = E / c

               B = 10⁻² / 3 10⁸

               B = 3 10⁻¹¹ T

d) Let's use a direct proportions rule if we fear 2.5 10¹⁸ photons in an area  A = 4π R² where R = 100 m how many photons are there in the area of ​​the detector r = 1 cm,   A’= 10⁻⁴ m²

             #_photons = 2.5 10¹⁸ A_detector / A_sphere

             #_photons = 2.5 1018 10-4 / 4π 10⁴

             #_photons = 2 10⁹ photons in the detector area

for the number of photons to decrease to 1, the radius of the sphere must be 2 10⁹ m

Answer:

Part A - 4.084 mJ

Part B - 0.908 mJ

Part C - 8.168 mJ

Explanation:

Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?

Since capacitors C₂ and C₃ are in series, their equivalent capacitance is C',

1/C' = 1/C₂ + 1/C₃      (Since C₁ = C₂ = C₃ = C)

1/C' = 1/C + 1/C

1/C' = 2/C

C' = C/2

Since C' is in parallel with C₁, the equivalent capacitance for the circuit is C" = C₁ + C' = C + C/2 = 3C/2

C" = 3C/2

The energy stored in the circuit, W = 1/2C"V² where C" = equivalent capacitance = 3C/2 and V = voltage = 15.0 V

W = 1/2C"V²

W = 1/2(3C/2)V²

W = 3CV²/4

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = 3CV²/4

W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/4

W = 3 × 24.2 × 10⁻⁶ F × 225 V²/4

W = 16335/4 × 10⁻⁶ FV²

W = 4083.75 × 10⁻⁶ J

W = 4.08375 × 10⁻³ J

W = 4.08375 mJ

W ≅ 4.084 mJ

Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?

If the capacitors are connected in series, their equivalent resistance is C'

and 1/C' = 1/C₁ + 1/C₂ + 1/C₃

Since C₁ = C₂ = C₃ = C

1/C' = 1/C + 1/C + 1/C

1/C' = 3/C

C' = C/3

The energy stored in the circuit, W = 1/2C'V² where C' = equivalent capacitance = C/3 and V = voltage = 15.0 V

W = 1/2C'V²

W = 1/2(C/3)V²

W = CV²/6

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = CV²/6

W = 24.2 × 10⁻⁶ F (15.0 V)²/6

W = 24.2 × 10⁻⁶ F × 225 V²/6

W = 5445/6 × 10⁻⁶ FV²

W = 907.5 × 10⁻⁶ J

W = 0.9075 × 10⁻³ J

W = 0.9075 mJ

W ≅ 0.908 mJ

Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?

If the capacitors are connected in parallel, their equivalent resistance is C'

and C' = C₁ + C₂ + C₃

Since C₁ = C₂ = C₃ = C

C' = C + C + C

C' = 3C

The energy stored in the capacitor network, W = 1/2C'V² where C' = equivalent capacitance = 3C and V = voltage = 15.0 V

W = 1/2C'V²

W = 1/2(3C)V²

W = 3CV²/2

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = 3CV²/2

W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/2

W = 3 × 24.2 × 10⁻⁶ F × 225 V²/2

W = 16335/2 × 10⁻⁶ FV²

W = 8167.5 × 10⁻⁶ J

W = 8.1675 × 10⁻³ J

W = 8.1675 mJ

W ≅ 8.168 mJ

[tex]\textsf{When an electron is hit by a }[/tex] [tex]\textsf{photon of light, it absorbs the quanta}[/tex] [tex]\textsf{of energy the photon was carrying}[/tex] [tex]\textsf{and moves to a higher energya}[/tex] [tex]\textsf{ state. Electrons therefore have to }[/tex] [tex]\textsf{jump around within the atom as }[/tex] [tex]\textsf{they either gain or lose energy. }[/tex]

When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. Electrons therefore have to jump around within the atom as they either gain or lose energy.

Hope this answer helps you..!!!

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Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.  

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

[tex] F = qv \times B = qvBsin(\theta) [/tex]     (1)          

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

[tex] K = \frac{1}{2}mv^{2} [/tex]

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]  

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!        

Answer: [tex]3.92\ m/s[/tex]

Explanation:

Given

Mass of the attached object is [tex]m=2.3\ kg[/tex]

Spring is stretched by [tex]A=0.5\ m[/tex]

Speed reaches zero after [tex]t=0.4\ s[/tex]

Speed is zero at the extremities of the S.H.M motion that is

[tex]\Rightarrow \dfrac{T}{2}=0.4\\\\\Rightarrow T=0.8\ s[/tex]

Time period of motion is [tex]0.8\ s[/tex] which can also be given by

[tex]\Rightarrow \omega T=2\pi\\\\\Rightarrow \omega=\dfrac{2\pi }{T}\\\\\Rightarrow \omega =\dfrac{2\pi }{0.8}\\\\\Rightarrow \omega=\dfrac{5\pi }{2}[/tex]

Maximum speed for S.H.M. is [tex]v_{max}=A\omega[/tex]

[tex]\Rightarrow v_{max}=0.5\times 2.5\pi\\\Rightarrow v_{max}=3.92\ m/s[/tex]

Answer:

The speed of proton is 2.1 x 10^5 m/s .

Explanation:

potential difference, V = 234 V

let the initial speed of the proton is v.

The kinetic energy of proton is

KE = q V

[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]

Answer:

The pressure near the surface of the pool will be less as compared that the bottom of the pool as water has weight. This is in relation to gravity

Explanation:

Answer:

The lowest possible frequency of sound for which this is possible is 1307.69 Hz

Explanation:

From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.

First, we will determine his distance from the second speaker using the Pythagorean theorem

l₂ = √(2.00²+5.00²)

l₂ = √4+25

l₂ = √29

l₂ = 5.39 m

Hence, the path difference is

ΔL = l₂ - l₁

ΔL = 5.39 m - 5.00 m

ΔL = 0.39 m

From the formula for destructive interference

ΔL = (n+1/2)λ

where n is any integer and λ is the wavelength

n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.

Then,

0.39 = (1+ 1/2)λ

0.39 = (3/2)λ

0.39 = 1.5λ

∴ λ = 0.39/1.5

λ = 0.26 m

From

v = fλ

f = v/λ

f = 340 / 0.26

f = 1307.69 Hz

Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.

Answer:

Magnetic Forces on Moving Charges. The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule.

The direction of the charge where it does not experience magnetic force is left and right. The correct option is C and D.

A magnetic field is a region in space where a magnetic force can be observed. It is created by moving electric charges, such as electrons, and is characterized by the direction and strength of the force it exerts on other magnetic materials or moving charges.

Magnetic field lines are used to visualize the direction and strength of the magnetic field. They represent the path that a small magnetic north pole would follow if placed in the magnetic field. The direction of the magnetic field is given by the direction in which the north pole of a compass needle would point if placed in the field.

Magnetic flux is the measure of the strength of the magnetic field passing through a surface. It is given by the product of the magnetic field strength and the area of the surface, as well as the cosine of the angle between the magnetic field and the surface normal. The unit of magnetic flux is Weber (Wb).

Magnetic flux is important in many applications, such as electric motors and generators, where the interaction between the magnetic field and moving charges produces electrical energy. It is also used in magnetic imaging techniques, such as MRI, to visualize the internal structures of the human body.

Here in the question,

Options A (up), B (down), E (into the screen), and F (out of the screen) are all perpendicular to the direction of the magnetic field and so will experience a magnetic force.

Therefore, options C and D are correct i. e left and right.as they are parallel and anti-parallel to the direction of the magnetic field, respectively.

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Answer:

Because SI system has fundamental units of MKS System

Answer:

Explanation: the unit of length ,mass , and time are same in both the system , thus, the SI system is the extended from of MKS system.

Answer:

b = frequency

Answer:

5572.8 N

Explanation:

Applying,

F  = ma.............. Equation 1

Where F = Force, m = mass of the car, a = acceleration.

We can find a by applying,

v² = u²+2as............. Equation 2

Where v = final velocity, u = initial velocity, a = acceleration,  = distance.

From the question,

Given: v = 0 m/s (come to rest), u = 1.7 m/s, s = 0.210 m

Substitute these value into equation 2

0² = 1.7²+2×0.21×a

a = -1.7²/(2×0.21)

a = -2.89/0.42

a = -6.88 m/s²

Also given: m = 810 kg

Substitute these value into equation 1

F = 810(-6.88)

F = -5572.8 N

Hence the force on the bumber is 5572.8 N

Answer:

............................................

Explanation:

Explanation:

Those tools that helps to make our work easier ,faster and more convenient in our daily life it is called simple Machine.

Answer:

2400kgm²

Explanation:

Rotational inertia=mass x radius²

Answer:

D) Quadruple.

Explanation:

We will use the second equation of motion to solve this problem:

[tex]s = v_it + \frac{1}{2}gt^2[/tex]

where,

s = distance travelled by the rock

vi = initial speed of rock = 0 m/s

t = time taken

g = acceleration due to gravity on the surface of the moon

Therefore,

[tex]s = (0\ m/s)t+\frac{1}{2}gt^2\\\\s =\frac{1}{2}gt^2[/tex]----------- equation (1)

Now, we double the time:

[tex]s' = \frac{1}{2}g(2t)^2\\\\s' = 4(\frac{1}{2}gt^2)[/tex]

using equation (1):

s' = 4s

Hence, the correct option is:

D) Quadruple.