What is the wavelength of the radio waves from an FM station operating at a frequency of 99.5 MHz

Answer:

a) Magnitude of the average velocity from t = 0 to t = 2 s = 2.24 m/s

b) Magnitude of the average velocity from t = 0 to t = 5 s = 3.16 m/s

Explanation:

a) Velocity is rate of change of position.

A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0

A particle's position coordinates (x, y) are (5.0 m, 5.0 m) at t = 2.0 s

Displacement = (5-1)i + (5-3)j = 4i + 2j

Change in time = 2s

Velocity

      [tex]v=\frac{4i+2j}{2}=2i+j[/tex]

Magnitude of velocity

      [tex]v=\sqrt{2^2+1^2}=2.24m/s[/tex]

b) Velocity is rate of change of position.

A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0

A particle's position coordinates (x, y) are (14.0 m, 12.0 m) at t = 5.0 s

Displacement = (14-1)i + (12-3)j = 13i + 9j

Change in time = 5s

Velocity

      [tex]v=\frac{13i+9j}{5}=2.6i+1.8j[/tex]

Magnitude of velocity

      [tex]v=\sqrt{2.6^2+1.8^2}=3.16m/s[/tex]

Answer:

21828 m

Explanation:

[tex]v_{ground}[/tex] = speed of sound through ground = 6 km/s = 6000 m/s

[tex]v_{air}[/tex] = speed of sound through air = 343 m/s

t = time taken for the vibrations to arrive

t' = time taken for sound to arrive = t + 60

d = distance of the point of explosion

time taken for the vibrations to arrive is given as

[tex]t = \frac{d}{v_{ground}}[/tex]                            eq-1

time taken for the sound to arrive is given as

[tex]t' = \frac{d}{v_{air}}[/tex]

[tex]t + 60 = \frac{d}{v_{air}}[/tex]

using eq-1

[tex]\frac{d}{v_{ground}} + 60 = \frac{d}{v_{air}}[/tex]

[tex]\frac{d}{6000} + 60 = \frac{d}{343}[/tex]

d = 21828 m

Answer:

12 m

Explanation:

f = 0.2 Hz, v = 2.4 m/s

v = f x λ

Where, λ is the wavelength

λ = v / f = 2.4 / 0.2 = 12 m

The wavelength is defined as the distance between two consecutive crests or troughs.

So, the distance between two consecutive crests is 12 m.

Answer:

A) weight=mass X gravitational force

10Kg X 10 M/S 2

100 KgM/S2

100Newton

B) Force =Mass X gravitational force

10Kg X 10 M/S2

100 Newton

The weight of a 10kg stone sitting at rest on the ground is 98 Newtons, and the normal force acting on it is also 98 Newtons.

The weight of an object is calculated by multiplying its mass by the acceleration due to gravity. In this case, the mass of the stone is 10kg and  The standard acceleration due to gravity on Earth is approximately 9.8 m/s2.  Therefore, the weight of the stone would be 10 kg x 9.8 m/s^2 = 98 Newtons.

b) The normal force acting on the stone is the force exerted by the surface of the ground on the stone in the upward direction. When the stone is sitting on the ground, the normal force is equal in magnitude and opposite in direction to the weight of the stone. Therefore, the normal force is also 98 N.

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Answer: false

Explanation: the air will increase if it is compressed by an adiabatic compressor

Answer:

a)

600 N

b)

1.2 x 10⁵ Pa

Explanation:

(a)

d₁ = diameter of small piston = 8 cm = 0.08 m

d₂ = diameter of large piston = 40 cm = 0.40 m

F₂ = force applied to large piston = 15000 N

F₁ = force applied to small piston = ?

Using pascal's law

[tex]\frac{F_{1}}{(0.25)\pi d_{1}^{2}} = \frac{F_{2}}{(0.25)\pi d_{2}^{2}}[/tex]

Inserting the values

[tex]\frac{F_{1}}{(0.08)^{2}} = \frac{15000}{(0.40)^{2}}[/tex]

F₁ = 600 N

b)

Pressure applied is given as

[tex]P = \frac{F_{1}}{(0.25)\pi d_{1}^{2}}[/tex]

[tex]P = \frac{(600)}{(0.25)(3.14) (0.08)^{2}}[/tex]

P = 1.2 x 10⁵ Pa

To determine the force needed on the small piston of a hydraulic lift, the ratios of the pistons' areas are used in combination with the load on the large piston, applying Pascal's Principle. The pressure in the hydraulic fluid is then found by dividing the force by the area of the piston where the force is applied.

Given the diameters of the pistons, we can find the areas by using the formula for the area of a circle, A = πr², where r is the radius of the piston.

For the provided diameters of 8.0 cm and 40 cm, the areas of the pistons are:

With a load of 15,000 N on the large piston:

Once we perform these calculations, we can determine the required force to apply to the small piston and the pressure applied to the fluid within the lift.

Answer:

42.3 MV

Explanation:

d = diameter of the metal sphere = 2.15 m

r = radius of the metal sphere

diameter of the metal sphere is given as

d = 2r

2.15 = 2 r

r = 1.075 m

Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C

Potential near the surface is given as

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

[tex]V = \frac{(9\times 10^{9})(5.05\times 10^{-3})}{1.075}[/tex]

V = 4.23 x 10⁷ volts

V = 42.3 MV

Answer:

The final temperature is 79.16°C.

Explanation:

Given that,

Heat [tex]Q=1.50\times10^{5}\ J[/tex]

Temperature = 20.0°C

Entropy = 465 J/k

We need to calculate the average temperature

Using relation between entropy and heat

[tex]\Delta S=\dfrac{\Delta Q}{T}[/tex]

[tex]T=\dfrac{\Delta Q}{\Delta S}[/tex]

Where, T = average temperature

[tex]\Delta Q[/tex]= transfer heat

[tex]\Delta S[/tex]= entropy

Put the value into the formula

[tex]T=\dfrac{1.50\times10^{5}}{465}[/tex]

[tex]T=322.58\ K[/tex]

We need to calculate the final temperature

Using formula of average temperature

[tex]T = \dfrac{T_{i}+T_{f}}{2}[/tex]

[tex]T_{f}=2T-T_{i}[/tex]....(I)

Put the value in the equation (I)

[tex]T_{f}=2\times322.58-293[/tex]

[tex]T_{f}=352.16\ K[/tex]

We convert the temperature K to degrees

[tex]T_{f}=352.16-273[/tex]

[tex]T_{f}=79.16^{\circ}\ C[/tex]

Hence, The final temperature is 79.16°C.

Answer:

55.79 minutes

Explanation:

Volume of water = 5200 L = 5.2 m^3

Diameter = 3 cm

radius, r = 1.5 cm = 0.015 m

v = 2.2 m/s

The volume of water coming out in 1 second = velocity x area

                                                       = 2.2 x 3.14 x 0.015 x 0.015

                                                       = 1.55 x 10^-3 m^3

1.55 x 10^-3 m^3 water flows = 1 second

5.2 m^3 water flows = 5.2 / (1.55 x 10^-3) = 3345.57 second

                                                                    = 55.79 minutes

Answer:

Explanation:

Comet is made by dust particles, icy particles, gases etc.

A comet has a fixed time to complete a revolution around the sun.

As a comet comes nearer to the sun, due to the heat of the sun the vapour and the icy particles becomes gases and due to the radial pressure of energy od sun, we observe a tail of comet which has vapours mainly. SI the comet is visible easily.

Answer:

The average power is 619 W.

(B) is correct option

Explanation:

Given that,

Weight = 62.0 kg

Height = 4.28 m

Time t = 4.20 s

We need to calculate the work done

Work done by woman

[tex]W=mgh[/tex]

Where,

m = mass

g = acceleration due to gravity

t = time

Put the value into the formula

[tex]W=62\times9.8\times 4.28[/tex]

[tex]W=2600.528 J[/tex]

We need to calculate the power

Using formula of power

[tex]P=\dfrac{W}{t}[/tex]

Where,

W = work done

t = time

Put the value in to the formula

[tex]P=\dfrac{2600.528}{4.20}[/tex]

[tex]P=619\ W[/tex]

Hence, The average power is 619 W.

The average power supplied by the woman is calculated by first determining the work done against gravity to move up the stairs, and then dividing this by the time taken. After performing these calculations, the most accurate answer is approximately 629 W.

The question is asking to calculate the average power supplied by a woman while running up the stairs. Power is defined as the rate at which work is done. In this scenario, the work is the woman's movement against the gravitational force, which is calculated using the formula W = mgh, where m is mass, g is the acceleration due to gravity and h is the height of the stairs. Substituting the given values, we get W = 62.0 Kg * 9.8 m/s² * 4.28 m = 2650.368 Joules. Power is calculated by dividing the work done by the time taken, so P = W/t = 2650.368 J / 4.20 s = 630.8 Watts, which is closest to option (d) 629 W.

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Explanation:

It is given that,

Mass of SUV, m = 900 kg

It is moving with a constant speed of 1.44 m/s due south. We need to find the total force on the vehicle. The second law of motion gives the magnitude of force acting on the object. It is given by :

F = m a

Since, [tex]a=\dfrac{dv}{dt}[/tex] i.e. the rate of change of velocity

As SUV is travelling at a constant speed. This gives acceleration of it as zero.

So, F = 0

So, the total force acting on SUV is 0 N. Hence, this is the required solution.

Answer:

The total force acting on the vehicle is 0 N.

Explanation:

Given that,

Mass of SUV = 900 kg

Velocity = 1.44 m/s

SUV is travelling at a constant speed of 1.44 m/s due north.

We need to calculate the force on the vehicle.

Using newton's second law

[tex]F = ma[/tex]....(I)

We know that,

The acceleration is the first derivative of the velocity of the particle.

[tex]a = \dfrac{dv}{dt}[/tex]

Now,put the value of a in equation (I)

[tex]F=m\dfrac{dv}{dt}[/tex]

SUV is traveling at a constant speed it means acceleration is zero.

So, The force will be zero.

Hence, The total force acting on the vehicle is 0 N.

Answer:

09m

Explanation:

yes the current flow in clockwise direction

Explanation:

Given that,

Diameter of the metal ring, d = 4.2 cm

Radius, r = 2.1 cm

Initial magnetic field, B = 1.12 T

The magnetic field is decreasing at the rate of, [tex]\dfrac{dB}{dt}=0.24\ T/s[/tex]

Due to change in magnetic field, an emf is induced in it. And hence, electric field is induced. It is given by :

[tex]\int\limits {E.dl}=\dfrac{d}{dt}(\int\limits{B.dA)}[/tex]

[tex]E.(2\pi r)=\dfrac{dB}{dt}(\pi r^2)[/tex]

[tex]E=\dfrac{dB}{dt}\times \dfrac{r}{2}[/tex]

[tex]E=0.24\times \dfrac{2.1\times 10^{-2}}{2}[/tex]

[tex]E=0.00252\ N/C[/tex]

So, the magnitude of the electric field induced in the ring has a magnitude of 0.00252 N/C.

The direction of electric field will be counter clock wise direction as viewed by someone on the south pole of the magnet.

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

[tex]F_{g}[/tex] = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

[tex]T[/tex] = Tension force in upward direction

[tex]F_{d}[/tex] = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

[tex]F_{g}[/tex] - [tex]F_{d}[/tex] - [tex]T[/tex] = 0

7000 - 1800 - [tex]T[/tex] = 0

[tex]T[/tex] = 5200 N

[tex]T[/tex] = 5.2 x 10³ N

Part B)

[tex]F_{g}[/tex] = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

[tex]T[/tex] = Tension force in upward direction

[tex]F_{d}[/tex] = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

[tex]T[/tex]  - [tex]F_{g}[/tex] - [tex]F_{d}[/tex] = 0

[tex]T[/tex] - 7000 - 1800  = 0

[tex]T[/tex] = 8800 N

[tex]T[/tex] = 8.8 x 10³ N

Answer:

(a) 6.42 x 10^-18 N

(b) 9.73 x 10^8 m/s^2

Explanation:

v = 760 m/s, B = 0.034 T, m = 6.6 x 10^-27 kg, q = 3.2 x 10^-19 C, theta = 51 degree

(a) F = q v B Sin theta

F = 3.2 x 10^-19 x 760 x 0.034 x Sin 51

F = 6.42 x 10^-18 N

(b) Acceleration, a = Force / mass

a = (6.42 x 10^-18) / (6.6 x 10^-27)

a = 0.973 x 10^9

a = 9.73 x 10^8 m/s^2

Answer:

The new force F' will be same of the original force F.

Explanation:

Given that,

Charges = 0.3

We need to calculate the force between the charges

Suppose that the distance between the charges is r.

The force between the charges

[tex]F =\dfrac{kq_{1}q_{2}}{r^2}[/tex]

Put the value into the formula

[tex]F=\dfrac{k\times0.3\times0.3}{r^2}[/tex]

[tex]F=\dfrac{k\times(0.09)}{r^2}[/tex]

If the charges are doubled, and the distance between them increased by 100%.

So, The charges are 0.6 and the distance is 2r.

Then,

The force between the charges

[tex]F'=\dfrac{k\times0.6\times0.6}{(2r)^2}[/tex]

[tex]F'=\dfrac{k\times0.09}{r^2}[/tex]

[tex]F'=F[/tex]

Hence, The new force F' will be same of the original force F.

Answer:

The total elongation for the tension member is of 0.25mm

Explanation:

Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:

[tex]\sigma=E*\epsilon[/tex] (1)

Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:

[tex]\delta L=L*\epsilon[/tex] (2)

Here L is the member extension and δL the change total longitudinal elongation.  

Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:

[tex]\sigma=P/A[/tex]  

[tex]\sigma=22.44N / 1290 mm^2[/tex]  

[tex]\sigma=0.0174 N/mm^2[/tex]  

Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:

[tex]=\sigma=E*\epsilon[/tex]

[tex]\epsilon=\sigma/E[/tex]

[tex]\epsilon=0.0174 \frac{N}{mm^2}/\ 204 \frac{N}{mm^2} [/tex]

[tex]\epsilon=8.53*10^-{5}[/tex]

Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):

[tex]\delta L=3048 mm * 8.53*10^{-5} [/tex]  

[tex]\delta L= 0.25 mm [/tex]

Answer:

The acceleration of the block is 6.35 m/s².

Explanation:

It is given that,

A force is applied to a block to move it up a 30° incline. The applied force is, F = 65 N

Mass of the block, m = 5 kg

We need to find the acceleration of the block. From the attached figure, it is clear that.

[tex]F_x=ma_x[/tex]

[tex]F\ cos\theta-mg\ sin\theta=ma_x[/tex]

[tex]a_x=\dfrac{F\ cos\theta-mg\ sin\theta}{m}[/tex]

[tex]a_x=\dfrac{65\ N\ cos(30)-5\ kg\times 9.8\ m/s^2\ sin(30)}{5\ kg}[/tex]

[tex]a_x=6.35\ m/s^2[/tex]

So, the acceleration of the block is 6.35 m/s². Hence, this is the required solution.

Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².The magnitude of the acceleration of the block will be 6.35 m/sec².

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and accelertaion in the different components and balancing the equation gets.Components in the x-direction.

The given data in the problem,

F is the applied force =65 N

Θ is the angle of inclined plane=30°

m is the mass of the block= 5 kg

We need to find the acceleration of the block in the x-direction

[tex]\rm F_x=ma_x\\\\\rm F_x=Fcos\theta-mgsin\theta\\\\\rm Fcos\theta-mgsin\theta=ma_x\\\\\rm a_x=\frac{Fcos\theta-mgsin\theta}{m}\\\\ \rm a_x=\frac{65\timescos30^0-mgsin30^0}{5} \\\\\rm a_x=6.35 m/sec^2[/tex]

Hence the magnitude of the acceleration of the block will be 6.35 m/sec².

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Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = [tex]\frac{1}{3}\pi r^2 h[/tex]

thus,

volume of the cone = [tex]\frac{1}{3}\pi 1.2^2\times 4[/tex] = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the [tex]\frac{1}{4}[/tex]times the total height

thus,

center of mass lies at,  h' = [tex]\frac{1}{4}\times4=1m[/tex]

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

The work against gravity required to build a cone of height 4 m and base of radius 1.2 m out of a material of density 600 kg/m3 is 35,467.278 J.

Given to us

Height of the cone = 4m

The radius of the cone = 1.2 m

Density of the material = 600 kg/m³

We know that the work gone against the gravity is given as,

[tex]W = mgh'[/tex]

where W is the work, m is the mass, g is the acceleration due to gravity, and h' is the center of mass.

Also, the mass of an object is the product of its volume and density, therefore,

[tex]m = v \times \rho[/tex]

We know the value of the volume of a cone,

[tex]m = \dfrac{1}{3}\pi r^2 h \times \rho[/tex]

The center of mass of a cone lies at the center of its base at 1/4 of the total height from the base,

[tex]h' = \dfrac{h}{4}[/tex]

Substitute all the values we will get,

[tex]W = mgh'\\\\W = (v \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi r^2h \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi (1.2)^2 \times4 \times 600) \times 9.81 \times \dfrac{4}{4}\\\\W = 35467.278\rm\ J[/tex]

Hence, the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lightweight material of density 600 kg/m3 is 35,467.278 J.

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Answer:

(a) 2.23 m/s

(b) 0.9 m

Explanation:

h = 1 m, t = 0.1 second

horizontal component of initial velocity, ux = 2 m/s

vertical component of initial velocity, uy = 0  

(a) Let v be the velocity after 0.1 seconds. Its vertical component is vy and horizontal component is vx.

The horizontal component of velocity remains constant as in this direction, acceleration is zero.

vx = ux = 2 m/s

Use first equation of motion in Y axis direction.

vy = uy + g t

vy = 0 + 9.8 x 0.1 = 0.98 m/s

Resultant velocity after 0.1 second

v^2 = vx^2 + vy^2

v^2 = 2^2 + 0.98^2

v = 2.23 m/s

(b) Let it takes time t to land.

Use second equation of motion along Y axis

h = uy t + 1/2 g t^2

1 = 0 + 1/2 x 9.8 x t^2

t = 0.45 second

Let it lan at a distance x.

so, x = ux x t

x = 2 x 0.45 = 0.9 m

A book that slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m, has a coefficient of kinetic friction of 0.166.

A book slides across a level, carpeted floor with an initial speed (u) of 3.25 m/s. It comes to rest (final speed = v = 0m/s) after 3.25 m (s).

Since this is a uniformly decelerated rectilinear motion, we can calculate the acceleration (a) using the following kinematic expression.

[tex]v^{2} = u^{2} + 2as\\\\(0m/s)^{2} = (3.25m/s)^{2} + 2a(3.25m)\\\\a = -1.63 m/s^{2}[/tex]

The negative sign only explains that the friction opposes the movement. We can calculate the force exerted by the friction (F) using Newton's second law of motion.

[tex]F = m \times a[/tex]    [1]

where,

We can also calculate the force of friction using the following expression.

[tex]F = k \times N = k \times m \times g[/tex]    [2]

where,

Given [1] = [2], we get,

[tex]m \times a = k \times m \times g\\\\k = \frac{1.63 m/s^{2} }{9.81 m/s^{2}} = 0.166[/tex]

A book that slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m, has a coefficient of kinetic friction of 0.166.

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This question involves the concepts of the equations of motion, Newton's Second Law, and Frictional Force.

The coefficient of kinetic friction is found to be "0.15".

First, we will use the third equation of motion to find out the acceleration of the book:

[tex]2as = v_f^2-v_i^2\\[/tex]

where,

a = acceleration = ?

s = distance covered = 3.25 m

vf = final speed  = 0 m/s

vi = initial speed = 3.25 m/s

Therefore,

[tex]2a(3.25\ m) = (0\ m/s)^2-(3.25\ m/s)^2\\\\a=\frac{-10.56\ m^2/s^2}{7\ m/s}[/tex]

a = - 1.51 m/s² (negative sign indicates deceleration)

Now the force can be calculated using Newton's Second Law of motion:

F = ma

This force is also equal to the frictional force:

F = μmg

comparing both forces, we get:

ma = μmg

a = μg

[tex]\mu = \frac{a}{g} = \frac{1.51\ m/s^2}{9.81\ m/s^2}[/tex]

μ = 0.15

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The attached picture shows the equations of motion.

Answer:

a) 800 keV

b) 25 km

Explanation:

[tex]Strength\ of\ Electric\ field=2\times 10^6\ V/m\\a)\ Potential\ Difference=Strength\ of\ Electric\ field\times Distance\\\Rightarrow Potential\ Difference=Kinetic\ Energy\ =2\times 10^6\times 0.4\\\therefore Energy=0.8\times 10^6\ eV=800\ keV\\[/tex]

[tex]b)\ Potential\ difference=50\ GeV=50\times 10^9\ eV\\Distance=\frac{Potential\ difference}{electric\ field}\\\Rightarrow Distance=\frac{50\times 10^9}{2\times 10^6}\\\Rightarrow Distance=25\times 10^3\ m\\\therefore Distance=25\ km[/tex]

Answer:

a) 800 keV

b)  24.996 km.

Explanation:

(a) we have

[tex]\large \Delta K.E=q\Delta V[/tex]  .............(1)

where,

[tex]\large \Delta K.E[/tex] = Change in kinetic energy

[tex]q[/tex] = charge of an electron

[tex]\Delta V[/tex] = Potential difference

also

[tex]\large E=\frac{V}{d}[/tex]       .......(2)

E = electric field

d = distance traveled

Now from (1) and (2) we have,

[tex]\large \Delta K. E=qV=qEd[/tex]

substituting the values in the above equation, we get

[tex]\large \Delta K. E=(1.6\times 10^{-19}C)(2\times 10^6V/m)(0.400m)(\frac{1eV}{1.6\times 10^{-19}J})(\frac{1keV}{1000eV})[/tex]

[tex]\large \Delta K. E=800keV[/tex]

Thus, the energy gained by the electron is 800 keV if it is accelerated over a distance of 0.400 m.

(b) Using the equation (1), we have

[tex]\large d=\frac{\Delta K.E}{qE}[/tex]

[tex]\large d=\frac{(50\times 10^9eV)}{(1.6\times 10^{-19C})(2\times 10^6V/m)}(\frac{1.6\times 10^{-19}J}{1eV})[/tex]

or

[tex]\large d=2.4996\times 10^4m[/tex]

or

[tex]\large d=24.996\times 10^3m=24.996km[/tex]

Thus, to gain 50.0 GeV of energy the electron must be accelerated over a distance of 24.996 km.

Answer:

Part a)

[tex]k = 1632 J[/tex]

Part b)

[tex]KE = 47 J[/tex]

Part c)

[tex]m = 7.9 kg[/tex]

Part d)

[tex]v = 2.57 m/s[/tex]

Part e)

[tex]KE = 26.1 J[/tex]

Part f)

[tex]PE = 20.9 J[/tex]

Explanation:

Total Mechanical energy is given as

[tex]E = 47.0 J[/tex]

Its maximum displacement from mean position is given as

[tex]A = 0.240 m[/tex]

Part a)

Now from the formula of energy we know that

[tex]E = \frac{1}{2}kA^2[/tex]

[tex]47 = \frac{1}{2}k(0.240)^2[/tex]

[tex]k = 1632 J[/tex]

Part b)

At the mean position of SHM whole mechanical energy will convert into kinetic energy

so it is given as

[tex]KE = 47 J[/tex]

Part c)

As per the formula of kinetic energy we know that

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

[tex]47 = \frac{1}{2}m(3.45^2)[/tex]

[tex]m = 7.9 kg[/tex]

Part d)

As we know by the equation of the speed of SHM is given as

[tex]v = \sqrt{\frac{k}{m}(A^2 - x^2)}[/tex]

[tex]v = \sqrt{\frac{1632}{7.9}(0.24^2 - 0.16^2)}[/tex]

[tex]v = 2.57 m/s[/tex]

Part e)

As we know by the formula of kinetic energy

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

[tex]KE = \frac{1}{2}(7.9)(2.57^2)[/tex]

[tex]KE = 26.1 J[/tex]

Part f)

As per energy conservation we know

KE + PE = Total energy

[tex]26.1 + PE = 47[/tex]

[tex]PE = 20.9 J[/tex]

(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]

(b) The KE at equilibrium point =47j

(c) The mass =7.9kg

(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]

(e)The KE of block 0.160m will be 26.1

(f) The PE will be 20.9j

(a) for finding spring constant

KE=47 j

A=0.240m

By using the formula

[tex]E=\dfrac{1}{2} kx^2[/tex]

[tex]47=\dfrac{1}{2} k(0.240)^2[/tex]

[tex]k=1632\frac{N}{m^2}[/tex]

(b) At the mean position the whole mechanical energy will be equal to KE so

KE=47j

(c) The mass of the system

[tex]KE =\dfrac{1}{2} mv^2[/tex]

[tex]47=\dfrac{1}{2} m(3.45^2)[/tex]

[tex]m=7.9kg[/tex]

(d)Now the speed of the block

[tex]v=\sqrt{\dfrac{k}{m} (A^2-x^2)}[/tex]

[tex]v=\sqrt{\dfrac{1632}{7.9} (0.24^2-0.16^2)}[/tex]

[tex]v=2.57\frac{m}{s}[/tex]

(e) The KE of the block

[tex]KE=\dfrac{1}{2} mv^2=\dfrac{1}{2} 7.9(2.57)^2=26.1J[/tex]

(f) The PE of the system

[tex]Total Energy = KE+PE[/tex]

[tex]PE= 47-26.1 =20.9J[/tex]

Thus

(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]

(b) The KE at equilibrium point =47j

(c) The mass =7.9kg

(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]

(e)The KE of block 0.160m will be 26.1

(f) The PE will be 20.9j

To know more about the spring-mass system follow

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Explanation:

A satellite orbiting the Earth is a characteristic example of the uniform circular motion, where the direction of the velocity vector [tex]\vec{V}[/tex] is perpendicular to the radius [tex]r[/tex] of the trajectory.  Hence, the velocity is changing although the speed is constant.

On the other hand,  acceleration [tex]\vec{a}[/tex] is directed toward the center of the circumference (that is why it is called centripetal acceleration).  

Now, according to Newton's 2nd law, the force [tex]\vec{F}[/tex] is directly proportional and in the same direction as the acceleration:  

[tex]\vec{F}=m.\vec{a}[/tex]  

Therefore, the net force on the satellite resulting from its circular motion points towards the center of the circle (where the Earth is in the Earth-satellite system).

Answer:

1)  The acceleration of the car = 1.17 m/s²

2) The time required to reach 55 mi/h = 7.59 seconds.

Explanation:

1)  Initial speed of motorist, u = 35 mph = 15.56 m/s

   Final speed of motorist, v = 55 mph = 24.44 m/s

   Distance traveled, s = 500ft = 152.4 m

   We have v² = u² + 2as

                  24.44² = 15.56² + 2 x a x 152.4

                  a = 1.17 m/s²

  The acceleration of the car = 1.17 m/s²  

2) We have v = u + at

                   24.44 = 15.56 + 1.17 x t

                           t = 7.59 s

   The time required to reach 55 mi/h = 7.59 seconds.  

It takes 225 Earth days for Venus to orbit the Sun. Interestingly, Venus spins on its axis so slowly that its day (243 Earth days) is longer than its year. The Sun takes 117 Earth days to return to the same place in Venus' sky.

The time it takes for Venus to make one orbit around the sun, also known as its orbital period, is actually 225 Earth days. This is quite different compared to Earth which takes 365.25 days to orbit the Sun. Additionally, Venus spins on its axis very slowly, with its rotational period being 243 Earth days. As a result, a day on Venus - considering its rotation - is longer than its year! Also, this leads to an unusual phenomenon where the sun takes 117 Earth days to return to the same place in the Venusian sky. This suggests that Venus' rotation and orbit display unique characteristics compared to other planets in our solar system, likely due to factors from its formation.

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Venus completes one orbit around the Sun in approximately 225 Earth days. This is shorter than Earth's orbital period due to Venus's closer distance to the Sun.

Orbit of Venus Around the Sun-

The distance between the Earth and the Sun is approximately 1.50 x 1011 meters, while Venus is closer, at 1.08 x 1011 meters. This difference in distance influences the orbital period of each planet. Venus takes about 225 Earth days to complete one full orbit around the Sun.

Venus has a nearly circular orbit and, being closer to the Sun than Earth, receives almost twice as much light and heat. The elliptical orbits mean that different planets have different orbital periods, with those closer to the Sun having shorter years.

The light leaves the glass at about 42° relative to its normal

[tex]\texttt{ }[/tex]

Let's recall Snell's Law about Refraction as follows:

[tex]\boxed{n_1 \sin \theta_1 = n_2 \sin \theta_2}[/tex]

where:

n₁ = incident index

θ₁ = incident angle

n₂ = refracted index

θ₂ = refracted angle

[tex]\texttt{ }[/tex]

Given:

incident angle = θ₁ = 30°

index of refraction of air = n₃ = 1.0

index of refraction of glass = n₂ = 1.5

index of refraction of water = n₁ = 1.33

Asked:

refracted angle = θ₃ = ?

Solution:

Surface between Glass - Water:

[tex]n_1 \sin \theta_1 = n_2 \sin \theta_2[/tex]

[tex]\sin \theta_2 = \frac{n_1}{n_2} \sin \theta_1[/tex] → Equation A

[tex]\texttt{ }[/tex]

Surface between Air - Glass:

[tex]n_2 \sin \theta_2 = n_3 \sin \theta_3[/tex]

[tex]\sin \theta_3 = \frac{n_2}{n_3} \sin \theta_2[/tex]

[tex]\sin \theta_3 = \frac{n_2}{n_3} (\frac{n_1}{n_2} \sin \theta_1)[/tex] ← Equation A

[tex]\sin \theta_3 = \frac{n_1}{n_3} \sin \theta_1[/tex]

[tex]\sin \theta_3 = \frac{1.33}{1.0} \times \sin 30^o[/tex]

[tex]\sin \theta_3 = \frac{1.33}{1.0} \times \frac{1}{2}[/tex]

[tex]\sin \theta_3 \approx 0.665[/tex]

[tex]\boxed{\theta_3 \approx 42^o}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Light

The angle at which the light leaves the glass is approximately 19.5°.

When a light ray crosses from one medium to another, such as from water to glass, it undergoes refraction, which causes the ray to bend. The angle between the incident ray and the normal to the surface of the glass is called the angle of incidence. To find the angle of incidence, we can use Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the indices of refraction of the two media.

In this case, the index of refraction of water (n1) is 1.33 and the index of refraction of glass (n2) is 1.5.

The angle of incidence (θ1) is given as 30°. We can use the formula: n1*sin(θ1) = n2*sin(θ2) to solve for θ2, the angle at which the light leaves the glass.

Plugging in the values, we get: 1.33*sin(30°) = 1.5*sin(θ2).

Solving for θ2 gives us θ2 = sin^(-1)((1.33*sin(30°))/1.5).

Evaluating this expression, we find that θ2 is approximately 19.5°.

Answer:

d = 0.50 m

Explanation:

Let say the speed at the top and bottom of the window is

[tex]v_1 \: and \: v_2[/tex] respectively

now we have

[tex]d = \frac{v_1 + v_2}{2}t[/tex]

[tex]1.90 = \frac{v_1 + v_2}{2} (0.380)[/tex]

[tex]v_1 + v_2 = 10 [/tex]

also we know that

[tex]v_2 - v_1 = 9.8(0.380)[/tex]

[tex]v_2 - v_1 = 3.72[/tex]

now we have from above equations

[tex]v_2 = 6.86 m/s[/tex]

[tex]v_1 = 3.14 m/s[/tex]

now the distance from which it fall down is given as

[tex]v_f^2 - v_i^2 = 2ad[/tex]

[tex]3.14^2 - 0^2 = 2(9.8)d[/tex]

[tex]d = 0.50 m[/tex]

Answer:

a) There are 100 centimeters in 1 meter.

b) [tex]\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}[/tex]

Explanation:

a) We have the conversion

         1 m = 100 cm

   So there are 100 centimeters in 1 meter.

b) 1 inch = 2.54 cm

    [tex]1cm=\frac{1}{2.54}inch[/tex]

   [tex]\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}[/tex]

Answer:

The magnitude of the average angular acceleration of the disk is 2761.1 rad/s².

Explanation:

Given that,

Angular velocity of optical disk= 1045 rad/s

Angular velocity of particular disk = 646.1 rad/s

Time = 0.234 sec

We need to calculate the average angular acceleration

Using formula of  angular acceleration

[tex]\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}[/tex]

Where, [tex]\omega_{f}[/tex] = final angular velocity

[tex]\omega_{i}[/tex] = Initial angular velocity

t = time

Put the value in the equation

[tex]\alpha = \dfrac{0-646.1}{0.234}[/tex]

[tex]\alpha= -2761.1\ rad/s^2[/tex]

Negative sign shows the angular deceleration.

Hence, The magnitude of the average angular acceleration of the disk is 2761.1 rad/s².