Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but a much smaller diameter. Part A If you weigh 665 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 16.0 km ? Take the mass of the sun to be ms = 1.99×1030 kg , the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be g = 9.8 m/s2 . Express your weight wstar in newtons. View Available Hint(s)

Final answer:

The width of the slit is 1.0 μm.

Explanation:

When light passes through a single slit, it undergoes diffraction, which causes interference patterns. The width of the central peak in the diffraction pattern is related to the width of the slit and the wavelength of the light. In this case, the width of the central peak is given as 5.0 mm and the wavelength is given as 600 nm.

Using the formula for the width of the central peak, we can solve for the width of the slit:

Width of slit = (wavelength * distance to screen) / (number of the peak * distance to the peak)

Substituting the given values into the formula, we find that the width of the slit is 1.0 μm.

Answer:

-6sin(2t)

Explanation:

fayemioluwatomisin is correct until the end.

x(0)=0

[tex]x(t)=C_1cos(2t)+C_2sin(2t)\\0=C_1cos(2(0))+C_2sin(2(0))\\C_1=0[/tex]

x'(0)=-12

[tex]x'(t)=2C_2cos(2t)\\-12=2C_2cos(2(0))\\-12=2C_2\\C_2=-6[/tex]

[tex]x(t)=-6sin(2t)[/tex]

The mass-spring system's equation of motion can be determined using Hooke's Law to find the spring constant, and applying the conditions of simple harmonic motion (SHM).

The question relates to the physical concept of a spring-mass system and its simple harmonic motion (SHM). The force exerted on a spring and the resulting stretch can determine the spring constant (k), using Hooke's Law F = kx, where F is the force applied and x is the displacement from the equilibrium position. In this case, we're given that a force of 880 newtons stretches the spring 4 meters, which allows us to calculate the spring constant k = 880 N / 4 m = 220 N/m.

Now, considering that the mass (m) attached to the spring is 55 kilograms and it's released with an upward velocity (v) of 12 m/s from the equilibrium position, the initial conditions for the SHM are established. The equation of motion for a spring-mass system in SHM is x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency (ω = √(k/m)), and φ is the phase constant. The velocity is the first derivative of the displacement function, v(t) = -Aω sin(ωt + φ), and the acceleration is the second derivative of the function.

The maximum velocity and acceleration occur when the displacement is zero and maximum respectively. To fully determine the equation of motion for this particular situation, additional information about the phase constant or initial displacement may be needed, unless it is assumed that the spring is released from the equilibrium point with an upward velocity, in which case the initial displacement would be zero, and the phase constant φ could be determined.

Newton is used to measure force

The most used unit for the force measurement is "newton" which is the SI unit of the force

Answer:

newtons

Explanation:

Answer:

4000 N

Explanation:

First we calculate the acceleration

2as = vf^2 - vi^2

2xax0.1 = 0^2-40^2

0.2 x a = -1600

a = - 8000 m/s^2

F= ma

F= 0.5 x 8000

F 4000N

Explanation:

answer and explanation is in the picture

Answer:

Explanation:

loss of energy while passing through target by bullet

= 1/2 mu² - 1/2 mv² , m is mass of bullet , u is initial velocity and v is final velocity .

= 1/2 x m ( u² - v² )

= .5 x .023 x ( 1100² - 950² )

= 3536.25 J.

This loss is due to negative work done by  friction force

If friction force be F

Work done by friction force = F x .33

F x .33 = loss of kinetic energy

F x .33 = 3536.25

F = 10716 N

impulse of F

F X t , time period during which this force remains active

10716 x t = change in momentum of bullet

= .023 ( 1100 - 950 )

= 3.45

t = 3.45 / 10716

= 3.22 x 10⁻⁴ s.

Average force on the target = friction force created = 10716 N

Impulse by force on target = 10716 x 3.22 x 10⁻⁴

impulse on target = change in momentum of target

= mass of target x its velocity after impact

=  400 v

v = 10716 x 3.22 x 10⁻⁴ / 400

= 86.26 x 10⁻⁴ m /s

Answer:

3.53 N/C

Explanation:

Electric field = F / q where F is the force in N and q is charge on the electron

F = mass of an electron × a ( acceleration in m/s)

using equation of motion to solve for the acceleration

s ( distance ) = ut + 0.5 at² since the electron is starting from rest then ut = 0

2s / t² = a

F = me × ( 2s / t²)

E electric field = me × ( 2s / t²)  / q = me × 2s / ( t² × q)

me, mass of an electron = 9.11 × 10⁻³¹ kg

E = (9.11 × 10⁻³¹ kg × 2 × 0.038 m) / ( (3.5 × 10⁻⁷s)² × 1.6 × 10⁻¹⁹ C) = 0.0353 × 10² N/C = 3.53 N/C

Answer: The magnitude of the electric field is 3.53 N/C

Explanation: Please see the attachments below

Answer:

Explanation:

A )

When empty , H₀ length of barge is inside water .

volume of barge inside water = A x H₀

Weight of displaced water = AH₀ x ρ x g

Buoyant force = weight of displaced water = AH₀ ρg

B)

It should balance the weight of barge

Weight = buoyant force

Weight = AH₀ ρg

mass of barge = weight / g

weight / g = AH₀ ρ

= 550 x .55 x 1000

= 302500 kg

Answer:

Only 9% weaker

Explanation:

Because this is where most stuff that people do in space takes place. So, um, here we're at a radius of the earth plus 300 kilometers. You may already be seeing why this isn't going to have much effect if this were except the 6.68 times, 10 to the sixth meters. And so the value of Gout here. You know, Newton's gravitational constant times, the mass of the Earth divided by R squared for the location we're looking at. And so this works out to be 8.924 meters per second squared, which is certainly less than it is at the surface of the earth. However, this is only 9% less than acceleration for gravity at the surface. So the decrease in the gravity gravitational acceleration of nine percent not really going toe produces a sensation of weightlessness.

Answer:

The gravity at the altitude of 450 km above the earth's surface is (0.87 × the normal acceleration due to gravity); that is, 0.87g.

This shows that the gravity at this altitude isn't as weak as initially thought; it is still 87% as strong as the gravity on the surface of the earth.

The gravity is only 13% weaker at an altitude of 450 km above the earth's surface.

Explanation:

The force due to gravity that is usually talked about is basically the force of attraction between the planetary bodies such as the earth and objects on its surface.

The force is given through the Newton's gravitational law which explains that the force of attraction between two bodies is directly proportional to the product of the masses of both bodies and inversely proportional to the square of their distance apart .

Let the force of attraction between a body of mass, m on the surface of the earth and the earth itself, with mass M

F ∝ (Mm/r²)

F = (GMm/r²)

G = Gravitational constant (the constant of proportionality)

r = distance between the body and the earth, and this is equal.to the radius of the earth.

This force is what is now translated to force of gravity or weight of a body.

F = mg

where g = acceleration due to gravity = 9.8 m/s²

F = (GMm/r²) = (mg)

g = (GM/r²) = 9.8 m/s²

So, for a body at a distance that is 450 km above the earth's atmosphere, the distance between that body and the centre of the earth is (r + 450,000) m

Let that be equal to R.

R = (r + 450,000) m

Note that earth's radius is approximately 6400 km

r = 6400 km = 6,400,000 m

R = 6400 + 450 = 6850 km = 6,850,000 m

Normal acceleration due to gravity = 9.8 m/s²

9.8 = (GM/6,400,000²)

GM = 9.8 × 6,400,000²

Acceleration due to gravity at a point 450 km above the earth's surface

a = (GM/R²)

a = (GM/6,850,000²)

Note that GM = 9.8 × 6,400,000²

a = (9.8 × 6,400,000²) ÷ (6,850,000²)

a = 9.8 × 0.873 = 8.56 m/s²

a = 87% of g.

The gravity at this altitude above the earth's surface is (0.87 × the normal acceleration due to gravity)

Hope this Helps!!!

Answer:

B)

Explanation:

Hi there,

Notice in the prompt it is saying given mass, which means quantity of gas is constant. This eliminates option C, as this is Avogadro's Law (the more gas you have, the more space it takes up). It also states pressure remains constant, so it eliminates D, Boyle's Law (pressure is inversely related with volume), and A, The Combined Gas Law (combines Charles, Boyle's, and Gay-Lussac's Laws).

Charles' Law is telling us that as a gas is heated, the volume increases. This makes sense, as when gases are heated the average kinetic energy (energy of motion and movement) of the system is higher because of heat increase. When the system's kinetic energy is higher, it will have the tendency to occupy more space to balance out the increase in energy. Think of people who become more jittery in an already crowded space, they will push outward to accompany more space to balance.

So, answer is B.

If you liked this solution, hit Thanks or give a Rating!

thanks,

Answer:

6370 J

Explanation:

By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m

[tex]W = E_p = mgh [/tex]

where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference

[tex]W = 50*9.8*13 = 6370 J[/tex]

Answer:

68.46 m.

Explanation:

Given,

coefficient of friction,μ = 0.67

speed of truck, v = 30 m/s

distance travel by the truck to stop = ?

Now,

Calculation of acceleration

we know,

f = m a

and also

f = μ N = μ mg

Equating both the forces equation

m a = μ mg

a = μ g

a = 0.67 x 9.81

a = 6.57 m/s²

Now, using equation of kinematics

v² = u² + 2 a s

0 = 30² - 2 x 6.57 x s

s = 68.46 m

Hence, the minimum distance travel by the truck is equal to 68.46 m.

The truck would need to stop in a minimum distance of approximately 68.5 meters to prevent the box from sliding off, based on the given parameters and considering the law of inertia and the role of static friction.

To determine the minimum distance the truck can stop without the box sliding off, we need to consider the law of inertia and the role of static friction.

The force must meet or exceed the force exerted by the truck's deceleration to keep the box from sliding.

Given that the initial velocity of the truck is 30 m/s and the box has a mass of 500 kg, we can use the formula for static friction, which is F = μN, where F is the force of friction, μ is the coefficient of static friction, and N is the normal force.

In this case, N equates to the gravitational force on the box, so N = mg = 500 kg * 9.8 m/s² = 4900 N. Substituting these values in the formula for static friction, we get F = 0.67 * 4900 N = 3283 N.

The force of friction, in terms of motion, is also equivalent to mass times acceleration (F = ma), so we can set this equal to the static friction we calculated and solve for acceleration: 3283 N = 500 kg * a. Solving for a gives an acceleration of approximately 6.57 m/s².

Using motion equations, specifically v² = u² + 2as where v is the final velocity (0 m/s), u is the initial velocity (30 m/s), a is the acceleration (-6.57 m/s² because it's deceleration), and s is the distance, we can compute for the distance s which gives a value of approximately 68.5 meters.

Learn more about Physics of Static Friction here:

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The probable question is :-

A truck is moving along a straight horizontal road at 30 m/s. A box is placed on the bed of the truck, and the coefficient of static friction between the bed and the box is 0.67. What is the least distance in which the truck can come to a stop without causing the box to slide? Consider the deceleration of the truck due to braking.

Answer:

W=-1881.6J

Explanation:

we have that the change in the mass is

[tex]\frac{dm}{dt}=-c\\m(0)=60kg\\m(8)=60kg-6kg=54kg[/tex]

by solving the differential equation and applying the initial conditions we have

[tex]\int dm=-c\int dt\\m=-ct+d\\m(0)=-c(0) + d=60 \\m(8)=-8c+d=54[/tex]

by solving for c and d

d=60

c=0.75

The work needed is

W = m(t) gh

by integrating we have

[tex]dW_T= gh\int dm \\\\W_T=gh\int_0^8 -0.75dt\\\\W_T=(9.8\frac{m}{s^2})(32m)(-0.75(8))=-1881.6J[/tex]

hope this helps!!

Answer:

c.[tex]a_m[/tex]

Explanation:

We are given that

Acceleration due to gravity on the moon=[tex]a_m[/tex]

Acceleration due to gravity on the earth=[tex]a_e[/tex]

[tex]g_m=\frac{1}{6}g_e[/tex]

Net force due to am on an object on moon=[tex]F_{net}=ma_m[/tex]

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth=[tex]F=ma_e[/tex]

[tex]F=F_{net}[/tex](given)

[tex]ma_m=ma_e[/tex]

[tex]a_e=a_m[/tex]

Hence, option c is true.

Explanation:

We have,

Initial speed of an automobile, u = 71 km/h = 19.72 m/s

Diameter of the tie, d = 60 cm

Radius, r = 30 cm

(a) The angular speed of the tires about their axles is given by :

[tex]\omega=\dfrac{v}{r}\\\\\omega=\dfrac{19.72}{0.3}\\\\\omega=65.73\ rad/s[/tex]

(b) Final angular velocity of the wheel is equal to 0 as its stops. The angular acceleration of the wheel is given by :

[tex]\omega_f^2-\omega_i^2=2\alpha \theta[/tex]

[tex]\theta=40\ rev\\\\\theta=251.32\ rad[/tex]

[tex]0-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{\omega_i^2}{2\theta}\\\\\alpha =\dfrac{(65.73)^2}{2\times 251.32}\\\\\alpha =-8.59\ rad/s^2[/tex]

(c) Let the car move a distance d during the braking. So,

[tex]d=\theta r\\\\d=251.32\times 0.3\\\\d=75.39\ m[/tex]

Therefore, the above is the required explanation.

a. The angular speed is 65.73 rad/s.

b. The magnitude of the angular acceleration is -8.59 rad/s².

c. The distance should be 75.39m.

Since

Initial speed of an automobile, u = 71 km/h = 19.72 m/s

Diameter of the tie, d = 60 cm

Radius, r = 30 cm

a. Now the angular speed should be

= v/r

= 19.72/0.3

= 65.73 rad/s

b. Now the magnitude is

= 65.73^2/2*251.32

= -8.59 rad/s^2

c. The distance should be

= 251.32*0.3

= 75.39 m

Learn more about speed here: https://brainly.com/question/18212021

Answer: The approximate magnetic field is 1.38 × 10^-3T

Explanation: Please see the attachment below

The approximate magnetic field at a point within the solenoid is 4.4 × 10^-4 T.

To find the magnetic field strength at a point within a solenoid, we can use the equation B = µ0nI, where B is the magnetic field strength, µ0 is the permeability of free space, n is the number of turns per unit length, and I is the current. Given that the solenoid carries a current of 10.0 A and has 110.0 turns per meter of its length, we can substitute these values into the equation to calculate the magnetic field strength:

B = (4π × 10-7 T·m/A)(110.0 turns/m)(10.0 A) = 4.4 × 10-4 T

Therefore, the approximate magnetic field at a point within the solenoid is 4.4 × 10-4 T.

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Answer:

a)= 98kJ

b)=108kJ

c) = 10kJ

Explanation:

a. The work that is done by gravity on the elevator is:

Work = force * distance  

= mass * gravity * distance

= 1000 * 9.81 * 10  

= 98,000 J

= 98kJ

b)The net force equation in the cable

T - mg = ma

T = m(g+a)

T = 1000(9.8 + 10)

T = 10800N

The work done by the cable is

W = T × d

= 10800N × 10

= 108000

=108kJ

c) PE at 10m = 1000 * 9.81 * 10 = 98,100 J  

Work done by cable = PE +KE  

108,100 J = KE + 98,100 J  

KE = 10,000 J

= 10kJ

=

Answer:

A)Work done by gravity = -98 Kj

B) W_tension = 108 Kj

C) Final kinetic energy Kf = 10 Kj

Explanation:

We are given;

mass; m = 1000kg

Upward acceleration; a = 1 m/s²

Distance; d = 10m

I've attached a free body diagram to show what is happening with the elevator.

A) From the image i attached, the elevators weight acting down is mg.

While T is the tension of the cable pulling the elevator upwards.

Now, we know that,

Work done = Force x distance

Thus, W = F•d

We want to calculate the work done by gravity;

From the diagram I've drawn, gravity is acting downwards and in am opposite direction to the motion having an upward acceleration.

Thus, Force of gravity = - mg

So, Work done by gravity;

W_grav = - mgd

W_grav = -1000 x 9.8 x 10 = -98000 J = -98 Kj

B) Again, W = F.d

W_tension = T•d

Let's find T by summation of forces in the vertical y direction

Thus, Σfy = ma

So, T - mg = ma

Thus, T = ma + mg

T = m(a + g)

Plugging in values,

T = 1000(1 + 9.8)

T = 1000 x 10.8 = 10800 N

So, W_tension = T•d = 10800 N x 10m = 108000 J = 108 Kj

C) From work energy theorem,

Net work = change in kinetic energy

Thus, W_net = Kf - Ki

Where Kf is final kinetic energy and Ki is initial kinetic energy.

Now since the elevator started from rest, Ki = 0 because velocity at that point is zero.

Thus, W_net = Kf - 0

W_net = Kf

Now, W_net is the sum of work done due to gravity and work done due to another force.

Thus, in this case,

W_net = W_grav + W_tension

W_net = -98000 J + 108000 J

W_net = 10000J = 10Kj

So,since W_net = Kf

Thus, Final kinetic energy Kf = 1000J

Answer:

The volume is 2.22x10⁵m³

Explanation:

the solution is in the attached Word file

To calculate the volume of space containing 1.0 J of electromagnetic energy near the Earth, use the formula Volume = Energy / Intensity after converting the intensity from kW/m² to W/m².

The volume of space that contains 1.0 J of electromagnetic energy can be found using the formula:

Volume = Energy / Intensity

Given that the intensity is 1.35 kW/m², you need to convert it to W/m² (1 kW = 1000 W).

After converting the intensity, you can then calculate the volume using the given energy of 1.0 J.

The volume of space near Earth containing 1.0 J of electromagnetic energy can be found by first calculating the energy density using the given intensity and speed of light, and then dividing the amount of energy by energy density, resulting in a volume of 2.22  imes 10^5 m^3.

Finding the Volume of Space Containing 1.0 J of Electromagnetic Energy

To find the volume of space near Earth that contains 1.0 J of electromagnetic energy given the intensity (I) from the sun as 1.35 kW/m2, we can use the formula for energy density (u), which is given by the equation u = I/c, where c is the speed of light. Now, to find the volume (V) that contains energy (E), we use the equation V = E/u.

First, let's convert the intensity to watts per square meter: 1.35 kW/m2 = 1350 W/m2. Next, calculate the energy density (u):
u = 1350 W/m2 / (3.0 * 108 m/s) = 4.5 * 10-6 J/m3.

With the energy density known, we can calculate the volume (V) that contains 1.0 J of energy:
V = 1.0 J / (4.5 * 10-6 J/m3) = 2.22 * 105 m3.

Therefore, a volume of 2.22 * 105 m3 near Earth contains 1.0 J of electromagnetic energy.

Answer:

0.358 kg

Explanation:

From simple harmonic motion,

T = 2π√(m/k)................ Equation 1

Where T = period of the spring, k = spring constant of the spring, m = mass suspended, π = pie

make m the subject of the equation

m = kT²/4π².................. Equation 2

Given: k = 15 N/m, T = 0.97 s, π = 3.14

Substitute into equation 2

m = 15(0.97²)/(4×3.14²)

m = 14.1135/39.4384

m = 0.358 kg.

Hence mass suspended = 0.358 kg

0.356kg

The period, T, of a mass on a spring of spring constant, k, in simple harmonic motion can be calculated as follows;

T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]         --------------------(i)

Where;

m = mass

From the question;

T = 0.97s

k = 15N/m

Taking π = 3.142 and substituting the values of T and k into equation (i) as follows;

0.97 = 2 x 3.142 x [tex]\sqrt{\frac{m}{15} }[/tex]

0.97 = 6.284 x [tex]\sqrt{\frac{m}{15} }[/tex]

[tex]\frac{0.97}{6.284}[/tex] = [tex]\sqrt{\frac{m}{15} }[/tex]

0.154 = [tex]\sqrt{\frac{m}{15} }[/tex]

Square both sides

0.154² = [tex]\frac{m}{15}[/tex]

0.0237 = [tex]\frac{m}{15}[/tex]

m = 0.356

Therefore, the mass that could be suspended on the spring to have an oscillation period of 0.97s when in SHM is 0.356kg

Answer:

Velocity between points A and B will be 0.2344 m/s

Complete Question:

A tennis racket is thrown vertically into the air. The center of gravity G has a velocity of 5 m/s upwards. Angular velocity about the x - direction of 1 rad/s and angular velocity about the y - direction of 20 rad/s. Knowing that the horizontal distance between points A and G as well as G and B is 25 cm and knowing that the maximum width of the of the racket head is 30 cm, determine the velocity of Points A and B.

Answer:

a) [tex]v_{A} = 10 m/s[/tex]

b)

[tex]v_{B} = 3i + 0.15 j m/s\\v_{B} = \sqrt{3^{2} + 0.15^{2} }\\v_{B} = 3.004 m/s[/tex]

Explanation:

a) Velocity of point A

The velocity of point A is a combination of the translational and the rotational velocity of the racket.

[tex]v_{A} = v_{G} + v_{r}[/tex]

[tex]v_{G} = 5 m/s[/tex]

[tex]v_{r} = wr[/tex]

r = 25 cm = 0.25 m

w = 20 rad/s

[tex]v_{r} = 20 * 0.25\\v_{r} = 5 m/s[/tex]

[tex]v_{A} = 5 + 5\\v_{A} = 10 m/s[/tex]

b) Velocity of point B

At point B, the linear velocity is in the +ve z-direction while the rotational velocity is in the -ve z-direction:

[tex]v_{G} = 5 m/s[/tex]

[tex]v_{r} = -r w\\v_{r} = - 0.25 * 20\\v_{r} = - 5 m/s[/tex]

[tex]v_{Bz} = v_{G} + v_{r} \\v_{Bz} = 5 -5\\v_{Bz} = 0 m/s[/tex]

In the y - direction, r = 30/2 = 15 cm = 0.15 m

r = 0.15 m

[tex]w_{x} = 1 rad/s[/tex]

[tex]v_{By} = rw_{x} \\v_{By} = 0.15 * 1\\v_{By} = 0.15 rad/s[/tex]

In the x - direction, r = 0.15 m, [tex]w_{y} = 20 rad/s[/tex]

[tex]v_{Bx} = rw_{y} \\v_{Bx} = 0.15 * 20\\v_{Bx} = 3.0 rad/s[/tex]

[tex]v_{B} = 3 i +0.15 j\\[/tex] m/s

Answer:

The magnitude of the force exerted by the ball on the ground during the 0.5 s of contact = 90.16 N

Explanation:

Given,

Mass of mud ball = m = 2.40 kg

Height the ball is released from = y = 18 m

Total contact time of ball and the ground = t = 0.5 s

The Newton's second law of motion explains that the magnitude of the change of momentum is equal to the magnitude of a body's impulse.

Change in momentum = Magnitude of Impulse

Change in momentum = (final momentum) - (initial momentum)

Since the ball is dropped from rest, initial momentum = 0 kgm/s

But to calculate its final momentum, we need the ball's final velocity before hitting the ground.

Using the equations of motion,

u = initial velocity of the ball = 0 m/s (ball was dropped from rest)

v = final velocity of the ball = ?

g = acceleration due to gravity = 9.8 m/s²

y = vertical distance covered by the ball = 18 m

v² = u² + 2gy

v² = 0² + (2)(9.8)(18)

v² = 352.8

v = 18.78 m/s

Final momentum of the ball = (m)(v)

= (2.4) × (18.78) = 45.08 kgm/s

Change in momentum = 45.08 - 0 = 45.08 kgm/s

Impulse = Ft

Change in momentum = Magnitude of Impulse

45.08 = F × (0.5)

F = (45.08/0.5) = 90.16 N

Hope this Helps!!!

Answer:

90.14 N

Explanation:

according to the impulse momentum theorem,

Impulse = change in momentum

Where impulse = force × time and change in momentum = m ( v - u).

The object was initially at rest, hence it initial velocity is zero.

To get the final velocity, we use the formula below

v² = u² + 2gh

Where h = height of the cliff = 18.0m

v² = 2 × 9.8 × 18

v² = 352.8

v = √333.2

v = 18.78 m/s

At t = 0.50s and v = 18.78 m/s, we can get the average force of impact

F×0.50 = 2.4 (18.78 - 0)

F × 0.50 = 2.4 (18.78)

F × 0.50 = 45.072

F = 45.072 /0.50

= 90.14 N

The "it must be five times larger" current change if the energy stored in the inductor is to remain the same.

Explanation:

A current produced by a modifying magnetic field in a conductor is proportional to the magnetic field change rate named INDUCTANCE (L). The expression for the Energy Stored, that equation is given by:

[tex]U= \frac{1}{2} LI^2[/tex]

Here L is the inductance and I is the current.

Here, energy stored (U) is proportional to the number of turns (N) and the current (I).

[tex]L = \frac{\mu_0 N^2 *A}{l}[/tex]

mu not - permeability of core material

A -area of cross section

l - length

N - no. of turns in solenoid inductor

Now,given that the proportion always remains same:

[tex]\frac{N_2}{N_1} = \frac{I_1}{I_2}[/tex]

In this way the expression

[tex]\frac{1}{5} = \frac{I_1}{I_2}[/tex]

[tex]I_2 = I_1 \times 5[/tex]

Thus, it suggest that "it must be five times larger" current change if the energy stored in the inductor is to remain the same.

Answer:

AM radio, FM radio, microwaves, sodium light.

Explanation:

Electromagnetic radiation are waves from electromagnetic field which spread through the space or any other material medium and carries radiating energy. Examples includes X rays, radio waves, Gamma rays, etc. Exposure to high level of electromagnetic radiation could be harmful to human body, on the other hand, science as not been able to prove that exposing humans to low level electromagnetic radiation is harmful to our health.

Answer:2.7m/s^2

Explanation:

mass=2.3kg

Force=6.2Newton

Acceleration=force ➗ mass

Acceleration=6.2 ➗ 2.3

Acceleration=2.7m/s^2

The resulting acceleration of the object is  2.70 m/s^2.

Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).

Given that:

Mass of the object: m = 2.3 kg.

Force applied on it: F = 6.2 Newtons.

Now from Newton's 2nd law of motion: it can be stated that:

force = mass × acceleration

acceleration = force/mass

= 6.2 Newton/2.3 kg

= 2.70 m/s^2.

Hence,  the acceleration of the object is 2.70 m/s^2.

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Answer:

p = 4/3   f

Explanation:

The constructor station describes the position of the image object in relation to the focal length

         1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distances to the object and the image respectively

 the objent distance  is

         1 / p = 1 / f - 1 / q

they indicate that the image is located at q = 4f this distance is positive because the image is after the lens

        1 / p = 1 / f - 1 / 4f

        1 / p = 3 / 4f

         p = 4f / 3

the object is positive because it is to the left of the lens, according to the optical sign convention

The sign convention is that the distance to the objects is positive if it is on the left side of the lens and the image is positive if it is on the right side of the lens.

Answer:

Michelson

Explanation:

Answer:

Answer

Explanation:

The followings are the reasons for the reduction in  the death toll from Tornadoes occurence in the 2000s compared to 1950s;

(a) Improvement in technology and rapid natural occurence prediction. Through the use of advanced radar systems the early detection of natural occurence e.g tornadoes, providing  warning signal to the residents and tracking their possible pathways is made possible.

Fast spread of information through television channels, Radio, Social media etc., help in fast spreading of information, thereby sending early signals to people thus making them prepare for its occurence.

Improvement of protective measures, provision of housing infrastructure for the displaced. With increase in technology, the protective measures are also improving.

Quick Government and medical interventions for casualties.  Conducting Tornado drills help people to practice to take cover in a specified location during tornadoes.

TV meteorologist are able to inform viewers of the approaching tornadoes and hurricanes  which can be tracked from the time of its origin i.e., from a depression to a storm.  Advanced radar data help in delineating the possible pathways of the hurricane.

Tornadoes on the other hand form in a thunderstorm or during a frontal activity. In thunderstorms there are vortices formed which are pre-cursors of tornadoes.. A vortice when touches the ground forms a tornado. Once a tornado is on the ground it becomes easy for radars to measure windpeeds. Hence a TV meteorologist can report the intensity of the Tornado.

Explanation:

Mass, m = 0.464 kg

Compression in the spring, x = 0.646 m

(a) The net force acting on the spring is given by :

[tex]kx=mg[/tex]

k is spring constant

[tex]k=\dfrac{mg}{x}\\\\k=\dfrac{0.464\times 9.8}{0.646 }\\\\k=7.03\ N/m[/tex]

(b) The angular frequency of the spring mass system is given by :

[tex]\omega=\sqrt{\dfrac{k}{m}} \\\\\omega=\sqrt{\dfrac{7.03}{0.464 }} \\\\\omega=3.89\ rad/s[/tex]

The period of oscillation is :

[tex]T=\dfrac{2\pi}{\omega}\\\\T=\dfrac{2\pi}{3.89}\\\\T=1.61\ m/s[/tex]

(c) As the spring oscillates, its will reach to a height of 2(0.338) = 0.676 m

The spring constant is calculated to be 7.03 N/m using Hooke's Law. The period of oscillation for the block-spring system is found to be 1.74s utilizing the formula for simple harmonic motion. The height reached above the point of release by the block is 0.338m, based on the conservation of energy principle.

To find the spring constant, which can be denoted by k, we can use Hooke's Law (F = -kx). The force, F, in this case is the weight of the block (mass x gravity), mg = 0.464 kg x 9.8 m/s² = 4.5472 N. This force causes the spring to stretch 0.646 m. Therefore, k is calculated as F/x = 4.5472 N / 0.646 m = 7.03 N/m.

The period of oscillation, T, for a block-spring system executing simple harmonic motion is given by T = 2π√(m/k), where m is the mass and k is the spring constant. Substituting the given values, T = 2π√(0.464 kg / 7.03 N/m) = 1.74s.

For the height above the point of release that the block will reach, again we come back to conservation of energy. At the point of release, the block possesses only elastic potential energy, 1/2kA², where A is the amplitude. This energy will be entirely converted into gravitational potential energy, mgh, at the block's highest point. Therefore, h = (1/2kA²)/mg. So, h = (1/2 * 7.03 N/m * (0.338 m) ²) / (0.464 kg * 9.8 m/s²) = 0.338 m, above the point of release.

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The mass of the solid cylinder is calculated by setting up a triple integral over the volume of the cylinder using cylindrical coordinates with the given density function. The integral is calculated from the inside out, starting with the integral over z (height), followed by r (radius), and finally θ (angular measure).

The mass of a solid object in three-dimensions using a cylindrical coordinate system (r,θ,z) can be expressed as an integral over the volume of the object multiplied by the density function. In your case, given that the solid cylinder D equals {(r, θ, z) : 0 ≤ r ≤ 2, 0 ≤ z ≤ 10} with density function ρ(r, θ, z) = 1+ z/2, we calculate mass M as follows:

M=∫02π∫02∫010(1+ z/2)rdzdrdθ

Here, we start by calculating the innermost integral (∫010(1+ z/2)dz) first, then move to the middle integral (∫02dr) and finally the outermost integral (∫02πdθ). Each integral works on the outcome of the next inner integral until the entire mass of the cylinder is calculated.

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The mass of the solid cylinder is [tex]\(140\pi\).[/tex]

The mass of the solid cylinder is given by the triple integral

[tex]\[ \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{10} \left(1 + \frac{z}{2}\right) r \, dz \, dr \, d\theta. \][/tex]

To find the mass of the solid cylinder, we need to evaluate the triple integral of the density function over the volume of the cylinder in cylindrical coordinates. The density function is given by [tex]\(\rho(r, \theta, z) = 1 + \frac{z}{2}\).[/tex]

The solid cylinder is defined by the set [tex]\(D = \{(r, \theta, z) : 0 \leq r \leq 2, 0 \leq z \leq 10\}\)[/tex]. The limits of integration for[tex]\(r\), \(z\), and \(\theta\)[/tex]  are determined by the geometry of the cylinder:

- For[tex]\(r\)[/tex], the radial distance from the z-axis, the limits are from 0 to 2, since the radius of the cylinder is 2 units.

- For[tex]\(z\),[/tex] the height of the cylinder, the limits are from 0 to 10, since the height is 10 units.

- For [tex]\(\theta\),[/tex] the angle around the z-axis, the limits are from 0 to [tex]\(2\pi\),[/tex] since a full rotation around the z-axis is [tex]\(2\pi\)[/tex] radians.

The differential volume element in cylindrical coordinates is [tex]\(r \, dz \, dr \, d\theta\),[/tex] where the [tex]\(r\)[/tex]  factor accounts for the Jacobian of the transformation from Cartesian to cylindrical coordinates.

Now, we can set up the triple integral as follows:

[tex]\[ \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{10} \left(1 + \frac{z}{2}\right) r \, dz \, dr \, d\theta. \][/tex]

To evaluate this integral, we would first integrate with respect to [tex]\(z\)[/tex] from 0 to 10, then with respect to [tex]\(r\)[/tex] from 0 to 2, and finally with respect to [tex]\(\theta\)[/tex]  from 0 to[tex]\(2\pi\)[/tex] . The integral can be computed as follows:

1. Integrate with respect to [tex]\(z\)[/tex] first:

[tex]\[ \int_{0}^{10} \left(1 + \frac{z}{2}\right) dz = \left[z + \frac{z^2}{4}\right]_{0}^{10} = 10 + \frac{100}{4} = 10 + 25 = 35. \][/tex]

2. Next, integrate with respect to[tex]\(r\):[/tex]

[tex]\[ \int_{0}^{2} 35r \, dr = \left[\frac{35r^2}{2}\right]_{0}^{2} = \frac{35 \cdot 4}{2} = 70. \][/tex]

3. Finally, integrate with respect to [tex]\(\theta\):[/tex]

[tex]\[ \int_{0}^{2\pi} 70 \, d\theta = \left[70\theta\right]_{0}^{2\pi} = 70 \cdot 2\pi = 140\pi. \][/tex]

Answer: 140cm

Explanation:

Radius of curvature r = 2f

Where f is the focal point

f = 70cm, therefore

r =2 x 70 = 140cm

Answer:

The radius of curvature is 9.35 cm.

Explanation:

Given:

u = -5 cm

v = -70 cm

The radius of curvature of the mirror can be obtained from the expression:

[tex]\frac{1}{v} +\frac{1}{u} =\frac{2}{R} \\-\frac{1}{v} -\frac{1}{u} =\frac{2}{R} \\-\frac{1}{70} -\frac{1}{5} =\frac{2}{R} \\-0.014-0.2=\frac{2}{R} \\R=-9.35cm[/tex]