Calculate the deflection at point C of a beanm subjected to uniformly distributed load w 275 N/mm on span AB and point load P-10 kN at C. Assume that L = 5 m and EI = 1.50 × 107 N·m2

Final answer:

The question inquires about the time needed for 80% of Pb-209 to decay, knowing its half-life is 3.3 hours. By applying the exponential decay formula, we calculate that approximately 7.39 hours are required for 80% of Pb-209 to decay.

Explanation:

The question involves the concept of radioactive decay and specifically asks how long it will take for 80% of Pb-209 to decay, given that its half-life is 3.3 hours. To find the time required for 80% of the lead to decay, we use the half-life formula and the property that radioactive decay is an exponential process. Since 80% decay means 20% remains, we set up the equation based on the exponential decay formula: N = N0(1/2)(t/T), where N is the remaining amount of substance, N0 is the initial amount, t is the time elapsed, and T is the half-life of the substance.

Substituting the given values and solving for t, we find:

N0 = 1 gram (100% initially)

N = 0.2 grams (20% remains)

T = 3.3 hours

Thus, the equation becomes 0.2 = 1(1/2)(t/3.3). Solving for t gives us the time required for 80% decay.

After calculations, the result is that it takes approximately 7.39 hours for 80% of the Pb-209 to decay. This showcases the practical application of exponential decay and half-life in determining the amount of a radioactive substance that remains after a given period.

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram.  There are three forces acting on the car.  Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank.  If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

The magnitude of the charge on the second object is 0.025 μC and its sign is negative because it is required work to bring the two charges together, suggesting these are opposite charges and repel each other.

The problem can be solved using the formula for the work done on a charge moving in an electric field, which is determined by the formula W = k * q1 * q2 / r, where k is the Coulomb's constant (8.99 × 10⁹ N · m²/C²), q1 and q2 are the charges, and r is the distance between them. From the problem, we know W = 32J, q1 = +35 μC, and r = 2.00 mm (the difference in the y-coordinates). Solving for q2 gives q2 = W * r / (k * q1) = 32J * 2.00 x 10⁻³m / (8.99 × 10⁹ N · m²/C² * 35 x 10⁻⁶C) = approximately -0.025 μC. Therefore, the magnitude of the charge on the second object is 0.025 μC, and its sign is negative because it takes work to bring the two charges together.

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Answer:

4582 N

Explanation:

The initial speed of the test car is

u = 11.4 m/s

While the final speed is

v = 0

The displacement of the test car during the collision is

d = 0.78 m

So we can find the acceleration of the car by using the following SUVAT equation:

[tex]v^2 - u^2 = 2ad\\a=\frac{v^2-u^2}{2d}=\frac{0-(11.4)^2}{2(0.78)}=-83.3 m/s^2[/tex]

Now we can find the average force acting on the dummy by using Newton's second law:

F = ma

Where m = 55 kg is the mass. Substituting,

[tex]F=(55 kg)(-83.3 m/s^2)=-4582 N[/tex]

So the size of the average force is 4582 N.

Answer:

160 m/s

Explanation:

Momentum is conserved:

mu = mv + MV

(7.00) (200) = (7.00)v + (150) (1.8)

v = 160 m/s

161.43m/s

Using the principle of conservation of linear momentum i.e

Total momentum before impact is equal to total momentum after impact.

=> Momentum of bullet before impact + Momentum of tin before impact

                               =

   Momentum of bullet after impact + Momentum of tin after impact

   i.e

   [tex]m_{B}[/tex] [tex]u_{B}[/tex] + [tex]m_{T}[/tex] [tex]u_{T}[/tex] = [tex]m_{B}[/tex] [tex]v_{B}[/tex] + [tex]m_{T}[/tex] [tex]v_{T}[/tex]

Where;

[tex]m_{B}[/tex] = mass of bullet = 7.00g = 0.007kg

[tex]m_{T}[/tex] = mass of tin can = 150g = 0.15kg

[tex]u_{B}[/tex] = initial velocity of bullet before impact = 200m/s

[tex]u_{T}[/tex] = initial velocity of tin can before impact = 0m/s (since the can is stationary)

[tex]v_{B}[/tex] = final velocity of the bullet after impact

[tex]v_{T}[/tex] = final velocity of the tin can after impact = 180cm/s = 1.8m/s

Substitute these values into the equation above;

=>    [tex]m_{B}[/tex] [tex]u_{B}[/tex] + [tex]m_{T}[/tex] [tex]u_{T}[/tex] = [tex]m_{B}[/tex] [tex]v_{B}[/tex] + [tex]m_{T}[/tex] [tex]v_{T}[/tex]

=> (0.007 x 200) + (0.15 x 0) = (0.007 x [tex]v_{B}[/tex])  + (0.15 x 1.8)

=> 1.4 + 0 = 0.007[tex]v_{B}[/tex] + 0.27

=> 1.4 = 0.007[tex]v_{B}[/tex] + 0.27

=> 0.007[tex]v_{B}[/tex] = 1.4 - 0.27

=> 0.007[tex]v_{B}[/tex] = 1.13

Solve for [tex]v_{B}[/tex]

=> [tex]v_{B}[/tex] = 1.13 / 0.007

=> [tex]v_{B}[/tex] = 161.43m/s

Therefore, the speed of the bullet after impact (leaving the can) is 161.43m/s

Answer:

a)

16.33 Ω

b)

45009.18 A

Explanation:

a)

L = length of the line = 935 km = 935000 m

d = diameter of the line = 3.50 cm = 0.035 m

ρ = resistivity of the line = 1.68 x 10⁻⁸ Ω.m

Area of cross-section of the line is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.035)²

A = 0.000961625 m²

Resistance of the line is given as

[tex]R=\frac{\rho L}{A}[/tex]

inserting the values

R = (1.68 x 10⁻⁸) (935000)/(0.000961625)

R = 16.33 Ω

b)

V = potential difference across the line = 735 kv = 735000 Volts

i = current carried by the wire

Using ohm's law, current carried by the wire is given as

[tex]i=\frac{V}{R}[/tex]

i = 735000/16.33

i = 45009.18 A

Answer:

(a) 2.34 s

(b) 6.71 m

(c) 38.35 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by

[tex]T = \frac{2 u Sin\theta }{g}[/tex]

[tex]T = \frac{2 \times 20 \times Sin35 }{9.8}[/tex]

T = 2.34 second

(b) The formula for the maximum height is given by

[tex]H = \frac{u^{2} \times Sin^{2}\theta }{2g}[/tex]

[tex]H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}[/tex]

H  = 6.71 m

(c) The formula for the range is given by

[tex]R = \frac{u^{2} \times Sin 2\theta }{g}[/tex]

[tex]R = \frac{20^{2} \times Sin 2 \times 35}{9.8}[/tex]

R = 38.35 m

(d) It hits with the same speed at the initial speed.

Answer:

Acceleration due to gravity value

[tex]=9.89m/ {s}^{2} [/tex]

Explanation:

Time period of simple pendulum is given by the expression

[tex]T=2\pi\sqrt{\frac{l}{g}} \\ [/tex]

Here we have

T = 0.88 s

l = 19.4 cm = 0.194 m

Substituting

[tex]0.88=2\pi\sqrt{\frac{0.194}{g}}\\\\\frac{0.194}{g}=0.0196\\\\g=9.89m/s^2 \\ [/tex]

Acceleration due to gravity value

[tex]=9.89m/s^2 \\ [/tex]

The particle has an acceleration vector with one component directed toward the center of its orbit, and the other directing tangentially to its orbit. Call these components [tex]\vec a_c[/tex] ([tex]c[/tex] for center) and [tex]\vec a_t[/tex] ([tex]t[/tex] for tangent). Then its acceleration vector has magnitude

[tex]|\vec a|=\sqrt{\|\vec a_c\|^2+\|\vec a_t\|^2}[/tex]

We have

[tex]\|\vec a_c\|=\dfrac{\|\vec v\|^2}r[/tex]

where [tex]\|\vec v\|[/tex] is the particle's speed and [tex]r[/tex] is the radius of orbit, so

[tex]\|\vec a_c\|=\dfrac{\left(37.2\frac{\rm m}{\rm s}\right)^2}{21\,\rm m}=65.9\dfrac{\rm m}{\mathrm s^2}[/tex]

We're given that the particle's speed changes at a rate of 23.1 m/s^2. Its velocity vector points in the same direction as [tex]\vec a_t[/tex], i.e. perpendicular to [tex]\vec a_c[/tex], so

[tex]\|\vec a_t\|=23.1\dfrac{\rm m}{\mathrm s^2}[/tex]

Then the magnitude of the particle's acceleration is

[tex]\|\vec a\|=\sqrt{\left(65.9\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(23.1\dfrac{\rm m}{\mathrm s^2}\right)^2}=\boxed{69.8\dfrac{\rm m}{\mathrm s^2}}[/tex]

The magnitude of the acceleration of the particle is approximately 70.55 m/s^2, calculated by using the formulas for combined radial and tangential acceleration in circular motion.

In this physics problem, the particle not only moves around in a circle but is also experiencing an increase in speed which is a case of combined radial and tangential acceleration. Radial acceleration, known as centripetal acceleration (ar), is the result of the change in direction of the velocity vector, while tangential acceleration (at) comes from changes in speed.

The total acceleration of an object in circular motion is given by:

a = sqrt((ar^2) + (at^2))

Centripetal acceleration can be calculated using the formula ar = v^2 / r, where: v = speed (37.2 m/s), r = radius of the circle (21 m). This gives us ar = (37.2^2) / 21, which approximately equals 66.62 m/s^2.

The tangential acceleration is given in the problem: at = 23.1 m/s^2.

We therefore calculate the total acceleration using the formula above which gives us:

a = sqrt((66.62^2) + (23.1^2)) which approximately equals 70.55 m/s^2.

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Answer:

69.4 pA

Explanation:

n = number of sodium ions = 1473

e = magnitude of charge on each sodium ion = 1.6 x 10⁻¹⁹ C

t = time taken to flow across the membrane = 3.4 x 10⁻⁶ sec

Total Charge on sodium ions is given as

q = n e                                            eq-1

Current produced is given as

[tex]i = \frac{q}{t}[/tex]

Using eq-1

[tex]i = \frac{ne}{t}[/tex]

Inserting the values

[tex]i = \frac{(1473)(1.6\times 10^{-19})}{3.4\times 10^{-6}}[/tex]

i = 69.4 x 10⁻¹² A

i = 69.4 pA

Answer:

17 seconds

Explanation:

In the y direction:

y = y₀ + v₀ᵧ t + ½ gt²

0 = 140 + (120 sin 38) t + ½ (-9.8) t²

4.9 t² - 73.9 t - 140 = 0

Solve with quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

t = [ 73.9 ± √((-73.9)² - 4(4.9)(-140)) ] / 9.8

t = -1.7, 16.8

Since t can't be negative, t = 16.8.  Rounding to 2 sig-figs, the projectile lands after 17 seconds.

Answer:

Neither

Explanation:

In this situation, the net force acting on the toy car moving in the circle has two components:

- There is a component which is tangential (parallel) to the circle - we can understand this by the fact that the car is speeding up: this means that its tangential speed is changing, so it has a tangential acceleration, therefore there must be a component of the force tangential to the circle (parallel to the circle)

- There is a component which is radial to the circle, pointing towards the centre - this is called centripetal force. This is due to the fact that the car is constantly changing direction of motion: so, there must be a force that causes this change in direction of the car, and this force points towards the centre of the circle, and it is called centripetal force.

Explanation:

It is given that,

Mass of the football player, m = 92 kg

Velocity of player, v = 5 m/s

Time taken, t = 10 s

(1) We need to find the original kinetic energy of the player. It is given by :

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

[tex]k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2[/tex]

k = 1150  J

In two significant figure, [tex]k=1.2\times 10^3\ J[/tex]

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

i.e. [tex]P=\dfrac{W}{t}=\dfrac{\Delta K}{t}[/tex]

[tex]P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}[/tex]

P = 115 watts

In two significant figures, [tex]P=1.2\times 10^2\ Watts[/tex]

Hence, this is the required solution.  

Answer:A

Explanation:

Initial velocity, u = 2m/s

Final velocity, v = 4m/s

Distance covered, s = 12m

Acceleration, a = ?

Using

v² = u² + 2as

2as = v² - u²

a = v²-u²/2s

a = 4²-2²/2 x 12

a = 16-4/24

a = 12/24

a = 0.5m/s²

Answer:

D]  speed

Explanation:

Answer:

[tex]2.1\cdot 10^{-3} C[/tex]

Explanation:

The initial charge stored on the capacitor is given by

[tex]Q_0 =C_0 V[/tex]

where

[tex]C_0 = 6.0 \mu F = 6.0 \cdot 10^{-6}F[/tex] is the initial capacitance

V = 100 V is the potential difference across the capacitor

Solving the equation,

[tex]Q_0 = (6.0 \cdot 10^{-6}F)(100 V)=6.0 \cdot 10^{-4}C[/tex]

The charge stored in the capacitor when inserting the dielectric is

[tex]Q = k Q_0[/tex]

where

k = 4.5 is the dielectric constant

Substituting,

[tex]Q=(4.5)(6.0 \cdot 10^{-4}C)=2.7\cdot 10^{-3}C[/tex]

So the additional charge is

[tex]\Delta Q=Q-Q_0 = 2.7 \cdot 10^{-3}C - 6.0 \cdot 10^{-4}C=2.1\cdot 10^{-3} C[/tex]

The additional charge that flows into the capacitor once it is immersed in transformer oil (of dielectric constant 4.5) and kept connected to the source is 2100 μC.

The question involves a capacitor that is connected across a 100-V voltage source and then submerged in transformer oil. The capacitor initially charges to capacity while in the air. When a capacitor is then immersed in a material with a dielectric constant, its ability to store charge improves. Here, the dielectric constant of the transformer oil is given as 4.5, suggesting that the capacitor's capacity to store charge will increase 4.5 times as compared to when it was in air.

The additional charge that flows into the capacitor can be calculated using the formula for the charge in a capacitor, Q = CV. The initial charge (Q1) on the capacitor when it was in air would be Q1 = CV1 = 6.0 μF * 100 V = 600 μC. After immersing in transformer oil, the capacitance would increase by a factor of 4.5, giving a new capacitance C2 = 4.5 * 6.0 μF = 27.0 μF. The new charge (Q2) would be Q2 = CV2 = 27.0 μF * 100 V = 2700 μC. Hence the additional charge that's flown from the source would be Q2 - Q1 = 2700 μC - 600 μC = 2100 μC.

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Answer:

=1419.19 meters.

Explanation:

The time it takes for the shell to drop to the tanker from the height, H =1/2gt²

610m=1/2×9.8×t²

t²=(610m×2)/9.8m/s²

t²=124.49s²

t=11.16 s

Therefore, it takes 11.16 seconds for a free fall from a height of 610m

Range= Initial velocity×time taken to hit the tanker.

R=v₁t

Lets change 300 mph to kph.

=300×1.60934 =482.802 kph

Relative velocity=482.802 kph-25 kph

=457.802 kph

Lets change 11.16 seconds to hours.

=11.16/(3600)

=0.0031 hours.

R=v₁t

=457.802 kph × 0.0031 hours.

=1.41918 km

=1.41919 km × 1000m/km

=1419.19 meters.

Distance is a numerical representation of the distance between two objects or locations. The horizontal distance behind the tanker will be 1419.19 m.

Distance is a numerical representation of the distance between two objects or locations. The distance can refer to a physical length or an estimate based on other factors in physics or common use.

The given data in the problem is ;

v₁ is the horizontal flying velocity  = 300 m/h

H is the  altitude = 610 m

v₂ is he cruise velocity= 25km/h

If t is the time taken the shell drops to the tanker from the height h is found by the formula;

[tex]\rm H =\frac{1}{2} gt^2 \\\\ \rm t=\sqrt{\frac{H}{2g} } \\\\ \rm t=\sqrt{2gh} \\\\ \rm t=\sqrt{2\times 610 \times 9.81} \\\\ \rm t=11.16 \sec[/tex]

The velocity of bomber obtained after unit conversion;

[tex]V_{12}=300\times 1.60 = 482.802[/tex]

Relative velocity is defined as the velocity of an object with respect to the other object.

Relative velocity=482.802 kph-25 kph=457.802 kph

In one hour there are 3600 seconds then the conversion is found by;

On changing 11.16 seconds to hours we found;

[tex](\frac{11.16}{3600} )=0.0031 \ hours.[/tex]

The range is the horizontal distance which is given by ;

Range= Initial velocity×time taken to hit the tanker.

[tex]\rm R= v \times t \\\\ \rm R= 457.8\times 0.0031\\\\ \rm R= 1.41918\ Km \\\\ \rm R=1419.19 m[/tex]

Hence the horizontal distance behind the tanker will be 1419.19 m.

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Answer:

The speed of the center of mass of the two carts is 0.103 m/s

Explanation:

It is given that,

Mass of the air track cart, m₁ = 350 g = 0.35 kg

Velocity of air track cart, v₁ = 1.25 m/s

Mass of cart, m₂ = 280 g = 0.28 kg

Velocity of cart, v₂ = -1.33 m/s (it is travelling in opposite direction)

We need to find the speed of the center of mass of the two carts. It is given by the following relation as :

[tex]v_{cm}=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]

[tex]v_{cm}=\dfrac{0.35\ kg\times 1.25\ m/s+0.28\ kg\times (-1.33\ m/s)}{0.35\ kg+0.28\ kg}[/tex]

[tex]v_{cm}=0.103\ m/s[/tex]

Hence, this is the required solution.

Answer:

12353 V m⁻¹ = 12.4 kV m⁻¹

Explanation:  

Electric field between the plates of the parallel plate capacitor depends on the potential difference across the plates and their distance of separation.Potential difference across the plates V over the distance between the plates gives the electric field between the plates. Potential difference is the amount of work done per unit charge and is given here as 21 V. Electric field is the voltage over distance.

E = V ÷ d = 21 ÷ 0.0017 = 12353 V m⁻¹

Answer:

2.4 x 10^21

Explanation:

V = 9 V, R = 1.4 ohm, t = 1 minute = 60 second

Use Ohm's law

V = I R

I = V / R

I = 9 / 1.4

I = 6.43 A

Now use Q = I t

Q = 6.43 x 60 = 385.7 C

Number of electrons passing in 1 minute , n

= total charge in one minute / charge of one electron

n = 385.7 / (1.6 x 10^-19) = 2.4 x 10^21

Answer:

2.18 s

Explanation:

H = - 25 m ( downwards)

U = - 0.8 m/s

g = - 9.8 m/s^2

Let time taken is t.

Use second equation of motion

H = u t + 1/2 g t^2

- 25 = - 0.8 t - 1/2 × 9.8 × t^2

4.9 t^2 + 0.8 t - 25 = 0

By solving we get

t = 2.18 s

Answer:

The position function is [tex]s_{t}=2t^3+5t^2-5t+9[/tex].

Explanation:

Given that,

Acceleration [tex]a =12t+10[/tex]

Initial velocity [tex]v_{0} = -5\ cm/s[/tex]

Initial displacement [tex]s_{0}=9\ cm[/tex]

We know that,

The acceleration is the rate of change of velocity of the particle.

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

The velocity is the rate of change of position of the particle

[tex]v=\dfrac{dx}{dt}[/tex]

We need to calculate the the position

The acceleration is

[tex]a_{t} = 12t+10[/tex]

[tex]\dfrac{dv}{dt} = 12t+10[/tex]

[tex]a_{t}=dv=(12t+10)dt[/tex]

On integration both side

[tex]\int{dv}=\int{(12t+10)}dt[/tex]

[tex]v_{t}=6t^2+10t+C[/tex]

At t = 0

[tex]v_{0}=0+0+C[/tex]

[tex]C=-5[/tex]

Now, On integration again both side

[tex]v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt[/tex]

[tex]s_{t}=2t^{3}+5t^2-5t+C[/tex]

At t = 0

[tex]s_{0}=0+0+0+C[/tex]

[tex]C=9[/tex]

[tex]s_{t}=2t^3+5t^2-5t+9[/tex]

Hence, The position function is [tex]s_{t}=2t^3+5t^2-5t+9[/tex].

To find the position function of a particle given acceleration a(t) = 12t + 10, one integrates twice. The first integral gives the velocity function, the second gives the position or displacement function. These are determined to be v(t) = 6t^2 + 10t - 5 and s(t) = 2t^3 +5t^2 - 5t + 9, respectively.

The subject of this problem involves calculating the position function, or displacement, of a particle given an acceleration function, initial velocity, and initial position. In this case, acceleration is given by a(t) = 12t + 10.

To find the velocity function, v(t), you integrate the acceleration function: ∫a(t) dt = ∫(12t + 10) dt = 6t^2 + 10t + C1, where C1 is the constant of integration. We know that v(0) = −5 cm/s, so C1 is -5: the full velocity function is v(t) = 6t^2 + 10t - 5.

We then integrate the velocity function to find the position function, s(t): ∫v(t) dt = ∫(6t^2 + 10t - 5) dt = 2t^3 +5t^2 - 5t + C2. Given s(0) = 9 cm, we find C2 = 9. The position function s(t) is therefore s(t) = 2t^3 +5t^2 - 5t + 9.

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The planet's orbital period is about 388 days

[tex]\texttt{ }[/tex]

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

[tex]\texttt{ }[/tex]

Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of the planet = m = 6.75 × 10²⁴ kg

mass of the star = M = 2.75 × 10²⁹ kg

radius of the orbit = R = 8.05 × 10⁷ km = 8.05 × 10¹⁰ m

Unknown:

Orbital Period of planet = T = ?

Solution:

Firstly , we will use this following formula to find the orbital period:

[tex]F = ma[/tex]

[tex]G \frac{ Mm}{R^2}=m \omega^2 R[/tex]

[tex]G M = \omega^2 R^3[/tex]

[tex]\frac{GM}{R^3} = \omega^2[/tex]

[tex]\omega = \sqrt{ \frac{GM}{R^3}}[/tex]

[tex]\frac{2\pi}{T} = \sqrt{ \frac{GM}{R^3}}[/tex]

[tex]T = 2\pi \sqrt {\frac{R^3}{GM}}[/tex]

[tex]T = 2 \pi \sqrt {\frac{(8.05 \times 10^{10})^3}{6.67 \times 10^{-11} \times 2.75 \times 10^{29}}}[/tex]

[tex]T \approx 3.35 \times 10^7 \texttt{ seconds}[/tex]

[tex]T \approx 388 \texttt{ days}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

To calculate the orbital period of the planet, convert the radius to meters, substitute the mass of the star and the radius into Kepler's third law, solve for the orbital period squared, and convert to Earth days.

The question involves applying Kepler's third law of planetary motion to calculate the orbital period of a planet revolving around a star. The law relates the orbital period (T) to the radius (r) of the orbit and the mass (M) of the star around which the planet orbits. Kepler's third law can be expressed as:

T² = (4π²/GM) · r³,

where G is the gravitational constant (6.67430 × 10⁻¹¹ m³/kg · s²).

To find the orbital period of the planet:

The student will then have the orbital period of the planet in Earth days.

Answer:

Force required is 4387 N in the opposite direction of motion.

Explanation:

We have equation of motion v² = u² + 2as

      v = 0m/s , u = 15 m/s, s = 31.8 m

  Substituting

         0² = 15² + 2 x a x 31.8

          a = -3.54 m/s²

So, deceleration = 3.54 m/s²

Force = Mass x Acceleration

          = 1240 x -3.54 = -4387 N

So force required is 4387 N in the opposite direction of motion.

Answer:

2.0 s, 200 m

Explanation:

Time to hit the ground depends only on height.  Since the plane is at the same height, the weight lands at the same time as before, 2.0 s.

Since the plane is going twice as fast, the weight will travel twice as far (ignoring air resistance).  So it will travel a horizontal distance of 200 m.

Answer:

1) 2 seconds

2) 200 m

Explanation:

1) Fall time at initial speed [tex]s_{1}[/tex] = [tex]t_{1}[/tex]

  Fall time at final speed [tex]s_{2}[/tex] = [tex]t_{2}[/tex]

  Initial fall height [tex]h_{1}[/tex] at initial speed = Final fall height [tex]h_{2}[/tex] at final speed i.e [tex]h_{1}[/tex] = [tex]h_{2}[/tex]

s = speed

t = time

h = height

Therefore, since fall time depends on fall height where acceleration due to gravity (g) is constant,

Fall time at [tex]s_{1}[/tex] = Fall time at [tex]s_{2}[/tex]

i.e [tex]t_{1}[/tex] = [tex]t_{2}[/tex] = 2.0 s

Time taken to land = 2.0 s

2) Initial distance traveled ([tex]S_{1}[/tex]) at initial speed [tex]s_{1}[/tex] = 100 m

   Final speed [tex]s_{2}[/tex] is double initial speed i.e [tex]s_{2}[/tex] = [tex]2s_{1}[/tex]

Therefore, since distance traveled is directly proportional to speed,

Final distance traveled [tex]S_{2}[/tex] at final speed [tex]s_{2}[/tex] is double initial distance [tex]S_{1}[/tex]

i.e [tex]S_{2}[/tex] = [tex]2S_{1}[/tex]

    [tex]2S_{1}[/tex] = 2 x 100 m = 200 m

Distance traveled = 200 m

Answer:

[tex]8.6\cdot 10^{-18} m[/tex]

Explanation:

Initially, the electron is travelling undeflected at constant speed- this means that the electric force and the magnetic force acting on the electron are balanced. So we can write

q E = q v B

where

q is the electron's charge

[tex]E=8.3\cdot 10 V/m[/tex] is the electric field magnitude

v is the electron's speed

[tex]B=7.3\cdot 10^3 T[/tex] is the magnitude of the magnetic field

Solving for v,

[tex]v=\frac{E}{B}=\frac{8.3 \cdot 10 V/m}{7.3\cdot 10^3 T}=0.011 m/s[/tex]

Then the electric field is turned off, so the electron (under the influence of the magnetic field only) will start moving in a circle of radius r. Therefore, the magnetic force will be equal to the centripetal force:

[tex]qvB= m \frac{v^2}{r}[/tex]

where

[tex]q=1.6\cdot 10^{-19} C[/tex] is the electron's charge

[tex]m=9.11\cdot 10^{-31} kg[/tex] is the electron's mass

Solving for r, we find the radius of the electron's orbit:

[tex]r=\frac{mv}{qB}=\frac{(9.11\cdot 10^{-31} kg)(0.011 m/s)}{(1.6\cdot 10^{-19} C)(7.3\cdot 10^3 T)}=8.6\cdot 10^{-18} m[/tex]

Answer:

H = 4.12 m

Explanation:

As we know that horizontal range is the distance moved in horizontal direction

Since horizontal direction has no acceleration

so here we have

[tex]Range = v_x T[/tex]

here we know that

[tex]v_x = vcos32[/tex]

so from above formula

[tex]92 = (vcos32)(6.4)[/tex]

[tex]v = 16.95 m/s[/tex]

now we have maximum height is given as

[tex]H = \frac{(vsin32)^2}{2g}[/tex]

[tex]H = \frac{(16.95 sin32)^2}{2(9.8)}[/tex]

[tex]H = 4.12 m[/tex]

Answer:

10.19 m

Explanation:

Energy is conserved, so elastic energy stored in spring = gravitational energy of block.

1/2 kx² = mgh

h = kx² / (2mg)

h = (5200 N/m) (0.096 m)² / (2 × 0.240 kg × 9.8 m/s²)

h = 10.19 m

Answer with Explanation:

since the two blocks move (stick) together, the collision is inelastic, which does not conserve kinetic energy.  So do not use kinetic energy consideration.

Fortunately, in such a situation, momentum is still conserved.

Momentum of 1.0 kg block

= 1.0 * 3.0 = 3.0 kg-m/s

Momentum of second block

= 1.0 * 1.0 = 1.0 kg-m/s

Total mass after collision = 1.0+1.0 = 2.0 kg

Common velocity after collision

= total momentum / total mass

= (3.0+1.0)/2.0 = 2.0 m/s

Answer:

Option (A)

Explanation:

A ball bearing is a device which is use to reduce the friction.

The outer rim of the bearing is fixed with the part of machine and inner rim is fitted into shafts. Now the shafts rotates and only the small spheres in the bearing will rotate. The friction can be further reduced by apply the oil or grease to the bearing.