A 0.20 kg particle moves along the x axis under the influence of a stationary object. The potential energy is given by U(x) = 8x2 + 2x4, where U is in joules and x is in meters. If the particle has a speed of 5.0 m/s when it is at x = 1.0 m, its speed when it is at the origin is:

The bungee jumper experiences a downward acceleration of 4.09 m/s^2 at the point when the bungee cord has lengthened to 25m.

The subject of this problem pertains to physics, specifically, the principles of dynamics and Hooke's Law. At the instant when the bungee cord has stretched to a total length of 25m, it has undergone an extension of 5m since its natural length is 20m. According to Hooke’s Law, this extension causes a restoring force F = kx, where k is the spring constant and x is the change in length.

In this case, F = 80 N/m × 5 m = 400 N which is the upward force exerted by the cord. The weight of the 70 kg person is mg, or 70 kg × 9.8 m/s² = 686 N downward. The net force acting on the jumper is the difference between the weight and the restoring force. This gives a net force of 686 N – 400 N = 286 N downward.

From Newton’s second law, force equals mass times acceleration, F = ma. Therefore, the acceleration of the jumper at the instant would be a = F/m, or 286 N ÷ 70 kg = 4.09 m/s² downward to 2 significant figures.

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Answer:

48.6 N

Explanation:

rate of mass per second, dm/dt = 5 kg/s

Velocity, v = 35 km/hr = 9.72 m/s

Force acting on the plate

F = v x dm/dt

F = 9.72 x 5 = 48.6 N

Thus, the force acting on the plate is 48.6 N.

To prevent the plate from moving horizontally, a force of 48.6 N is required. The initial horizontal momentum of the water is 48.6 kg*m/s.

To calculate the force needed to prevent the plate from moving horizontally when a horizontal water jet strikes a vertical plate, we first need to determine the change in momentum of the water. The water strikes the plate at a rate of 5 kg/s with a velocity of 35 km/hr. Let's convert the velocity to meters per second:

35 km/hr = (35 * 1000) / 3600 = 9.72 m/s.

Next, we calculate the initial horizontal momentum of the water jet:

Initial momentum = mass flow rate * velocity = 5 kg/s * 9.72 m/s = 48.6 kg*m/s.

The water stream moves in the vertical direction after the strike, so its horizontal momentum is reduced to zero. The change in momentum is:

Change in momentum = Initial momentum - Final momentum = 48.6 kg*m/s - 0 kg*m/s = 48.6 kg*m/s.

We use the change in momentum (impulse) to find the force since force is the rate of change of momentum:

Force = Change in momentum / Time

The mass flow rate gives the change in momentum per second, which means the force can be directly calculated as:

Force = 48.6 N

Thus the force required to prevent the plate from moving horizontally, a force of 48.6 N

Answer:

given,

mass of uniform bar = 2.9 Kg

length of string = 3 m

mass of monkey = 1.45 Kg

monkey is sitting at the center

Weight of the beam will also consider to as the center

total downward force  = 2.9 g  + 1.45 g

now tension will balance the force

T₁ + T₂ = 2.9 g  + 1.45 g

T₁ + T₂ = 4.35 g

taking moment about T₂ string

now,

T₁ x (3-0.74) - 4.35 g x (1.5 - 0.74) = 0

T₁ x  2.26 = 32.3988

T₁ = 14.34 N

Answer:[tex]V_{net}=4\frac{kQ}{a}[/tex]    

Explanation:

Given

charge on each Particle is Q

Length of diagonal of the square is 2a

therefore distance between center and each charge is [tex]\frac{2a}{2}=a[/tex]

Electric Potential of charged Particle is given by

For First Charge

[tex]V_1=\frac{kQ}{a}[/tex]

[tex]V_2=\frac{kQ}{a} [/tex]

[tex]V_3=\frac{kQ}{a}[/tex]

[tex]V_4=\frac{kQ}{a}[/tex]

total Electric Potential At center is given by

[tex]V_{net}=V_1+V_2+V_3+V_4[/tex]

[tex]V_{net}=4\times \frac{kQ}{a}[/tex]

[tex]V_{net}=4\frac{kQ}{a}[/tex]            

The sphere's angular acceleration is [tex]2.064 rad/s^2[/tex], and it will take approximately 10.17 seconds to decrease its rotational speed by 21.0 rad/s when experiencing a constant friction force.

Part A: Calculating Angular Acceleration

To find the angular acceleration, we must first calculate the torque. The torque ( au) caused by the friction force (F) is the product of the force and the radius (r) at which the force is applied, and torque is given by [tex]\tau = r \times F[/tex]. Using the diameter (d) to find the radius, we have r = d / 2 = 4.50 cm / 2 = 2.25 cm = 0.0225 m. The given friction force is 0.0200 N. Therefore, [tex]\tau = 0.0225 m \times 0.0200 N = 0.00045 Nm[/tex].

Next, we use the moment of inertia (I) for a sphere, [tex]I=\frac{2}{5} mr^2[/tex], to calculate the angular acceleration ([tex]\alpha[/tex]). The mass (m) is 220 g which is 0.220 kg. So, [tex]I = \frac{2}{5} \times 0.220 kg \times (0.0225 m)^2 = 0.000218025 kg m^2[/tex]. Angular acceleration, [tex]\alpha = \frac{\tau}{I} = \frac{0.00045 Nm}{0.000218025 kgm^2} \longrightarrow \alpha= 2.064 rad/s^2[/tex].

Part B: Time to Decrease Rotational Speed

To find the time (t) to decrease the rotational speed by 21.0 rad/s, we can use the equation [tex]\omega = \omega_0 + \alpha \times t[/tex], where [tex]\omega_0[/tex] is the initial angular velocity and [tex]\omega[/tex] is the final angular velocity. Since the sphere is slowing down due to friction, the angular acceleration will be negative, [tex]\alpha = -2.064 rad/s^2[/tex]. We are looking for the time it takes for the speed to decrease by 21.0 rad/s, so if the initial speed is [tex]\omega_0[/tex] and the final is [tex]\omega_0 - 21.0 rad/s[/tex], then [tex]0 = \omega_0 - 21.0 rad/s + \alpha \times t[/tex]. Solving for t gives [tex]t = \frac{21.0 rad/s}{2.064 rad/s^2} \approx t = 10.17 s[/tex].

Answer:

[tex]T_2=630^{\circ}C[/tex]'

Explanation:

Original temperature of the gas, [tex]T_1=28^{\circ}C=301\ K[/tex]

From the ideal gas equation,

[tex]P_1V_1=nRT_1[/tex]

Since,

[tex]P_2=3P_1[/tex]

[tex]nRT_2=3(nRT_1)[/tex]

[tex]T_2=3T_1[/tex]

[tex]T_2=3\times 301[/tex]

[tex]T_2=903\ K[/tex]

or

[tex]T_2=630^{\circ}C[/tex]

So, the new temperature of the gas is 630 degree Celsius. Hence, this is the required solution.

Answer:

correct option is (b) 2.7cm

Explanation:

given data

n = 3/2 = 1.5

sides of length  = 8 cm

to find out

How far from one of the faces of the block that what is the image distance of the picture from the face

solution

we get here distance that is

distance d = [tex]\frac{d_o}{n}[/tex]      ......................1

put here value as do is [tex]\frac{8}{2}[/tex]  = 4 cm

so

distance d = [tex]\frac{d_o}{n}[/tex]  

distance d = [tex]\frac{4}{1.5}[/tex]  

distance d = 2.6666666 = 2.67

so correct option is (b) 2.7cm

As the image distance of the picture from the face is 2.7 cm, the correct answer is (b) 2.7cm.

The problem is related to optics, specifically the refraction of light through a medium with a given refractive index, here Lucite, which has a refractive index of n = 1.5.

The picture is physically at the center of the block. Since the block has sides of length 8 cm, the center is 4 cm from any face.

To find the apparent distance due to refraction, we use the formula for the apparent depth (d') from the real depth (d) and the refractive index (n) given as:

[tex]d' = \frac{d}{n}[/tex]

Substituting the given values, we get:

d' = [tex]\frac{4 cm}{1.5}[/tex] ≈ 2.67 cm.

Rounding to the closest option, the apparent distance is 2.7 cm from one of the faces of the block.

Answer:

  v₂ = v/1.5= 0.667 v

Explanation:

For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.

Initial moment before pushing

    p₀ = 0

Final moment after they have been pushed

    [tex]p_{f}[/tex] = m₁ v₁ + m₂ v₂

   p₀ =  [tex]p_{f}[/tex]

   0 = m₁ v₁ + m₂ v₂

   m₁ v₁ = - m₂ v₂

Let's replace

   M (-v) = -1.5M v₂

   v₂ = v / 1.5

  v₂ = 0.667 v

Conservation of momentum dictates that the total momentum before and after the push must be the same since there are no external forces. Therefore, student 2 and the cart will move in the opposite direction to student 1's movement, with a velocity of two-thirds of student 1's velocity.

This question relies on the concept of Conservation of Momentum. In this scenario, the frictionless carts allow us to use the principle of momentum conservation, because if there is no external force, the total momentum of the system remains constant.

Second Cart's Velocity Calculation:

Applying the principle, if student 1 and the cart (both with mass M) are initially at rest, the initial momentum of the system is 0. When student 1 pushes and goes in the opposite direction with velocity -v, the momentum of the system is still 0 (since no external forces are present). So, if student 2 (with mass 1.5M) is pushed in the opposite direction, for the total momentum to remain zero, the velocity (V2) of student 2 and the cart must be such that:
-M * v (student 1's momentum) + 1.5M * V2 (student 2's momentum) = 0.
Solving for V2 gives us V2 = -(-M * v) / 1.5M = 2v/3.

So, the velocity of student 2 and the cart is 2v/3 in the opposite direction to student 1's movement.

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Answer:

[tex]\lambda=56.5cm[/tex]

Explanation:

Wavelength is calculated as:

[tex]\lambda=\frac{V}{f}[/tex]

Replacing the given values:

[tex]\lambda=\frac{345}{610}[/tex]

[tex]\lambda=0.565m[/tex]

Converting the result into cm:

[tex]\lambda=56.5cm[/tex]

Answer:

Tracy’s constant acceleration during her time of contact with Tom is 1.25 m/s².

Explanation:

Given that,

Mass of Tracy = 49 kg

Mass of Tom = 77 kg

Distance = 0.4 m

Time = 0.8 s

We need to calculate Tracy’s constant acceleration during her time of contact with Tom

Using equation of motion

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

Where, s = distance

a = acceleration

t = time

Put the value into the formula

[tex]0.4=0+\dfrac{1}{2}a\times(0.8)^2[/tex]

[tex]a=\dfrac{0.4\times2}{(0.8)^2}[/tex]

[tex]a=1.25\ m/s^2[/tex]

Hence, Tracy’s constant acceleration during her time of contact with Tom is 1.25 m/s².

Tracy’s constant acceleration during her time of contact with Tom is 1.25 m/s².

Here we applied the motion equation i.e. shown below:

[tex]S = ut + \frac{1}{2}at^2[/tex]

Here s = distance

a = acceleration

t = time

So,

[tex]0.4 = \frac{1}{2} a \times (0.8)^2\\\\a = \frac{0.4\times 2}{(0.8)^2}[/tex]

=  1.25 m/s²

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Answer:

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m,   m2 = 4m,    v1=vf_p,    v2 = vf_alpha

The conservation momentum states that:

pi = pf      

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

      m1 =  m, m2 = 4m, q1 = e, q2 = 2e

      and   v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 =  k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

Answer:

It is necessary to know the radius of the planet.  

Explanation:

The speed of the spacecraft can be found by means of the equation of the Universal law of gravity:

[tex]F = G \frac{M.m}{r^{2}}[/tex]  (1)

Where F is the gravitational force, G is gravitational constant, M is the mass of the planet, m is the mass of the spacecraft and r is the orbital radius of the spacecraft.

Equation 1 can be express in terms of the speed by using Newton's second law and the equation for centripetal acceleration:

[tex]F = ma[/tex]  (2)

Replacing equation 2 in equation 1 it is gotten:

[tex]ma = G \frac{M.m}{r^{2}}[/tex] (3)

the centripetal acceleration is defined as:

[tex]a = \frac{v^{2}}{r}[/tex]  (4)

Replacing equation 4 in equation 3 it is gotten:

[tex]m\frac{v^{2}}{r} = G \frac{M.m}{r^{2}}[/tex] (5)

Then, v can be isolated from equation 5:

[tex]mv^{2} = G \frac{M.m}{r}[/tex]

[tex]v^{2} = G \frac{M.m}{rm}[/tex]

[tex]v^{2} = G \frac{M}{r}[/tex]

[tex]v = \sqrt{\frac{GM}{r}}[/tex]

However, the orbital radius of the spacecraft is obtained by the sum of the radius of the planet and the height of the spacecraft above the surface of the planet (r = R+h)

[tex]v = \sqrt{\frac{GM}{R+h}}[/tex]  (6)

Hence, by equation 6 it can be concluded that it is necessary to know the radius of the planet in order to calculate the speed of the spacecraft.

Answer:

α₂= 0.1  rad/s²

t= 62.8 s

Explanation:

Given that

For small wheel

r₁= 1.2 cm

α₁ = 3 rad/s²

For large wheel

r₂= 36 cm

Angular acceleration = α₂  rad/s²

The tangential acceleration for the both wheel will be same

a = α₁ r₁=α₂ r₂

Now by putting the values in the above equation

α₁ r₁=α₂ r₂

3 x 1.2 = 36 x α₂

α₂= 0.1  rad/s²

Given that

N = 60 rpm

Angular speed in rad/s ω

[tex]\omega = \dfrac{2\pi N}{60}[/tex]

[tex]\omega = \dfrac{2\pi \times 60}{60}[/tex]

ω = 6.28 rad/s

Time taken is t

ω = α₂ t

6.28 = 0.1 t

t= 62.8 s

The angular acceleration of the pottery wheel is 0.1 rad/s². It will rotate at 60 rpm after approximately 63 seconds.

The concept involved in the question relates to the principle of angular momentum conservation. The angular acceleration of the pottery wheel can be calculated from the given angular acceleration of the small wheel and the ratio of their radii. The equation that connects linear acceleration (a), angular acceleration (alpha), and radius (r) is a = alpha*r. Given that the small wheel has an angular acceleration, alpha1=3 rad/s² and radius r1=1.2 cm, while the pottery wheel has radius r2=36 cm, the linear acceleration of both is the same. Hence, the angular acceleration of the pottery wheel, alpha2 = a/r2 = (alpha1*r1)/r2 = (3 * 1.2 cm)/(36 cm) = 0.1 rad/s².

Now, for how long until the pottery wheel rotates at 60 rpm, first convert 60 rpm to rad/s. Note that 1 rev = 2π rad, and 1 min = 60 s. So, 60 rev/min = (60 * 2π rad)/(60 s) = 2π rad/s = final angular velocity, omega.

Then, use the formula for time in angular motion having constant angular acceleration: t = (omega - omega_initial) / alpha, and since the pottery wheel starts from rest, omega_initial = 0. Hence, t = omega/alpha2 = (2π rad/s)/(0.1 rad/s²) = 20π s, or approximately 63 seconds.

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Answer:

3. One Newton is the force that gives a 1 kg body an acceleration of 1 m/s².

Explanation:

The force is a vector magnitude that represents every cause capable of modifying the state of motion or rest of a body or of producing a deformation in it.

Its unit in the International System is the Newton (N). A Newton is the force that when applied on a mass of 1 Kg causes an acceleration of 1 m/s².

1 N = kg*(m/s²)

Answer:

a) [tex]r_{cm} = 4672 km[/tex]

b)r_cm = 0.733 R_{earth}

Explanation:

Location of center of mass of Earth- moon system is

[tex]r_{cm}= \frac{M_{earth}r_{earth}+M_{moon}r_{moon}}{M_{moon}+M{earth}}[/tex]

[tex]r_{cm}= \frac{6\times10^{24}(0)+7.34\times10^{22}384000\times1000}{6\times10^{24}+7.34\times10^{22}}[/tex]

=4.672×10^(6) m

a) [tex]r_{cm} = 4672 km[/tex]

b) [tex]\frac{r_{cm}}{R{earth}} = \frac{4672}{6371} = 0.733[/tex]

therefore

r_cm = 0.733 R_{earth}

The center of mass of the Earth-Moon system is the point around which the two bodies orbit and it is located within Earth but not at the center. The distance from the Earth's center to this point can be calculated using the masses of the Moon and Earth, and the distance between them. This distance expressed as a fraction of the Earth's radius is approximately 0.73.

The center of mass of the Earth-Moon system is the point about which the two bodies orbit, and it's located inside the Earth, not at the Earth's center. The distance from the Earth's center to the system's center of mass can be calculated using the formula: d = D*(m/(M+m)), where D is the distance between the Earth and the Moon, m is the Moon's mass, M is the Earth's mass.

To express the previous value as a fraction of the Earth's radius, you must divide it by the radius of the Earth. For instance, let's assume the distance turned out to be 4671 km and the Earth's radius is 6371 km. The resulting fraction would be 4671/6371 which simplifies to approximately 0.73.

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Answer:

4.86 seconds

Explanation:

Velocity of projection, u = 14 m/s

angle of projection, θ = 20°

Formula for the time of flight

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

For earth

Te = (2 x 14 x Sin 20) / 9.8

Te = 0.98 s

For moon

g' = g/6 = 1.64 m/s^2

Tm = ( 2 x 14 x Sin 20) / 1.64

Tm = 5.84 seconds

Tm - Te = 5.84 - 0.98 = 4.86 s

So, it takes 4.86 s more time of flight on moon than the earth.  

The golf ball hit by Alan Shepard was in flight approximately 4.87 seconds longer on the moon than on Earth.

To analyze the flight duration of a projectile under different gravitational conditions.

1. Initial Parameters:

Initial velocity of the golf ball, [tex]v_0 = 14 \, \text{m/s}[/tex]

Launch angle, [tex]\theta = 20^{\circ}[/tex]

Acceleration due to gravity on Earth, [tex]g = 9.81 \, \text{m/s}^2[/tex]

Acceleration due to gravity on the moon, [tex]g_{moon} = \frac{g}{6} = 1.635 \, \text{m/s}^2[/tex]

2. Determine the Vertical Component of Initial Velocity:
The vertical component of the initial velocity can be calculated using the formula:
[tex]v_{0y} = v_0 \sin(\theta)[/tex]

Substituting the known values:
[tex]v_{0y} = 14 \, \text{m/s} \cdot \sin(20^{\circ}) \approx 4.78 \, \text{m/s}[/tex]

3. Calculate Time of Flight on Earth:
The time of flight [tex]T[/tex] for a projectile launched and landing on a flat surface can be calculated using the formula:
[tex]T = \frac{2 v_{0y}}{g}[/tex]

For Earth:
[tex]T_{Earth} = \frac{2 \cdot 4.78 \, \text{m/s}}{9.81 \, \text{m/s}^2} \approx 0.975 \, \text{s}[/tex]

4. Calculate Time of Flight on the Moon:
Using the same formula but with the moon's gravity:
[tex]T_{Moon} = \frac{2 v_{0y}}{g_{moon}}[/tex]

Substituting the values:
[tex]T_{Moon} = \frac{2 \cdot 4.78 \, \text{m/s}}{1.635 \, \text{m/s}^2} \approx 5.84 \, \text{s}[/tex]

5. Calculate the Difference in Flight Time:
Finally, to find how much more time the ball was in flight on the moon compared to Earth:
[tex]\Delta T = T_{Moon} - T_{Earth}[/tex]
[tex]\Delta T = 5.84 \, \text{s} - 0.975 \, \text{s} \approx 4.87 \, \text{s}[/tex]

Answer:

178378378.37 Pa

0.01897

Explanation:

F = Force = [tex]6.6\times 10^4\ N[/tex]

A = Area = [tex]3.7\times 10^{-4}\ m^2[/tex]

Y = Young's modulus of bone under compression = [tex]9.4\times 10^{9}\ Pa[/tex]

[tex]\varepsilon[/tex] = Strain

Stress is given by

[tex]\sigma=\frac{F}{A}\\\Rightarrow \sigma=\frac{6.6\times 10^4}{3.7\times 10^{-4}}\\\Rightarrow \sigma=178378378.37\ Pa[/tex]

The maximum stress that this bone can withstand is 178378378.37 Pa

Compression force is given by

[tex]F=Y\varepsilon A\\\Rightarrow \varepsilon=\frac{F}{YA}\\\Rightarrow \varepsilon=\frac{6.6\times 10^4}{9.4\times 10^{9}\times 3.7\times 10^{-4}} \\\Rightarrow \varepsilon=0.01897[/tex]

The strain that exists under a maximum-stress condition is 0.01897

The maximum stress that the femur can withstand is 1.78 * 10^8 Pascals, and the strain under this maximum stress condition is approximately 0.012 or 1.2%.

The subject of the question is the stress and strain on the femur bone, which is integral to the field of Physics. To address part (a) of the question, we must find the maximum stress that the femur can withstand. Stress is defined as the average force per unit area, or the force divided by the area. In this case, we are given a compressional force of 6.60 * 10^4 Newtons and a cross-sectional area of the femur as 3.70 * 10^-4 m2. We divide the force by the area to find the stress:

Stress = Force / Area = (6.60 * 10^4 N) / (3.70 * 10^-4 m2) = 1.78 * 10^8 Pa (Pascals).

For part (b) of the question, the strain under a maximum-stress condition can be calculated by dividing the stress by Young's modulus. For bone, the modulus can vary, and bones are brittle with the elastic region small. However, if we take an average value, say 1.5 * 10^10 Pa, then, the strain will be:

Strain = Stress / Young's modulus = (1.78 * 10^8 Pa) / (1.5 * 10^10 Pa) = 0.012.

So under maximum stress, the strain is 0.012 (dimensionless) or 1.2%. Remember, this is an approximation as the actual modulus can vary based on numerous factors such as age, diet and lifestyle of the individual.

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Answer:

The mass of mars is [tex]6.61\times10^{23}\ kg[/tex]

Explanation:

Given that,

Time = 1.02 s

Height = 2.00 m

Radius of mars [tex]r= 3.39\times10^{6}\ m[/tex]

We need to calculate the acceleration due to gravity on mars

Using equation of motion

[tex]h = ut+\dfrac{1}{2}gt^2[/tex]

[tex]h = \dfrac{1}{2}gt^2[/tex]

[tex]g=\dfrac{2h}{t^2}[/tex]

Where, u = initial velocity

t = time

h = height

Put the value into the formula

[tex]g=\dfrac{2\times2.00}{(1.02)^2}[/tex]

[tex]g=3.84\ m/s^2[/tex]

We need to calculate the mass of mars

Using formula of gravity

[tex]g=\dfrac{Gm}{r^2}[/tex]

Put the value into the formula

[tex]3.84=\dfrac{6.67\times10^{-11}\times m}{(3.39\times10^{6})^2}[/tex]

[tex]m=\dfrac{3.84\times(3.39\times10^{6})^2}{6.67\times10^{-11}}[/tex]

[tex]m=6.61\times10^{23}\ kg[/tex]

Hence, The mass of mars is [tex]6.61\times10^{23}\ kg[/tex]

In order to find the mass of Mars, we can use a physics formula for gravity. The given values, such as the drop distance, time it took, the radius of Mars, and the universal gravitational constant, can be substituted into this formula. This, then, gives a theoretical value for the mass of Mars.

To calculate the mass of Mars, we use the formula to find gravity, which is defined as g = GM/r², where G is the universal gravitational constant. We can set this equal to something else: 2d / t², where d is the drop distance and t is the time it took for the marble to drop. We then can isolate M (the mass of Mars) on one side.

By substituting the given values into the formula, we get: M = (g * r²) / G = ((2 * 2.00 m) / (1.02 s)² * (3.39 × 10⁶ m)²) / (6.67 x 10⁻¹¹ N·m²/kg²).

This Netwonian calculation gives the mass of Mars, theoretically, based on the observed falling time of the marble.

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Answer:

Explanation:

Given

mass of balls [tex]m= 5 kg[/tex]

[tex]N=45.6 rev/s[/tex]

angular velocity [tex]\omega =2\pi N=286.55 rad/s[/tex]

Length of Rod [tex]2L=1.1 m[/tex]

Tension in the Second half of rod

[tex]T_2=m\omega ^2(2L)=2m\omega ^2L[/tex]

[tex]T_2=5\times (286.55)^2\times 1.1[/tex]

[tex]T_2=451.609 kN[/tex]

For First Part

[tex]T_1-T_2=m\omega ^2L[/tex]

[tex]T_1=T_2+m\omega ^2L[/tex]

[tex]T_1=3 m\omega ^2L[/tex]

[tex]T_1=3\times 5\times (286.55)^2\times 0.55[/tex]

[tex]T_1=677.41 kN[/tex]

Final answer:

The question deals with the tension in a rotating rod with attached masses. Tension in the first half of the rod is dictated by the centripetal force for one mass, while the second half must account for two masses. It's a problem in the field of Physics, specifically rotational dynamics at the college level.

Explanation:

The student's question revolves around the concepts of rotational motion and tension in a system consisting of a massless rod with balls of equal mass attached at different points. Given the angular speed and the positions of the masses, we are to find the tension in two parts of the rod during rotation.

The tension for the first half of the rod can be calculated by considering the centripetal force required to keep the ball rotating in a circle of radius L, the middle of the rod. Since the ball at L is the only mass in the first segment, we only need to consider its centripetal force requirement.

For the second segment of the rod, from L to 2L, the tension must accommodate the centripetal force for both masses, one at the middle and the other at the end. The ball at the end experiences more tension because it is further from the pivot point and thus has a larger radius of rotation.

Note that actual equations and calculations are not provided here, as the question seems to request conceptual explanations rather than specific numerical solutions.

Answer:

d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero

Explanation:

Affirmations

a) true. The orbits are accurate in the Bohr model and probabilistic in quantum mechanics

b) True. If both give the same results and use the same quantum number (n)

c) True. If in angular momentum it is quantized, in the Bohr model too but it does not justify it

d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero

Answer:

The acceleration is 3.62 m/s²

Explanation:

Step 1: Data given

mass of the shell = 1.65 kg

angle = 38.0 °

Step 2: Calculate the acceleration

We have 2 forces working on the line of motion:

⇒ gravity down the slope = m*g*sinα

       ⇒ provides the linear acceleration

⇒ friction up the slope = F

      ⇒ provides the linear acceleration and also the torque about the CoM.

∑F = m*a = m*g*sin(α) - F

I*dω/dt = F*R

The spherical shell with mass m has moment of inertia I=2/3*m*R² Furthermore a pure rolling relates  dω/dt and a through a = R dω/dt. So the two equations become

m*a = m*g sin(α) - F

2/3*m*a = F  

IF we combine both:

m*a = m*g*sin(α) - 2/3*m*a

1.65a = 1.65*9.81 * sin(38.0) - 2/3 *1.65a

1.65a + 1.1a = 9.9654

2.75a = 9.9654

a = 3.62 m/s²

The acceleration is 3.62 m/s²

The magnitude of the acceleration of the center of mass of the spherical shell is 3.62 m/s².

Newton’s second law of motion shows the relation between the force mass and acceleration of a body. It says, that the net force applied on the body is equal to the product of mass of the body and the acceleration of it.

It can be given as,

[tex]\sum F=ma[/tex]

Here, (m) is the mass of the body and (a) is the acceleration.

A hollow spherical shell with mass 1.65 kg rolls without slipping down a slope that makes an angle of 38.0 ∘ with the horizontal. The free-fall acceleration g is 9.80 m/s2 .

The total friction force is equal to the force of gravity acting downward of the slope.

[tex]\sum F=mg\sin(\theta)-F\\[/tex]

For the force acting on the rotating spherical shell is,

[tex]F=\dfrac{2}{3}ma[/tex]

Put this value in above equation,

[tex]\sum F=(1.65)(9.80)\sin(38)-\dfrac{2}{3}(1.65)a\\ma=(1.65)(9.80)\sin(38)-\dfrac{2}{3}(1.65)a\\(1.65)a=(1.65)(9.80)\sin(38)-\dfrac{2}{3}(1.65)a\\a=(9.80)\sin(38)-\dfrac{2}{3}a\\a=3.62\rm \; m/s^2[/tex]

Thus, the magnitude of the acceleration of the center of mass of the spherical shell is 3.62 m/s².

Learn more about the Newton’s second law of motion here;

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Final answer:

The ratio of relativistic kinetic energy to nonrelativistic kinetic energy depends on the speed of the particle. For speeds much less than the speed of light, the ratio is close to 1, but for speeds approaching the speed of light, the relativistic effect becomes significant and the ratio increases.

Explanation:

The ratio of relativistic kinetic energy to nonrelativistic kinetic energy can be calculated using the formula for relativistic kinetic energy KErel = mc2((1/(1 - v2/c2)1/2) - 1), where m is the particle mass, v is its velocity, and c is the speed of light. The nonrelativistic kinetic energy is given by KEnr = (1/2)mv2. Firstly, to determine the ratios at given speeds, we calculate the relativistic factor γ, which is (1/(1 - v2/c2)1/2), for each speed

For part (a) with v = 2.71 x 10-3c, the relativistic factor is close to 1, indicating that the speeds are nonrelativistic, and the ratio KErel/KEnr would be close to 1. For part (b), at v = 0.855c, the relativistic factor is significantly greater than 1, and the relativistic kinetic energy will be much larger than the classical value, yielding a ratio much greater than 1.

Answer:

The mass of [tex]_{53}^{131}I[/tex] is [tex]8.09\times10^{-9}\ g[/tex].

Explanation:

Given that,

Half life [tex]t_{\frac{1}{2}}=8.08\ days[/tex]

Sample emitting radiation = 1.00 mCi = [tex]3.7\times10^{7}\ dps[/tex]

We need to calculate the rate constant

Using formula of rate constant

[tex]\lambda=\dfrac{0.693}{t_{\frac{1}{2}}}[/tex]

[tex]\lambda=\dfrac{0.693}{8.08\times24\times60\times60}[/tex]

[tex]\lambda=9.92\times10^{-7}\ s^{-1}[/tex]

We need to calculate the numbers of atoms

Using formula of numbers of atoms

[tex]N_{0}=\dfrac{N}{\lambda}[/tex]

[tex]N_{0} =\dfrac{3.7\times10^{7}}{9.92\times10^{-7}}[/tex]

[tex]N_{0}=3.72\times10^{13}\ atoms[/tex]

We need to calculate the mass of [tex]_{53}^{131}I[/tex]

Using formula for mass

[tex]m=\dfrac{131\times3.72\times10^{13}}{6.023\times10^{23}}[/tex]

[tex]m=8.09\times10^{-9}\ g[/tex]

Hence, The mass of [tex]_{53}^{131}I[/tex] is [tex]8.09\times10^{-9}\ g[/tex].

The problem is related to the concept of radioactive decay, specifically of the isotope Iodine-131. The activity of the sample is calculated to be 37,000,000 decays/second. Through a series of calculations using formulas for decay constant, the number of atoms and finally the sample mass, we find the mass to be 9.3x10^-16 g/day.

The problem given relates to nuclear physics and utilizes the concept of radioactive decay. It specifically involves the isotope Iodine-131 (131 53I), which decays with a half-life of 8.08 days.

Assuming the activity of the sample is 1 mCi, which is equivalent to 37,000,000 decays/second (since 1 Ci=3.7x10^10 decays/sec, 1 mCi = 1/1000 Ci), we can use the activity formula which is given by A = λN, where A is activity, λ is the decay constant, and N is the number of atoms in the sample. We first need to calculate λ, which we can find using the formula λ = Ln(2)/T(1/2), where T(1/2) is the half-life.

λ = Ln(2)/8.08 days = 0.086/day. Therefore, N = A/ λ = 37,000,000 / 0.086 = 4.3x10^8 atoms/day.

The remaining calculation is the mass of the sample. The atomic mass of Iodine-131 is approximately 131 g/mol. Using Avogadro's number (6.02x10^23 atoms/mol), we can calculate the mass: M = N * (131 g/mol) / (6.02x10^23 atoms/mol) = 9.3x10^-16 g/day.

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The work done to compress an ideal gas in a cylinder to 1/6 of its initial volume at constant temperature is -78.3 Joules.

This question involves the principle of work done in compressing a gas, a core concept in thermodynamics. The work done on an ideal gas at constant temperature (isothermal process) is given by the formula W = PΔV, where W is the work done, P is the atmospheric pressure, and ΔV is the change in volume.

Given that the initial volume (V1) is 9.33 * 10^-4 m³ and the final volume (V2) is 1/6 of the initial volume, so V2 = V1/6. Thus, the change in volume ΔV = V2 - V1 = V1/6 - V1 = -5V1/6.

Substituting the values into the formula and solving, we get W = PΔV = (1.013 * 10^5 Pa)(-5V1/6) = -5/6 * 1.013 * 10^5 Pa * 9.33 * 10^-4 m³ = -78.3 Joules.

Note that the work done is negative, which in this context means that work is done on the system (gas), not by it.

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Answer:

  [tex]v_{f}[/tex] = 3,126 m / s

Explanation:

In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.

the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s

Before the crash

    p₀ = m v₁₀ + M v₂₀

After the inelastic shock

    [tex]p_{f}[/tex]= m [tex]v_{1f}[/tex] + M [tex]v_{2f}[/tex]

    p₀ =  [tex]p_{f}[/tex]

    m v₀ + M v₂₀ = m [tex]v_{1f}[/tex]  + M [tex]v_{2f}[/tex]

We cleared the end of the train

     M [tex]v_{2f}[/tex] = m (v₁₀ - v1f) + M v₂₀

Let's calculate

     3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45

     [tex]v_{2f}[/tex]  = (-3.9 + 8.82) /3.60

      [tex]v_{2f}[/tex]  = 1.36 m / s

As we can see, this speed is lower than the speed of the car, so the two bodies are joined

set speed must be

      m v₁₀ + M v₂₀ = (m + M) [tex]v_{f}[/tex]

      [tex]v_{f}[/tex]  = (m v₁₀ + M v₂₀) / (m + M)

      [tex]v_{f}[/tex]  = (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)

      [tex]v_{f}[/tex] = 3,126 m / s

To solve this problem it is necessary to apply the kinematic equations of description of the movement in which it is understood that the velocity is the travel of a particle in a fraction of time, that is to say

[tex]v = \frac{x}{t}[/tex]

Where,

x = Displacement

t = time

In our case the speed is equivalent to that of the Light, and the distance is necessary to reach the moon by the asteroid.

[tex]v = 3.8*10^8m/s[/tex]

[tex]x = 3.84*10^5km = 3.84*10^8m[/tex]

Re-arrange to find t,

[tex]t = \frac{x}{V}[/tex]

[tex]t = \frac{3.84*10^8}{3.8*10^8}[/tex]

[tex]t = 1.28s[/tex]

Therefore will take 1.28 s for the light arrive on Earth.

The light from the asteroid collision on the Moon would take approximately 1.28 seconds to reach Earth.

The speed of light in a vacuum is approximately 3 x 10^8 meters per second. Since the Moon is 3.84 x 10^5 km away from Earth, we can calculate the time it takes for light to travel from the Moon to Earth using the formula:

Time = Distance / Speed

Converting the distance from km to meters, we get 3.84 x 10^8 meters. Plugging in the values:

Time = (3.84 x 10^8 meters) / (3 x 10^8 meters per second) = 1.28 seconds

So, it would take approximately 1.28 seconds for the light from the asteroid collision on the Moon to reach Earth.

Answer:

The average gauge pressure inside the vein is 110270.58 Pa

Explanation:

This question can be solved using the Bernoulli's Equation. First, in order to determine the outlet pressure of the needle, we need to find the total pressure exerted by the atmosphere and the fluid.

[tex]P_f: fluid's\ pressure\\P_f= \rho g h=1025\frac{kg}{m^3} \times 9.8 \frac{m}{s^2} \times 0.9 m=9040.5 Pa \\P_T: total\ pressure\\P_T=P_{atm}+P_f\\P_T=101325 Pa + 9040.5 Pa=110275.5 Pa\\[/tex]

Then, we have to find the fluid's outlet velocity with the transversal area of the needle, as follows:

[tex]S: transversal\ area \\S= \pi r^2=\pi (0.200 \times 10^{-3})^2=5.65 \times 10^{-7} m^2\\v=\frac{F}{S}=\frac{5.55 \times 10^{-8} \frac{m^3}{s}}{5.65 \times 10^{-7} m^2}=0.98\times 10^{-1} \frac{m}{s}[/tex]

As we have all the information, we can complete the Bernoulli's expression and solve to find the outlet pressure as follows:

[tex]P_T-P_{out}=\frac{1}{2} \rho v^2\\P_{out}=P_T-\frac{1}{2} \rho v^2=110275.5 Pa-\frac{1}{2} 1025\frac{kg}{m^3} (0.98\times 10^{-1} \frac{m}{s})^2=110275.5 Pa-4.92 Pa =110270.58 Pa[/tex]

Answer:

[tex]v=2.28m/s[/tex]

Explanation:

For the first mass we have [tex]m_1=3.5kg[/tex], and for the second [tex]m_2=6.0kg[/tex]. The pulley has a moment of intertia [tex]I_p=0.0352kgm^2[/tex] and a radius [tex]r_p=0.125m[/tex].

We solve this with conservation of energy.  The initial and final states in this case, where no mechanical energy is lost, must comply that:

[tex]K_i+U_i=K_f+U_f[/tex]

Where K is the kinetic energy and U the gravitational potential energy.

We can write this as:

[tex]K_f+U_f-(K_i+U_i)=(K_f-K_i)+(U_f-U_i)=0J[/tex]

Initially we depart from rest so [tex]K_i=0J[/tex], while in the final state we will have both masses moving at velocity v and the tangential velocity of the pully will be also v since it's all connected by the string, so we have:

[tex]K_f=\frac{m_1v^2}{2}+\frac{m_2v^2}{2}+\frac{I_p\omega_p^2}{2}=(m_1+m_2+\frac{I_p}{r_p^2})\frac{v^2}{2}[/tex]

where we have used the rotational kinetic energy formula and that [tex]v=r\omega[/tex]

For the gravitational potential energy part we will have:

[tex]U_f-U_i=m_1gh_{1f}+m_2gh_{2f}-(m_1gh_{1i}+m_2gh_{2i})=m_1g(h_{1f}-h_{1i})+m_2g(h_{2f}-h_{2i})[/tex]

We don't know the final and initial heights of the masses, but since the heavier, [tex]m_2[/tex], will go down and the lighter, [tex]m_1[/tex], up, both by the same magnitude h=1.25m (since they are connected) we know that [tex]h_{1f}-h_{1i}=h[/tex] and [tex]h_{2f}-h_{2i}=-h[/tex], so we can write:

[tex]U_f-U_i=m_1gh-m_2gh=gh(m_1-m_2)[/tex]

Putting all together we have:

[tex](K_f-K_i)+(U_f-U_i)=(m_1+m_2+\frac{I_p}{r_p^2})\frac{v^2}{2}+gh(m_1-m_2)=0J[/tex]

Which means:

[tex]v=\sqrt{\frac{2gh(m_2-m_1)}{m_1+m_2+\frac{I_p}{r_p^2}}}=\sqrt{\frac{2(9.8m/s^2)(1.25m)(6.0kg-3.5kg)}{3.5kg+6.0kg+\frac{0.0352kgm^2}{(0.125m)^2}}}=2.28m/s[/tex]

Final answer:

To solve this problem, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system (kinetic energy + potential energy) remains constant if no external forces are acting on the system.

Explanation:

Let's denote the initial position of the masses as position 1 and the final position as position 2. At position 1, the 3.5 kg mass is at a height h above the ground, and the 6.0 kg mass is at a height 1.25 m - h above the ground. At position 2, both masses are at a height 1.25 m above the ground.

The potential energy (PE) of the system at position 1 is:

PE1 = m1 * g * h + m2 * g * (1.25 m - h)

The potential energy (PE) of the system at position 2 is:

PE2 = (m1 + m2) * g * 1.25 m

Since the system is released from rest, the initial kinetic energy (KE) of the system is zero. The final kinetic energy (KE) of the system is:

KE2 = (1/2) * (m1 + m2) * v^2

where v is the speed of the masses after they have moved through 1.25 m.

According to the principle of conservation of mechanical energy, the total mechanical energy at position 1 is equal to the total mechanical energy at position 2:

PE1 + KE1 = PE2 + KE2

Substituting the expressions for PE1, PE2, and KE2, we get:

m1 * g * h + m2 * g * (1.25 m - h) = (m1 + m2) * g * 1.25 m + (1/2) * (m1 + m2) * v^2

Solving for v^2, we get:

v^2 = 2 * g * (m1 * h + m2 * (1.25 m - h) - (m1 + m2) * 1.25 m)

Substituting the given values for g, m1, m2, and h, we get:

v^2 = 2 * 9.8 m/s^2 * (3.5 kg * 1.25 m + 6.0 kg * (1.25 m - 1.25 m) - (3.5 kg + 6.0 kg) * 1.25 m)

Solving for v, we get:

v ≈ 2.3 m/s

So, the speed of the masses after they have moved through 1.25 m is approximately 2.3 m/s.

The radius R of a geosynchronous satellite's orbit is 42.2 x 10⁶ meters, or approximately 6.63 times the radius of the Earth.

The radius R of the orbit of a geosynchronous satellite that circles the Earth can be computed using the principles of circular motion and the law of universal gravitation, considering a geostationary satellite takes 23 hours 56 minutes and 4 seconds to complete one orbit. The gravitational force provides the necessary centripetal force to keep the satellite in orbit. Given the mass of the Earth (me = 5.98 x 10²⁴kg), the mean radius of the earth (R2 = 6.37 x 10⁶ m), and the universal gravitational constant (G = 6.67 x 10⁻¹¹ N m²kg⁻²), the radius R is already provided as 42.2 x 10⁶ m. This is the distance from the center of the Earth to the satellite. When expressed in terms of Earth's radii, this is approximately 6.63 Earth radii, since the radius of the Earth is 6.37 x 10⁶ m.

Answer:

  y = 2.74 m

Explanation:

The linear thermal expansion processes are described by the expression

         ΔL = α L ΔT

Where α the thermal dilation constant for concrete is 12 10⁻⁶ºC⁻¹, ΔL is the length variation and ΔT the temperature variation in this case 20ªc

If the bridge is 250 m long and is covered by two sections each of them must be L = 125 m, let's calculate the variation in length

        ΔL = 12 10⁻⁶ 125 20

        ΔL = 3.0 10⁻² m

Let's use trigonometry to find the height

The hypotenuse     Lf = 125 + 0.03 = 125.03 m

Adjacent leg           L₀ = 125 m

       cos θ = L₀ / Lf

       θ = cos⁻¹ (L₀ / Lf)

       θ = cos⁻¹ (125 / 125.03)

       θ = 1,255º

We calculate the height

       tan 1,255 = y / x

       y = x tan 1,255

       y = 125 tan 1,255

       y = 2.74 m

When a bridge is subjected to a temperature increase, the concrete spans rise or buckle due to thermal expansion. The height to which they rise can be calculated using the formula for change in length. In this case, the height is 0.42 m.

When a bridge is subjected to a temperature increase, it expands due to thermal expansion. In this case, the two concrete spans of the bridge are placed end to end without allowing any room for expansion. Therefore, when the temperature increases by 20.0°C, the spans rise or buckle. The height to which they rise can be calculated using the formula:

change in length = coefficient of linear expansion x original length x change in temperature

In this case, the change in length is 0.84 m. To calculate the height y to which the spans rise, divide this change in length by the number of spans, which is 2. Therefore, the height is 0.42 m.

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