At the equator, near the surface of Earth, the magnetic field is approximately 80.0 μT northward and the electric field is about 150 N/C downward in fair weather. Find the gravitational, electric, and magnetic forces on an electron with instantaneous velocity of 6.00 106 m/s directed to the east in this environment.

Answer:

Speed of the cars after the collision is 3.34 m/s.

Explanation:

It is given that,

Mass of one car, m₁ = 1500 kg

Velocity of this car, v₁ = + 30 m/s ( in east )

Mass of other car, m₂ = 3000 kg

Velocity of other car, v₂ = - 20 m/s (in south)

The two cars stick together after the collision. It is a case of inelastic collision. Let v is the speed of cars after collision. It can be calculated using the conservation of linear momentum as :

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]

[tex]v=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}[/tex]

[tex]v=\dfrac{1500\ kg\times 30\ m/s+3000\ kg\times (-20\ m/s)}{1500\ kg+3000\ kg}[/tex]

v = -3.34 m/s

So, the speed of the cars after the collision is 3.34 m/s. Hence, this is the required solution.

Answer:

1 Hz

Explanation:

For rope fixed on both ends the length corresponds to λ/2  (λ is wavelength)\

 Thus L =  λ/2

=> λ = 16 m

We know that frequency and wavelength are related as

 f x λ = v   where f is frequency and v is speed of the wave

thus f = v/λ

        f = 16/16 =1 Hz

The fundamental frequency of an 8 m rope tied at both ends with waves traveling at a speed of 16 m/s is 1 Hz, calculated using the formula for the fundamental frequency of a standing wave.

The fundamental frequency of a string fixed at both ends (like an 8 m rope with waves traveling at 16 m/s) can be found using the formula for the fundamental frequency of a standing wave, which is given by f = v / (2L), where f is the fundamental frequency, v is the speed of the wave, and L is the length of the string. In this case, L = 8 m and v = 16 m/s. Plugging these values into the formula gives us:

f = 16 m/s / (2 × 8 m) = 1 Hz.

Therefore, the fundamental frequency of the 8 m rope tied at both ends with a wave speed of 16 m/s is 1 Hz.

Answer:

1454.54 kJ/h or 0.404 kW

Explanation:

Given:

Removal of heat by the refrigerator = 800 kJ/h

Coefficient of performance, COP of the refrigerator = 2.2

given that the  refrigerator runs one-fourth of the time only

therefore, refrigerator removes the heat, Q = [tex]4\times 800 kJ/h[/tex] = 3200 kJ/h

Now, the Power drawn by the refrigerator while running = [tex]\frac{Q}{COP}[/tex]

=  [tex]\frac{3200 kJ/h}{2.2}[/tex]

= 1454.54 kJ/h

Hence, the power drawn by the refrigerator while running is = 1454.54 kJ/h

or 0.404 kW      (as 1 kW = 3600 kJ/h)

To determine the power the refrigerator draws when running, use the formula for the coefficient of performance (COP) and substitute the given values

To determine the power the household refrigerator draws when running, we need to use the formula for the coefficient of performance (COP) of a refrigerator. The COP is given by the ratio of the heat removed from the cold reservoir to the work done on the engine's working substance. In this case, the COP is given as 2.2.

The power drawn by the refrigerator can be calculated using the formula:

Power = COP x Heat removed from the cold reservoir / Time

Given that the refrigerator runs one-fourth of the time and removes heat from the food compartment at an average rate of 800 kJ/h, we can substitute these values into the formula to find the power drawn by the refrigerator.

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Answer:

Speed of the car after the collision is 1.2 m/s

Explanation:

It is given that,

Mass of first car, m₁ = 2 kg

Velocity of first car, v₁ = 6 m/s

Mass of second car, m₂ = 3 kg

Velocity of second car, v₂ = -2 m/s (it is travelling in opposite direction)

The two cars lock together and move along the road. This shows an inelastic collision. Let their common velocity is V. On applying the conservation of momentum as :

[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]

[tex]V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}[/tex]

[tex]V=\dfrac{2\ kg\times 6\ m/s+3\ kg\times (-2\ m/s)}{5\ kg}[/tex]

V = 1.2 m/s

So, after collision the speed of the cars is 1.2 m/s. Hence, this is the required solution.

Answer:

W = 907963.50 J = 907.96 J

Explanation:

Note: Refer to the figure attached

Now, from the figure we have similar triangles ΔAOB and ΔCOD

we have

[tex]\frac{5}{4}=\frac{x}{r}[/tex]

or

[tex]r=\frac{4x}{5}[/tex]

Now, the work done to empty the tank can be given as:

[tex]W = \int\limits^4_0 {(5-x)\rho\times g A} \, dx[/tex]

or

[tex]W = \int\limits^4_0 {(5-x)1080\times 9.8 (\pi r^2)} \, dx[/tex]

or

[tex]W = \int\limits^4_0 {(5-x)\times10584\times (\pi (\frac{4x}{5})^2)} \, dx[/tex]

or

[tex]W = 6773.76\pi\int\limits^4_0 {(5-x)x^2)} \, dx[/tex]

or

[tex]W = 6773.76\pi[\frac{5}{3}x^3-\frac{1}{4}x^4]^4_0 [/tex]

or

[tex]W = 6773.76\pi[\frac{128}{3}] [/tex]

or

W = 907963.50 J = 907.96 J

The work required to empty the tank by pumping the hot chocolate over the top of the tank is 907.96 J.

Work done is the force applied on a body to move it over a distance. The work required to lift a body through a height h is given as,

[tex]W=Fh[/tex]

Here, (F) is the magnitude of force and (f) is the height.

It can be rewritten as,

[tex]W=mgh[/tex]

Here, (m) is the mass of the body.

The tank is in the shape of an inverted right circular cone with height 5 meters and radius 4 meters. As, It is filled with 4 meters of hot chocolate.

Here the ratio of height to radius should be equal to the height at another point (say x), to the radius of that point. Thus,

[tex]\dfrac{5}{4}=\dfrac{x}{r}\\r=\dfrac{4x}{5}[/tex]

For the distance x the work done to empty the cone can be given as,

[tex]W=\int^4_0 (5-x)\rho g \pi r^2 dx\\W=\int^4_0 (5-x)1080\times9.81 \pi (\dfrac{4x}{5})^2 dx\\W=22280.4\int^4_0(5-x)x^2 dx\\W=22280.4[\dfrac{128}{3}]\\W=907.96\rm J[/tex]

Thus, the work required to empty the tank by pumping the hot chocolate over the top of the tank is 907.96 J.

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Answer:

18.73 m/s^2

Explanation:

f = 2662 rpm = 2662 / 60 rps

r = 6.725 cm = 0.06725 m

Acceleration, a = r w

a = r x 2 x pi x F

a = 0.06725 × 2 × 3.14 × 2662 / 60

a = 18.73 m/s^2

Answer:

Part a)

[tex]\Delta U = 4\times 10^5eV[/tex]

Part b)

[tex]\Delta U = -4\times 10^5eV[/tex]

Explanation:

Part a)

Change in potential energy of a charge is given as

[tex]\Delta U = q\Delta V[/tex]

here we know that

[tex]q = e[/tex] for proton

also we have

[tex]\Delta V = 400 kV[/tex]

now we have

[tex]\Delta U = e(400 kV)[/tex]

[tex]\Delta U = 4\times 10^5eV[/tex]

Part b)

Change in potential energy of a charge is given as

[tex]\Delta U = q\Delta V[/tex]

here we know that

[tex]q = -e[/tex] for proton

also we have

[tex]\Delta V = 400 kV[/tex]

now we have

[tex]\Delta U = -e(400 kV)[/tex]

[tex]\Delta U = -4\times 10^5eV[/tex]

Answer:

The ball's maxium height after 1.49 seconds will be 103.52 feet.

Explanation:

Vi= 48 ft/sec

hi= 32 ft

hf= ?

time of maxium height :  

t= Vi/g    t= 48 ft/sec  /  32.17 ft/sec²

t= 1.49 sec

maxium height:

hf= hi + Vi * t

hf= 32ft + 48 ft/sec * 1.49 sec

hf= 103.52 ft

Answer:

option (b), (c), (g)

Explanation:

When the battery is disconnected, the charge on the plates of a capacitor remains same.

As the capacitance of the capacitor is directly proportional to the dielectric constant.

C = k C0

Now the charge remains same, So

Q = Q0

Q = k C0 x V0 / k

So, potential between the plates is V0 / k.

Energy, E = 1/2 x C X V^2 = 1/2 x k C0 x V0^2 / k^2 = E0 / k

So, energy becomes E0 / k.

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

[tex]r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}[/tex]

here we have

[tex]m_1 = 10.2 kg[/tex]

[tex]m_2 = 4.6 kg[/tex]

[tex]r_1 = (0, 0)[/tex]

[tex]r_2 = (8.1cm, 0)[/tex]

now we have

[tex]r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}[/tex]

[tex]r_{cm} = {(37.26, 0)}{14.8}[/tex]

[tex]r_{cm} = (2.52 cm, 0)[/tex]

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

[tex]r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}[/tex]

here we have

[tex]m_1 = 10.2 kg[/tex]

[tex]m_2 = 4.6 kg[/tex]

[tex]r_1 = (0, 0)[/tex]

[tex]r_2 = (8.1cm, 0)[/tex]

now we have

[tex]r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}[/tex]

[tex]r_g_{cm} = {(37.26, 0)}{14.8}[/tex]

[tex]r_g_{cm} = (2.52 cm, 0)[/tex]

So center of mass is same as center of gravity because value of gravity is constant here

The center of mass and center of gravity for the two given masses are both located at x = 2.49 cm from the origin. In typical situations where gravitational forces are uniform, the center of gravity will coincide with the center of mass.

The location of the center of mass (xcom) and the center of gravity (xcog) for two bodies or particles can be calculated using the following formula: xcom = m1x1 + m2x2 / (m1 + m2) where m1, m2 are the masses and x1, x2 are their respective positions. In this case, m1 = 10.2 kg, x1 = 0 cm (since it's at the origin), m2 = 4.6 kg, and x2 = 8.1 cm.

Using these values, the formula becomes xcom = (10.2 kg * 0 cm + 4.6 kg * 8.1 cm) / (10.2 kg + 4.6 kg) = 36.86 cm / 14.8 kg = 2.49 cm. So, the center of mass is at x = 2.49 cm from the origin.

The concept of center of gravity is a specific application of the center of mass in the presence of a gravitational field. In everyday circumstances, where gravitational forces are uniform, the center of gravity coincides with the center of mass. Therefore, for this case, the center of gravity (xcog) will also be at x = 2.49 cm from the origin.

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Answer:

6538.8 Angstrom

Explanation:

work function, w = 1.9 eV = 1.9 x 1.6 x 10^-19 J = 3.04 x 10^-19 J

Let the longest wavelength is λ.

W = h c / λ

λ = h c / W

λ = (6.626 x 10^-34 x 3 x 10^8) / (3.04 x 10^-19)

λ = 6.5388 x 10^-7 m = 6538.8 Angstrom

Thus, the longest wavelength is 6538.8 Angstrom.

Answer:

10.83 m/s

Explanation:

Case 1:

m = mass attached to spring = 2 kg

x = compression of the spring = 40 cm = 0.40 m

k = spring constant

v₀ = maximum speed = 10 m/s

Using conservation of energy

maximum spring potential energy = maximum kinetic energy

(0.5) k x² = (0.5) m v₀²

k (0.40)² = (2) (10)²

k = 1250 N/m

Case 2 :

m = mass attached to spring = 2 kg

A = amplitude = 50 cm = 0.50 m

x = compression of the spring at half amplitude = A/2 = 50/2 = 25 cm = 0.25 m

k = spring constant = 1250 N/m

v = speed = ?

Using conservation of energy

maximum spring potential energy = spring potential energy + kinetic energy

(0.5) k A² = (0.5) k x² + (0.5) m v²

(1250) (0.50)² = (1250) (0.25)² + (2) v²

v = 10.83 m/s

Answer:

Total solar energy reaches the entire United States per hour is[tex]2.0864\times 10^{16} [/tex] kilo Joules.

Explanation:

Amount of energy reaching 1 square meter of area = 2278 kJ/hour

Total area of the United States = [tex]9,158,960 km^2=9,158,960\times 10^6 m^2[/tex]

[tex](1 km^2=10^6 m^2)[/tex]

Amount of energy reaching [tex]9,158,960\times 10^6 m^2[/tex] of area:

[tex]2278 kJ/hour\times 9,158,960\times 10^6 kJ/hour[/tex]

[tex]=2.0864\times 10^{16} kJ/Hour[/tex]

Total solar energy reaches the entire United States per hour is[tex]2.0864\times 10^{16} [/tex] kilo Joules.

Final answer:

The total solar energy that reaches the entire United States per hour is 20,866,319,680 MJ.

Explanation:

To calculate the total solar energy reaching the entire United States per hour, given that 2278 kJ of energy reaches a square meter in one hour, we would first convert the entire area of the United States into square meters. Since the area is given as 9,158,960 km², we convert this to square meters by multiplying by (1000 m/km)²:

9,158,960 km² * (1000 m/km)² = 9,158,960,000,000 m²

Next, we multiply the area in square meters by the energy received per square meter:

9,158,960,000,000 m² * 2278 kJ/m² = 20,866,319,680,000 kJ

To generate an accurate answer, we can express this in megajoules (MJ) by dividing by 1,000 (since 1 MJ = 1,000 kJ):

20,866,319,680,000 kJ / 1,000 = 20,866,319,680 MJ

Therefore, the total solar energy that reaches the entire United States per hour is 20,866,319,680 MJ.

Final answer:

The tangential speed of the tack is 0.724 m/s, calculated using Vt = rω formula with the given values.

Explanation:

The tangential speed of the tack can be calculated using the formula:

Vt = rω

where Vt is the tangential speed, r is the distance from the rotation axis (0.357 m), and ω is the angular speed (2.03 times/s).

Substitute the values:

Vt = 0.357 m * 2.03 times/s = 0.724 m/s.

Answer:

explained below

Explanation:

solar system consist of sun, planets, asteroids, stars etc.

Milky Way is  a galaxy.

Combination of  many solar systems then forms galaxy.

Our solar system is a part of milky way galaxy.

If you consider marbles as planets and balls as stars.

so, we can say that in milky way is consist of 100-400 billions balls and 100 million marbles.

The force of the ramp on the block is 2.45 N.

To find the force of the ramp on the block, we need to consider the forces acting on the block. The only force acting on the block along the ramp is the component of the weight of the block that is parallel to the ramp. This force can be calculated using the equation:

F = mg sin(theta)

where F is the force, m is the mass of the block, g is the acceleration due to gravity, and theta is the angle of the ramp. Plugging in the values, we get:

F = (2.0 kg)(9.8 m/s^2) sin(30 degrees) = 4.9 N sin(30 degrees) = 2.45 N

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Answer:

y and length is directly relation

Explanation:

Given data

A single-slit diffraction pattern is formed on a distant scree

angles involved = small

to find out

what factor will the width of the central bright spot on the screen change

solution

we know that  for single slit screen formula is

mass ƛ /area = sin θ and y/L = sinθ

so we can say mass ƛ /area =  y/L

and y = mass length  ƛ / area       .................1

in equation 1 here we can see y and length is directly relation so we can say from equation 1 that  the width of the central bright spot on the screen change if the distance from the slit to the screen is doubled

Answer:

impulse = 6.09 kg m/s

Force = 6090 N

Explanation:

As we know that the impulse is defined as the change in momentum of the ball

so here we will have

[tex]\Delta P = mv_f - mv_i[/tex]

now we know that

[tex]v_f = 25 m/s[/tex]

initial speed is given as

[tex]v_i = -17 m/s[/tex]

now impulse is given as

[tex]\Delta P = 0.145(25 - (-17))[/tex]

[tex]\Delta P = 6.09 kg m/s[/tex]

Now we also know that average force is defined as the rate of change in momentum

[tex]F = \frac{\Delta P}{\Delta t}[/tex]

so we have

[tex]F = \frac{6.09}{1 \times 10^{-3}}[/tex]

[tex]F = 6.09 \times 10^3 N[/tex]

The magnitude of the impulse delivered by the bat to the ball is 6.09 kg*m/s. The magnitude of the average force exerted by the bat on the ball during a contact time of 1.0 ms is 6090 N.

The question relates to the concept of impulse and average force in a collision scenario. Impulse is given by the change in momentum of the object, and it can also be calculated as the average force (F) multiplied by the time duration (t) of the impact.

Considering the baseball scenario where the ball has a mass of 0.145 kg, enters the strike zone at 17.0 m/s and leaves in the opposite direction at a speed of 25.0 m/s, we can calculate the magnitude of the impulse delivered by the bat as follows:

To find the magnitude of the average force exerted by the bat on the ball, we use the time duration of the contact:

So, the magnitude of the average force exerted by the bat on the ball is 6090 N.

Answer:

[tex]E = 9.66\times 10^{-6} N/C[/tex]

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

[tex]F = qE[/tex]

so here the force is tension force on it

[tex]F = 6.57 \times 10^{-2} N[/tex]

[tex] Q = 6.80 \times 10^3 C[/tex]

now we have

[tex]6.57 \times 10^{-2} = (6.80 \times 10^3)E[/tex]

[tex]E = 9.66\times 10^{-6} N/C[/tex]

direction is Horizontal

The magnitude of the electric field on the charged sphere in this scenario is approximately 1.17 x 10^-5 N/C. The direction of the electric field is horizontal, which is the same direction as the tension in the thread.

To start, we can use the equilibrium condition where the tension in the thread is equal to the force due to the electric field and gravity on the sphere. The formula to calculate the electric force is F = qE, and the gravitational force is F = mg, where F is the force, q is the electric charge, E is the electric field, m is the mass of the object, and g is the gravity constant.

Tension - electric force - gravitational force equals zero: T - F_electric - F_gravity = 0. We fill in the previous formulas: T - qE -mg = 0. This can be rearranged to E = (T + mg) / q.

In this case, the sphere's mass m is 0.018 kg, the tension T is 6.57 x 10^-2 N, and the sphere's charge q is 6.80 x 10^3 C, and we use g = 9.81 m/s². So, E = ((6.57 x 10^-2) + (0.018 * 9.81)) / 6.80 x 10^3.

This leads to an electric field magnitude of approximately 1.17 x 10^-5 N/C. The direction of the electric field is the same as the direction of the tension, which is horizontal due to the thread being horizontal.

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C. I = 4.2A.

The magnetic field inside a solenoid is given by the equation:

B = μ₀NI/L

Clearing I for the equation above.

I = BL/μ₀N

With B = 1.0T, L = 3 x 10⁻²m, μ₀ = 4π x 10⁻⁷T.m/A and N = 5748turns

I = [(1.0T)(3 x 10⁻²m)]/[(4π x 10⁻⁷T.m/A)(5748turns)]

I = 4.15 ≅ 4.2A

Answer:

QC = 122 KJ

QH = 2.64 x 122 = 322 KJ

Explanation:

TH = 500 Degree C = 500 + 273 = 773 K

TC = 20 degree C = 20 + 273 = 293 K

W cycle = 200 KJ

Use the formula for the work done in a cycle

Wcycle = QH - QC

200 = QH - QC    ..... (1)

Usse

TH / TC = QH / QC

773 / 293 = QH / QC

QH / QC = 2.64

QH = 2.64 QC     Put it in equation (1)

200 = 2.64 QC - QC

QC = 122 KJ

So, QH = 2.64 x 122 = 322 KJ

Answer:

The correct option is : (a) Create a strong magnetic field.

Explanation:

An electromagnet is a substance which produces magnetic field when electric current is passed through it. The magnetic field produced by them disappears when the electric current is turned off.

A Ferromagnetic substance is a substance that gets magnetized when kept in an external magnetic field. Such substances remain magnetized even after the external magnetic field is removed.

When an electromagnet and a ferromagnet is combined, it results in the production of a strong magnetic field.

Answer:

[tex]B = \mu_0(\frac{1}{3} \frac{J_0}{r_0} r^2)[/tex]

Explanation:

As the current density is given as

[tex]J = \frac{J_0}{r_0}r[/tex]

now we have current inside wire given as

[tex]i = \int J(2\pi r)dr[/tex]

[tex]i = \int \frac{J_0}{r_0} r(2\pi r)dr[/tex]

[tex]i = 2\pi \frac{J_0}{r_0} \int r^2 dr[/tex]

[tex]i = \frac{2}{3} \pi \frac{J_0}{r_0} r^3[/tex]

Now by Ampere's law we will have

[tex]\int B. dl = \mu_0 i[/tex]

[tex]B. (2\pi r) = \mu_0(\frac{2}{3} \pi \frac{J_0}{r_0} r^3)[/tex]

[tex]B = \mu_0(\frac{1}{3} \frac{J_0}{r_0} r^2)[/tex]

Answer:

D. half as much

Explanation:

let m and M be the mass of the planets and r be the distance between them.

then: the force of attraction between them is given by,

F = G×m×M/(r^2)

if we keep one mass constant and double the other and also double the distance between them.

the force of attraction becomes:

F1 = 2G×m×M/[(2×r)^2]

   = 2G×m×M/[4×(r)^2]

   = (1/2)G×m×M/(r^2)

   = 1/2×F

therefore, when you double one mass and keep the other mass constant and double the distance between the masses you decrease the force by a factor of 1/2.

a) For an EM wave traveling in a vacuum, this equation holds true:

c = fλ

c is the speed of light in a vacuum, f is the frequency, and λ is the wavelength.

Given values:

c = 3×10⁸m/s

λ = 6m

Plug in the values and solve for f:

3×10⁸ = f(6)

f = 50MHz

b) The direction of an EM wave's Poynting vector determines the direction of the wave's propagation.

S = 1/μ₀(E×B)

S is the Poynting vector, μ₀ is the magnetic constant, E is the electric field vector, and B is the magnetic field vector. Note that we are taking the cross product between E and B, not taking the product of two scalar quantities.

Since S depends on the cross product of E and B, you may use the right hand rule in the following way to determine the direction of B:

You can conclude that B must point along the z axis, so you can represent B with the k unit vector.

a) The frequency f of the wave is 50 MHz.

b) The direction of the magnetic field associated with the wave is along Z-axis with the k unit vector.

The Poynting vector is a measurement that expresses the strength and direction of the energy flow in electromagnetic waves.

Mathematically;

S = 1/μ₀(E×B)

Where: S =  Poynting vector,  μ₀ = magnetic constant, E = electric field vector, and B = the magnetic field vector

a) For an EM wave traveling in a vacuum, it can be written that:

c = fλ

Where c is the speed of light in a vacuum, f is the frequency, and λ is the wavelength.

Speed of light in vacuum: c = 3×10⁸m/s

Wavelength of the given EM Wave: λ = 6m

Hence, the frequency of the EM wave: f =  3×10⁸m/s/  6m = 50MHz.

b) The EM wave travels in the x direction, therefore S points in the x direction.

Electric field acts along the y axis.

Hence, The direction of the magnetic field associated with the wave is along Z-axis with the k unit vector.

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Answer:

Part a)

W = 16.7 N

Part b)

r = 2.45 R

Explanation:

Part a)

As we know that acceleration due to gravity on the surface of moon is 1/6 times the gravity on the surface of earth

So the force due to gravity will decrease by the factor of 6

so we will have

[tex]W_{moon} = \frac{1}{6}W_{earth}[/tex]

[tex]W_{moon} = \frac{1}{6}(100)[/tex]

[tex]W_{moon} = 16.7 N[/tex]

Part b)

For the same value of the weight as the surface of moon the acceleration due to gravity of earth must be 1/6 times

so we have

[tex]\frac{GM}{r^2} = \frac{GM}{6R^2}[/tex]

[tex]r^2 = 6R^2[/tex]

[tex]r = 2.45 R[/tex]

The object must be [tex]\( \sqrt{6} \)[/tex]  Earth radii from the center of Earth to weigh the same as it does on the Moon.

The weight of an object on the Moon's surface is approximately 1/6 of its weight on Earth's surface due to the Moon's lower gravitational acceleration. Therefore, the object that weighs 100 N on Earth will weigh:

[tex]\[ W_{\text{Moon}} = \frac{1}{6} \times W_{\text{Earth}} = \frac{1}{6} \times 100 \text{ N} = 16.67 \text{ N} \][/tex]

For the second part of the question, we need to find the Earth radius multiple where the gravitational acceleration is the same as that of the Moon. The gravitational acceleration g on Earth is approximately [tex]\( 9.81 \, \text{m/s}^2 \)[/tex], and on the Moon, it is [tex]\( \frac{9.81}{6} \, \text{m/s}^2 \)[/tex]. The gravitational acceleration ( g' ) at a distance r from the center of Earth is given by:

[tex]\[ g' = g \left( \frac{R_{\text{Earth}}}{r} \right)^2 \][/tex]

where [tex]\( R_{\text{Earth}} \)[/tex] is the radius of Earth. We want to find r such that ( g' ) is equal to the gravitational acceleration on the Moon:

[tex]\[ g \left( \frac{R_{\text{Earth}}}{r} \right)^2 = \frac{g}{6} \][/tex]

Solving for r:

[tex]\[ \left( \frac{R_{\text{Earth}}}{r} \right)^2 = \frac{1}{6} \] \[ \frac{R_{\text{Earth}}}{r} = \sqrt{\frac{1}{6}} \] \[ r = R_{\text{Earth}} \sqrt{6} \][/tex]

Answer:

390 N

Explanation:

The net acceleration is:

a² = aₓ² + aᵧ²

a² = (900 m/s²)² + (870 m/s²)²

a = 1250 m/s²

So the net force is:

F = ma

F = (0.31 kg) (1250 m/s²)

F = 390 N

Answer:

25.6 V

Explanation:

The kinetic energy of electron associated with its potential difference is given by eV which is equal to the 1/2 mv^2.

m = 9.1 x 10^-31 kg, v = 3 x 10^6 m/s, e = 1.6 x 10^-19 C

eV = 1/2 m v^2

V = mv^2 / 2 e

V = (9.1 x 10^-31) x (3 x 10^6)^2 / (2 x 1.6 x 10^-19)

V = 25.6 V

The player's foot must be in contact with the ball for a distance of 1.48 meters to increase the ball's speed from 2.00 m/s to 6.00 m/s.

To increase the ball's speed from 2.00 m/s to 6.00 m/s, we can use the equation:

v^2 = u^2 + 2as

where v is the final velocity (6.00 m/s), u is the initial velocity (2.00 m/s), a is the acceleration, and s is the distance.

Rearranging the equation, we have:

s = (v^2 - u^2) / (2a)

Plugging in the values, we get:

s = (6.00^2 - 2.00^2) / (2 * 40.0)

s = 1.48 m

Therefore, the player's foot must be in contact with the ball for a distance of 1.48 meters.

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The physics concepts of force, mass, acceleration, and work were used to determine the distance over which the soccer player's foot needs to be in contact with the soccer ball in order to increase its speed from 2.00 m/s to 6.00 m/s. The resulting distance was approximately 0.09 m or 9 cm.

This question involves the relationship between force, mass, and acceleration, as well as the concept of work. Great! For the student to understand this, we need to look at a couple of formulas from physics. The force exerted on an object equals its mass times its acceleration (F = ma). Also, work done on an object is the force applied to it times the distance over which the force is applied (W = Fd).

Firstly, we need to determine the acceleration of the soccer ball when it is kicked. Using the first formula, we rearrange to find a = F/m. Here, F is the force applied (40.0 N) and m is the mass of the soccer ball (0.42 kg). Substituting these values, we get a = 40.0N / 0.42kg = approximately 95.2 m/s².

The initial speed of the ball is 2.00 m/s and the final speed we want is 6.00 m/s. The change in speed (which is also the change in velocity, as the direction doesn't change) is therefore 6.00 m/s - 2.00 m/s = 4.00 m/s. Using the formula for acceleration, which is a change in velocity divided by time (a = Δv / t), we can plug in our known values to solve for the time of contact: t = Δv / a = 4.00 m/s / 95.2 m/s² = approximately 0.042 s.

With the time of contact known, we can now determine the distance over which the soccer player's foot must be in contact with the ball using the formula for distance traveled during uniformly accelerated motion: d = vt + 0.5at². The initial velocity v is 2.00 m/s, the time of contact t is 0.042 s, and the acceleration a is 95.2 m/s². Substituting these values, we find that the distance is approximately 0.09 m or 9 cm.

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Answer:

t = 206 sec

Explanation:

m = mass of the ocean liner = 39000 Mg = 39000 x 10⁶ g = 3.9 x 10⁷ kg

F = constant force applied by the tugboat = 210 kN = 210000 N

v₀ = initial velocity of the liner = 4 km/h = 1.11 m/s

v = final velocity of the liner = 0 m/s

a = acceleration of the liner

Acceleration of the liner is given as

[tex]a = - \frac{F}{m}[/tex]

[tex]a = - \frac{210000}{3.9\times 10^{7}}[/tex]

a = - 0.0054 m/s²

Using the equation

v = v₀ + at

0 = 1.11 + (- 0.0054) t

t = 206 sec