The weight of the mass added to the hanger is equal to the extra force on the gas, but what area should we use to calculate the added pressure from this mass?

With some of the experimental details out of the way, let's think a bit about what we expect to see in our data if the ideal gas law is a good model for our gas. We would like to verify the ideal gas law PV = nRT

Final answer:

In an inelastic collision where two objects of equal mass and speed but opposite directions collide, they will stick together and remain stationary. The correct option is c.

Explanation:

If two objects of the same mass are colliding at the same speed but in opposite directions, in an inelastic collision, the situation that will happen is (c) the objects will collide and stay stationary. This is because for a perfectly inelastic collision, the two objects stick together and their combined momentum is zero since they have equal mass and speed but opposite directions. The conservation of momentum dictates that because the initial momenta of the objects cancel each other out, the final momentum also has to be zero, thus the objects remain stationary after the collision.

Answer:

0.69s

Explanation:

10 cm = 0.1 m

Let t be the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude. At that time we have the angular velocity would be

[tex]\omega = \alpha t = 2.1 t[/tex]

And so the radial acceleration is

[tex]a_r = \omega^2 r = (2.1t)^2 r = 2.1^2 t^2 * 0.1= 0.441 t^2 m/s^2[/tex]

The tangential acceleration is always the same since angular acceleration is constant:

[tex]a_t = \alpha * r = 2.1 * 0.1 = 0.21 m/s^2[/tex]

For these 2 quantities to be the same

[tex]a_r = a_t[/tex]

[tex]0.441 t^2 = 0.21[/tex]

[tex]t^2 = 0.21/0.441 = 0.4762[/tex]

[tex]t = \sqrt{0.4762} = 0.69 s[/tex]

The value of t whereby the radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude is; t = 0.69 s

Let the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude be denoted as t. Thus, angular velocity at time (t) is;

ω = αt

where;

α is angular acceleration

We are given;

Thus, ω = 2.1t

Now, we can find the radial acceleration from the formula;

α_r = ω²r

Thus;

α_r =  (2.1t)² × 0.1

α_r = 0.441 t² m/s²

The tangential acceleration is gotten from the formula is;

α_t = α × r

α_t = 2.1 × 0.1

α_t = 0.21 m/s²

The condition in the question implies that the tangential acceleration is equal to the radial acceleration. Thus;

α_t = α_r

0.21 = 0.441 t²

t = √(0.21/0.441)

t = 0.69 s

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Answer:

Mass of asteroid will be [tex]M=3\times 10^{10}kg[/tex]

Explanation:

We have given diameter of the asteroid d = 250 miles

So radius [tex]R=\frac{250}{2}=125miles=125\times 1609.34=201167.5m[/tex]

Gravitational field strength [tex]g=4.95\times 10^{-11}m/sec^2[/tex]

Gravitational constant [tex]G=6.67\times 10^{-11}Nm^2/kg^2[/tex]

We know that [tex]g=\frac{GM}{R^2}[/tex], here M is the mass of asteroid

So [tex]4.95\times 10^{-11}=\frac{6.67\times 10^{-11}\times M}{201167.5^2}[/tex]

[tex]M=3\times 10^{10}kg[/tex]

To determine the mass of the asteroid, the radius is calculated from the diameter and then the formula for gravitational field strength is used, rearranging it to solve for mass. After substituting the gravitational field strength, radius, and gravitational constant into this formula, the asteroid's mass is found to be approximately 5.91 x 10^15 kg.

To find the mass, we will use the formula for gravitational field strength at the surface of a spherical body, which is given as g = GM/r², where G is the gravitational constant (6.67 x 10^-11 Nm²/kg²), M is the mass, and r is the radius of the body. We are given g (the gravitational field strength) as 4.95 x 10^-11 N/kg and the diameter of the asteroid as 250 miles (but we need the radius in meters, to match the units of G).

First, convert 250 miles to kilometers (1 mile = 1.60934 km) and then to meters. Then, divide this number by 2 to get the radius, which is about 201168 meters.

Next, rearrange the equation to solve for M: M = gr²/G. Substitute the given values into the equation: M = (4.95 x 10^-11 N/kg * (201168 m)²) / 6.67 x 10^-11 Nm²/kg². On doing the calculation, the asteroid's mass is approximately 5.91 x 10^15 kg.

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Explanation:

Using Newtons second law on each block

F = m*a

Block 1

[tex]T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1[/tex]

Block 2

[tex]T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2[/tex]

Block 3

[tex]- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3[/tex]

Solving Eq1,2,3 simultaneously

Divide 1 and 2

[tex]\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)} \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} = \frac{M_{1} *T_{2} }{M_{2} } .... Eq4[/tex]

Put Eq 4 into Eq3

[tex]T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } } ...Eq5[/tex]

Put Eq 5 into Eq2 and solve for a

[tex]a = \frac{M_{3}*g -u*g*(M_{1} + M_{2}) }{M_{1} + M_{2} + M_{3} } .... Eq6[/tex]

Substitute back in Eq2 and use Eq4 and solve for T2 & T1

[tex]T_{2} = M_{2}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\T_{1} = M_{1}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\[/tex]

Answer:

c. -14,000

Explanation:

Workdone by gas is the product of the pressure and the volume where there is a change of volume.

If v1 is the initial volume of the gas and v2 is the final volume of gas, the work done

= p(v2 - v1)

where p is the pressure

and p = 70,000 Pa

Given that volume decrease by 0.2m^3, v2 - v1 = -0.2

Workdone = 70000 ( -0.2)

Workdone = -14,000 J

Option c. -14,000

Answer:

Q₁ = Q₂ = 8.84 x 10⁻⁹ C

Explanation:

given,

mass of ball, m = 0.16 g = 1.6 x 10⁻⁴ Kg

ball each other, r = 6.8 cm

Weight of the ball

F_w = m g

F_w = 1.6 x 10⁻⁴ x 9.8

F_w = 1.56 x 10⁻³ N

The tension in each string is a force directed along the length of the string and is the hypotenuse of a right triangle.

we have to find the horizontal component of the forces.

The length of the string,L is 35 cm so, it will be the hypotenuse.

θ be the angle made with imaginary vertical line and the string.

now,

[tex]sin \theta = \dfrac{r\2}{L}[/tex]

[tex]sin \theta = \dfrac{3.4}{35}[/tex]

   θ = 5.57°

horizontal component of the force = ?

vertical component of force,F_v = 1.56 x 10⁻³ N

[tex]tan\theta = \dfrac{F_H}{F_v}[/tex]

[tex]tan(5.57^0) = \dfrac{F_H}{1.56\times 10^{-3}}[/tex]

 F_h = 1.52 x 10⁻⁴ N

now, each ball will be repelled by

F = 1.52 x 10⁻⁴ N

now calculation of charges

[tex]F = \dfrac{kQ_1Q_2}{r^2}[/tex]

Q₁ = Q₂ because both charge are same

[tex]1.52\times 10^{-4} = \dfrac{9\times10^9Q^2}{0.068^2}[/tex]

    Q² = 7.809 x 10⁻¹⁷

   Q = 8.84 x 10⁻⁹ C

hence the change on the balls were Q₁ = Q₂ = 8.84 x 10⁻⁹ C

Final answer:

The time needed for the rocket to reach 100 meters, given an acceleration function a = (6 + 0.02s) m/s², requires integrating the acceleration to get velocity and then position as a function of time, considering the initial conditions v = 0, s = 0, and t = 0.

Explanation:

To solve the problem of determining the time needed for a rocket to reach an altitude of 100 meters when its acceleration is given by a = (6 + 0.02s) m/s², we will integrate the acceleration to find the velocity as a function of position and then the position as a function of time. Since we have the initial conditions of starting from rest (v = 0) and starting at the ground (s = 0) when t = 0, we can use calculus to carry out the integration for motion under variable acceleration.

First, we integrate the acceleration to get velocity:

Then, we use the velocity function to find the time taken to reach 100 meters. In this scenario, as the question relates to variable acceleration, we are dealing with non-uniformly accelerated motion, which makes it more complex than just using basic kinematic equations.

Unfortunately, without specific guidance on the integration technique or an appropriate kinematic equation that takes into account variable acceleration, we cannot solve this problem directly. However, generally, to integrate acceleration to get velocity, we would apply the fundamental theorem of calculus and then integrate the velocity function to get the position over time. From there, we can find the time needed to reach a particular altitude.

The time needed for the rocket to reach an altitude of 100 meters is approximately 187 seconds.

Establish the relationship between acceleration and distance:

The given acceleration is a function of distance: a = 6 + 0.02s. We know that acceleration is the derivative of velocity with respect to time (a = dv/dt), and velocity is the derivative of position with respect to time (v = ds/dt). Using the chain rule of calculus, we get:

[tex]a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \cdot \frac{dv}{ds}[/tex]
[tex]6 + 0.02s = v \frac{dv}{ds}[/tex]

Separate variables and integrate:

[tex]\int v \, dv = \int (6 + 0.02s) \, ds[/tex]

Integrating both sides, we get:

[tex]\frac{v^2}{2} = 6s + 0.01s^2 + C[/tex]

Given the initial conditions, at s = 0, v = 0, so C = 0. Therefore, the equation simplifies to:

[tex]v^2 = 12s + 0.02s^2[/tex]

Express velocity as a function of s:

[tex]v = \sqrt{12s + 0.02s^2}[/tex]

Use the relationship between velocity and time:

Since v = ds/dt, we can write:

[tex]dt = \frac{ds}{\sqrt{12s + 0.02s^2}}[/tex]

Integrate both sides with respect to their respective variables:

[tex]t = \int \frac{ds}{\sqrt{12s + 0.02s^2}}[/tex]

This integral can be solved using appropriate methods or a substitution trick (depending on algebraic techniques or a table of integrals):

[tex]\int \frac{ds}{\sqrt{12s + 0.02s^2}} = \frac{1}{\sqrt{0.02}} \int \frac{ds}{\sqrt{s + \frac{12}{0.02}}}[/tex]
[tex]t = \frac{1}{\sqrt{0.02}} \left[ \frac{2}{2} \sqrt{s + \frac{12}{0.02}} \right][/tex]

After evaluating the definite integral from s = 0 to s = 100 m, we obtain:

[tex]t = \frac{1}{0.1414} \left[ \sqrt{100 + 600} - \sqrt{0} \right][/tex]
[tex]t = 7.07 \sqrt{700}[/tex]
[tex]t = 7.07 \times 26.46[/tex]
[tex]t \approx 187 \text{ seconds}[/tex]

Answer:The density of the metal sphere  is [tex]8.14g/cm^3[/tex]

Explanation:

Density is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{Volume}[/tex]

Given : Mass of metal sphere = 1.497 kg = 1497 g   (1kg=1000g)

Density of the metal sphere = ?[tex]g/cm^3[/tex]

Volume of the metal sphere = [tex]\frac{4}{3}\times \pi\times r^3=\frac{4}{3}\times 3.14\times (3.53)^3cm^3=184cm^3[/tex]

Putting in the values we get:

[tex]Density=\frac{1497g}{184cm^3}[/tex]

[tex]Density=8.14g/cm^3[/tex]

Thus the density of the metal sphere  is [tex]8.14g/cm^3[/tex]

The density of the metal in the solid sphere is 8.34 g/cm³.

The density of an object can be calculated by dividing the mass of the object by its volume. In this case, the mass of the metal sphere is given as 1.497 kg and the volume can be calculated using the formula V = 4/3 × π × (radius)^3. Plugging in the values, we get:

V = 4/3 × π × (3.53 cm)^3 = 179.5942 cm^3

Now, we can substitute the values in the density formula:

Density = mass / volume = 1.497 kg / 179.5942 cm^3 = 8.34 g/cm^3

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Answer: 0.44831m

Explanation:

Unwanted horizontal spring oscillator=2.65kg

Spring constant =38.5N/M

Speed=2.92m/s

Amplitude of oscillation=?

Potential energy=m*v2/2

=2.65*2.92/2

=3.8695J

Potential energy=kinetic energy

Potential energy=1/2kx^2

3.869=1/2*38.5*x^2

3.869=19.25x^2

Dividing both sides by 19.25

3.869/19.25=x^2

So therefore, x^2=√0.200987

x=0.44831m

Answer:

0.47556 m

Explanation:

[tex]F_B[/tex] = Child supplying force = 500 N

W = Weight of foam

F = Force on air

g = Acceleration due to gravity = 9.81 m/s²

By balancing the forces we get

[tex]F+W=F_B\\\Rightarrow F_B=500+W\\\Rightarrow F_B-W=500\\\Rightarrow \dfrac{4}{3}\pi R^3(\rho_{a}-\rho_{f})g=500\\\Rightarrow R=(\dfrac{500\times 3}{4\pi (\rho_{a}-\rho_{f})g})^{\dfrac{1}{3}}\\\Rightarrow R=(\dfrac{500\times 3}{4\pi (1000-95)\times 9.81})^{\dfrac{1}{3}}\\\Rightarrow R=0.23778\ m[/tex]

The diameter of the ball is [tex]2\times 0.23778=0.47556\ m[/tex]

In this question, the child experienced how density and buoyant force play key roles in the effort needed to submerge objects in water. Using Archimedes' Principle, we can calculate the diameter of the Styrofoam ball by equating the weight needed to submerge the ball with the buoyant force. After getting the volume of the ball, we can use the formula for the volume of a sphere to find the radius and consequently, the diameter.

The physics principle applicable here is Archimedes' Principle, which states that the buoyant force on an object submerged in fluid is equal to the weight of the fluid displaced by that object. The weight needed to submerge the ball is equal to the buoyant force, and we know the weight to be 5.00×102 N. Using the equation for buoyant force (Fb = ρfluid * g * Vobject), we can find out the volume of the water displaced by the Styrofoam ball.

Setting 5.00×102 N equal to the equation involving the density of water (1.000×103 kg/m3), the gravity (9.81 m/s2), and the volume of the Styrofoam ball, we can solve for the volume of the ball. After finding the volume of the ball, we can use the equation for the volume of a sphere,  V = 4/3 * π * r3, to find the radius, and consequently the diameter of the ball. So, the child in your scenario is experiencing the effects of density and buoyant force, underscoring why a larger force is required to push bigger objects underwater.

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Answer:

t = 0.0105 s

Explanation:

given,

mass of the baseball = 435 g = 0.435 Kg

initial speed of ball, u = 36 m/s

final speed of ball, v = 49 m/s

contact force , F = 3500 N

we know,

Impulse is equal to change in momentum

I = m v - m u

I = 0.435 (49-(-36))

I = 0.435 x 85

I = 36.975 kg.m/s

we also know that

I = F x t

36.975 = 3500 x t

t = 0.0105 s

time of contact is equal to 0.0105 s

Answer:

t = 3/2T

To find how long it takes to cover a total distance of 6A, we need to find the time it takes to cover a distance A then multiply by 6.

The step to the solution is given below in the attachment.

Explanation:

Thank you for reading

Final answer:

The object takes 3 times the period (3T) to travel a total distance of 6A in simple harmonic motion.

Explanation:

The time taken for the object to travel a total distance of 6A can be calculated using the formula for the period of simple harmonic motion (T). The period is the time it takes for one complete oscillation. Since the object is attached to a spring and is set in simple harmonic motion with an amplitude (A) and period (T), we can use the formula T = 2π√(m/k), where m is the mass of the object and k is the spring constant.

In this case, we need to find the time it takes for the object to travel a distance of 6A. A full oscillation covers a distance of 2A. Therefore, to cover 6A, the object needs to complete 3 full oscillations. So, the total time taken would be 3 times the period (3T).

Therefore, the object takes 3 times the period (3T) to travel a total distance of 6A.

Answer:

[tex]v_2(t)=10sin[31t+53.13^{\circ}]\ V[/tex]

Explanation:

Given in the question

[tex]\omega[/tex] = Angular frequency = 31 rad/s

[tex]V_2=(6+j8)V[/tex]

[tex]V_2=\sqrt{6^2+8^2}tan^{-1}\dfrac{8}{6}\\\Rightarrow V_2=10, 53.13^{\circ}[/tex]

Now,

[tex]v_2(t)=rsin[\omega t+\theta]\\\Rightarrow v_2(t)=10sin[31t+53.13^{\circ}]\ V[/tex]

The required function is

[tex]\mathbf{v_2(t)=10sin[31t+53.13^{\circ}]\ V}[/tex]

Final answer:

The sinusoid corresponding to the phasor V₂ = 6 + j8 V and ω = 31 rad/s is v₂(t) = 10 cos(31t + 53.13°) V, with the magnitude positive and the phase angle within -180° to 180° range.

Explanation:

To find the sinusoid corresponding to the phasor V₂ = 6 + j8 V and ω = 31 rad/s, we first need to determine the magnitude and phase of the phasor. The magnitude (V) is the square root of the sum of the squares of the real part and the imaginary part, which gives us:

V = √(6² + 8²) = √(36 + 64) = √100 = 10 V

To find the phase angle (θ), we use the arctangent of the imaginary part over the real part:

θ = arctan(± ÷ 6) = arctan(4÷ 3) ≈ 53.13 degrees

Now, we can write the sinusoidal function using the magnitude and phase as:

v₂(t) = 10 cos(31t + 53.13°) V

However, if the question specifies that the angle should be between -180° and 180°, and the angle provided here is already in that range, we do not need to adjust our answer.

Final answer:

The cake's temperature decreases according to Newton's Law of Cooling, and we can use this law to calculate the time it takes for the cake to reach a certain temperature.

Explanation:

The cake is removed from a 350°F oven and placed in a 70°F room. After 30 minutes, the cake's temperature decreases to 200°F. To find out when it will be 100°F, we can use Newton's Law of Cooling.

According to Newton's Law of Cooling, the rate of change of the temperature of an object is proportional to the difference between the object's temperature and the temperature of its surroundings. This can be expressed as:

T' = -k(T - Ts)

where T' is the rate of change of temperature with respect to time, T is the temperature of the object, Ts is the temperature of the surroundings, and k is the cooling constant. In this case, we can rearrange the equation to solve for time:

t = (1/k) * ln((T - Ts) / (T0 - Ts))

where t is the time, T0 is the initial temperature of the object, and ln is the natural logarithm.

Plugging in the values from the problem:

t = (1/k) * ln((200 - 70) / (350 - 70))

We can find the value of the cooling constant, k, by using the given information. Since we know the temperature dropped from 350°F to 200°F in 30 minutes, we can use this to find k:

-k = (T' / (T - Ts)) = (200 - 70) / (350 - 70) / 30

simplifying, we get:

k = -((200 - 70) / (350 - 70)) / 30

Now we can substitute the value of k into the equation for time:

t = (1 / -((200 - 70) / (350 - 70)))) * ln((200 - 70) / (350 - 70))

Calculating this equation will give us the approximate time it takes for the cake to reach 100°F.

Answer:

For real gas the volume of a given mass of gas will increase with increase in temperature.

Explanation:

With the piston head locked in place and place above the fire,the volume of the gas will increase,because the volume of a given mass of gas increases with increase temperature.

Final answer:

With the piston head locked in place, the volume of the gas will stay the same when placed above a flame, but pressure will increase due to the rise in kinetic energy of the gas molecules.

Explanation:

When the piston head is locked in place and then placed above a flame, the volume of the gas will stay the same because the piston cannot move to allow for an increase in volume. According to Charles's Law, which describes the direct relationship between the temperature and volume of a gas at constant pressure, if a gas is heated and can expand, its volume will increase. However, since the piston in this scenario is locked and prevents the gas from expanding, the volume cannot change. The effect of heating the gas while the piston is locked will result in an increase in the gas pressure instead. This occurs because as the gas molecules are heated, they gain kinetic energy and move faster, colliding with the walls of the container more frequently and with greater force, which leads to increased pressure.

Answer: q = 7.06 × 10^-21 C

the magnitude of the charges that make up the dipole is 7.06 × 10^-21 C

Explanation:

Given:

Torque, τ = 6.60×10^−26N⋅m

Angle made by p with a uniform electric field, θ = 90° (perpendicular)

Electric field, E = 8.50×10^4N/C

Length between dipole r = 1.10×10^−10 m

Torque acting on the dipole is given by the relation,

τ = pE sinθ....1

But,

p = qr .....2

Substituting equation 1 to 2

τ= qrEsinθ ....3

Making q the subject of formula

q = τ/rEsinθ .....4

Where;

q = magnitude of the charges that make up the dipole.

Substituting the given values into equation 4:

q = 6.60×10^−26N⋅m/(1.10×10^−10 m × 8.50×10^4N/C × sin90°)

q = 0.70588 × 10^-20 C

q = 7.06 × 10^-21 C

The magnitude of the charge making up the electric dipole, given a torque of 6.60×10−26N⋅m when the dipole moment is perpendicular to an electric field of 8.50×104N/C, and a separation of 1.10×10−10m between the charges, is approximately 7.05×10−21 C.

The torque (τ) on a dipole in a uniform electric field is given by the equation τ = pEsinθ, where p is the dipole moment, E is the electric field strength, and θ is the angle between the dipole moment and the electric field. In this case, the dipole moment is perpendicular to the electric field, so θ= 90 degrees, and sinθ= 1. The dipole moment, p, is the product of the magnitude of the charge (q) and the separation (d) between the charges, so p = qd.

Given τ = 6.60×10−26N⋅m, E = 8.50×104N/C, and d = 1.10×10−10m, we can first use the torque equation to find the dipole moment: p = τ / E = 6.60×10−26N⋅m / 8.50×104N/C = 7.76×10−31 C⋅m. Then, use p = qd to calculate the charge: q = p / d = 7.76×10−31 C⋅m / 1.10×10−10m = 7.05×10−21 C.

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If air resistance is negligible and the ball is moving at a constant velocity, the net force acting on the ball is zero because all forces are balanced. The ball is not accelerating; hence, it is either at rest or moving with a constant velocity.

The question pertains to the net force acting on a 0.6 kg ball flying through the air at low speed, where it is stated that air resistance is negligible. Since air resistance is negligible and no other forces are mentioned, we can infer that the only force acting on the ball is gravity. However, because the ball is in motion and not accelerating, the net force on the ball must be zero. If the ball is not accelerating, it means that it is either at rest or moving with a constant velocity, which indicates that all the forces acting on the ball are balanced. In this scenario, if the only force considered is gravity, without any other force like air resistance or applied force countering it, the ball would indeed be accelerating downward due to gravity.

Answer:

The woman's distance from the right end is 1.6m = (8-6.4)m.

The principles of moments about a point or axis running through a point and summation of forces have been used to calculate the required variable.

Principle of moments: the sun of clockwise moments must be equal to the sun of anticlockwise moments.

Also the sun of upward forces must be equal to the sun of downward forces.

Theses are the conditions for static equilibrium.

Explanation:

The step by step solution can be found in the attachment below.

Thank you for reading this solution and I hope it is helpful to you.

The woman is standing 6.4 meters from the left end of the board. Since the board is 8 meters long, she is standing [tex]\( 8 \, \text{m} - 6.4 \, \text{m} = 1.6 \, \text{m} \)[/tex] from the right end of the board.

To solve this problem, we need to apply the principles of static equilibrium to the board. The board is in equilibrium because it is not moving, which means the sum of the forces acting on it must be zero, and the sum of the torques (or moments) about any point must also be zero.

The total weight of the board and the woman is [tex]\( 500 \, \text{N} + 100 \, \text{N} = 600 \, \text{N} \)[/tex]. This total weight is balanced by the support forces at the ends of the board. Therefore, the sum of the support forces is equal to the total weight:

[tex]\[ F_L + F_R = 600 \, \text{N} \][/tex]

Substituting [tex]\( F_R = 3F_L \)[/tex] into the equation, we get:

[tex]\[ F_L + 3F_L = 600 \, \text{N} \] \[ 4F_L = 600 \, \text{N} \] \[ F_L = \frac{600 \, \text{N}}{4} \] \[ F_L = 150 \, \text{N} \][/tex]

Now we can find [tex]\( F_R \)[/tex]:

[tex]\[ F_R = 3F_L \] \[ F_R = 3 \times 150 \, \text{N} \] \[ F_R = 450 \, \text{N} \][/tex]

Next, we need to consider the torques about one of the support points. Let's choose the left end as our pivot point. The torque due to the woman's weight is the product of her weight and her distance from the left end, which we will call [tex]x[/tex]. The torque due to the board's weight acts at the center of the board (since the board is uniform), which is 4 meters from either end. The torque due to the support force [tex]\( F_R \)[/tex] acts at the right end.

Setting the sum of the torques equal to zero, we have:

[tex]\[ -F_R \times 8 \, \text{m} + 500 \, \text{N} \times x + 100 \, \text{N} \times 4 \, \text{m} = 0 \][/tex]

Substituting [tex]\( F_R = 450 \, \text{N} \)[/tex] and rearranging terms, we get:

[tex]\[ -450 \, \text{N} \times 8 \, \text{m} + 500 \, \text{N} \times x + 100 \, \text{N} \times 4 \, \text{m} = 0 \] \[ -3600 \, \text{N} \cdot \text{m} + 500 \, \text{N} \times x + 400 \, \text{N} \cdot \text{m} = 0 \] \[ 500 \, \text{N} \times x = 3600 \, \text{N} \cdot \text{m} - 400 \, \text{N} \cdot \text{m} \] \[ 500 \, \text{N} \times x = 3200 \, \text{N} \cdot \text{m} \] \[ x = \frac{3200 \, \text{N} \cdot \text{m}}{500 \, \text{N}} \] \[ x = 6.4 \, \text{m} \][/tex]

Therefore, the woman is standing 6.4 meters from the left end of the board. Since the board is 8 meters long, she is standing [tex]\( 8 \, \text{m} - 6.4 \, \text{m} = 1.6 \, \text{m} \)[/tex] from the right end of the board.

Answer:

Explanation:

Given

Three blocks are placed over each other at a certain distance.

Center of gravity of each block is at distance of 0.5 L from one end of block.

First We consider block 1 and 2

Block 1 center of gravity will try to tumble the block 1 if center of gravity torque goes beyond 0.5 L of second block.

i.e. maximum distance up to which block 1 is placed over block 2 is [tex]x=0.5 L[/tex]

combined center of gravity of 1 and 2 is

Center of gravity [tex]x=\frac{0.5L+L}{2}=\frac{3L}{4}[/tex]

Now consider block 2 and 3

Combined center of gravity of  block 1 and 2 will tumble over when their Center of gravity goes beyond edge of block 1

i.e. maximum value of [tex]d=\frac{3L}{4}[/tex]

(b) As the no of blocks increases center of gravity increases so maximum value of [tex]d\rightarrow \infty[/tex]

In the static equilibrium physics problem, the maximum overhang distance d for three stacked bricks without toppling is initially calculated for two bricks, then for three bricks. The problem scales with an infinite number of bricks to reveal an overhang approaching half the brick's length.

The question pertains to the physics concept of static equilibrium, specifically torque and center of mass in systems with multiple stacked objects. For part (a), considering the top two bricks first, the maximum overhang achievable without the bricks tumbling is one-fourth of the length of one brick. When adding the third brick, the maximum overhang distance d will increase, but calculation requires a step-by-step process taking into account the center of mass of the bricks in the stack and the fulcrum point. For part (b), with an infinite number of bricks, optimal stacking achieves a maximum distance that approaches half the length of a single brick, as per the harmonic series solution to this classic physics problem.

Answer:

Average recoil force experienced by machine will be 200 N

Explanation:

We have give mass of each bullet m = 50 gram = 0.05 kg

There are 4 bullets

So mass of 4 bullets = 4×0.05 = 0.2 kg

Initial speed of the bullet u = 0 m/sec

And final speed of the bullet v = 1000 m/sec

So change in momentum [tex]P=m(v-u)=0.2\times (1000-0)=200kgm/sec[/tex]

Time is given per second so t = 1 sec

We know that force is equal to rate of change of momentum

So force will be equal to [tex]F=\frac{200}{1}=200N[/tex]

So average recoil force experienced by machine will be 200 N

The average recoil force experienced by the machine gun is 100N.

The impulse-momentum theorem states that the impulse applied to an object will be equal to the change in its momentum.

Force (F) * change in time (Δt) = change in momentum = mass (m) * velocity (v)

FΔt = mv

m = 50 g = 0.05 kg

F = mv / Δt

F = (0.05kg * 1000 m/s * 4 bullets)/ 1 second

F = 100 N

The average recoil force experienced by the machine gun is 100N.

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The problem can be covered from different methods for development. I will approximate it by the proximity method. We know that the person breathes about 6 liters per minute, that is [tex]6 * 10 - 3 m ^ 3 / min[/tex] (Recall that [tex]1L = 1 * 10 - 3 m ^ 3[/tex])

Given the conditions, we have that at atmospheric pressure with a temperature of 20 ° C the air density is [tex]1.24kgm ^ 3[/tex]

Therefore, from the density ratio, the mass would be

[tex]\rho = \frac{m}{V}\rightarrow m = \rho \dot{V}[/tex]

Here,

m = mass per time unit

V = Volume per time unit

[tex]\rho[/tex] = Density

We have

[tex]m = (6*10^{-3}m^3 / min )(1.24kg/m^3 )[/tex]

[tex]m= 7.44*10^{-3} kg/ min[/tex]

Using the conversion factor from minutes to days,

[tex]m= 7.44*10^{-3} kg/ min(\frac{60min}{1hour})(\frac{24 hours}{1day })[/tex]

[tex]m = 10.7136kg/day[/tex]

Therefore he mass of air in kilograms that a person breathes in per day is 10.7136kg

Answer:

3,3*10^-11 J

Explanation:

A=Uq=2,08*10^8 V * 1,6*10^-19 C=3,3*10^-11J

Answer:

Change in electric potential energy ∆E = 365.72 kJ

Explanation:

Electric potential energy can be defined mathematically as:

E = kq1q2/r ....1

k = coulomb's constant = 9.0×10^9 N m^2/C^2

q1 = charge 1 = -2.1C

q2 = charge 2 = -5.0C

∆r = change in distance between the charges

r1 = 420km = 420000m

r2 = 160km = 160000m

From equation 1

∆E = kq1q2 (1/r2 -1/r1) ......2

Substituting the given values

∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)

∆E = 94.5 × 10^9 (3.87 × 10^-6) J

∆E = 365.72 × 10^3 J

∆E = 365.72 kJ

Answer:

The railing is at 56.4 m above the ground when the camera reaches the ground.

Explanation:

Hi there!

Let´s find how much time it takes the camera to reach the ground. The equation of the height of the camera is the following:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h = height at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

The initial height of the camera is 19 m and we need to find at which time its height is zero. Since the camera is dropped while the balloon is rising, the initial velocity of the camera is the same as the velocity of the balloon:

h = h0 + v0 · t + 1/2 · g · t²

When the camera hits the ground, h = 0

0 = 19 m + 11 m/s · t - 1/2 · 9.8 m/s² · t²

0 = 19 m + 11 m/s · t - 4.9 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 3.4 s (The other value is rejected because it is negative and time can´t be negative).

Since the balloon rises at constant speed, the equation of height of the railing is as follows:

h = h0 + v · t

To find the height of the railing 3.4 s after it was at 19 m, we have to solve the equation with h0 = 19 m and t = 3.4 s:

h = 19 m + 11 m/s · 3.4 s

h = 56.4 m

The railing is at 56.4 m above the ground when the camera reaches the ground.

Answer:

The magnitude of the tension in the rope on the right is 500 N.

Explanation:

Given that,

Weight of scaffold= 400 N

Weight of first painter = 500 N

Weight of second painter = 400 N

Tension in the rope = 800 N

According to figure,

We need to calculate the magnitude of the tension in the rope on the right

Using balance equation

[tex]T_{r}+T=W_{p}+W_{sp}+W_{s}[/tex]

Where, [tex]T_{r}[/tex]=Tension on left side

[tex]W_{r}[/tex]=weight of first painter

[tex]W_{sr}[/tex]=weight of second painter

[tex]W_{r}[/tex]=weight of scaffold

Put the value in the equation

[tex]800+T=400+500+400[/tex]

[tex]T=1300-800[/tex]

[tex]T=500\ N[/tex]

Hence, The magnitude of the tension in the rope on the right is 500 N.

The tension in the right rope supporting the scaffold and the painters can be found by using the equilibrium principle, which results to be 500N.

In physics, we often deal with problems like this using the principle of equilibrium, stating that the sum of all the forces in a system is zero. In this case, the scaffold is stationary, not moving up or down. Therefore, the sum of the forces on it should be zero.

The total weight of the scaffold and the painters is 400N (scaffold) + 500N (first painter) + 400N (second painter) = 1300N. It's being held by two ropes, the left having tension 800N and the right of which we are seeking the tension.

Using the principle of equilibrium, the sum of the tensions in the ropes should equal to total weight. So, if 800N (left rope) + Attention (right rope) = 1300N. This yields the equation, Tension (right rope) = 1300N - 800N, which equates to 500N, the tension in the right rope.

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Answer :

New force becomes, F' = 1.83 N

Explanation:

Let two point charges exert a force of 7.35 N force on each other. The electric force between two charges is given by :

[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]

[tex]q_1\ and\ q_2[/tex] are charges

r is the distance between charges if the distance between them is increased by a factor of 2, r' = 2r

New force is given by :

[tex]F'=\dfrac{kq^2}{r'^2}[/tex]

[tex]F'=\dfrac{kq^2}{(2r)^2}[/tex]

[tex]F'=\dfrac{1}{4}\dfrac{kq^2}{r^2}[/tex]

[tex]F'=\dfrac{1}{4}\times 7.35[/tex]

F' = 1.83 N

So, the new force between charges will be 1.83 N. Therefore, this is the required solution.          

Answer:

The stitches and dimples around a baseball  and a golf ball respectively, disturbs the air drag on the balls once they are in motion, allowing the them to travel more easily.

Explanation:

The stitches on a baseball disturbs the air drag on the ball when the ball is in motion, allowing the ball to travel more easily. Depending on the orientation of the ball in flight, the drag changes as the flow is disturbed by the stitches.  

A smooth ball with no stitches or dimples has more air drag that opposes the motion.

A golf ball is smooth ball with dimples to create a thin turbulent boundary layer of air that clings to the ball's surface. This allows the smoothly flowing air to follow the ball's surface a little farther around the back side of the ball, thereby decreasing the size of the wake, and allowing the ball to travel more easily.

To solve this problem we will use the concepts of the moment of rotational inertia, angular acceleration and the expression of angular velocity.

The rotational inertia is expressed as follows:

[tex]I = \sum mr^2[/tex]

Here,

m = Mass of the object

r = Distance from the rotational axis

The rotational acceleration in terms of translational acceleration is

[tex]\alpha = \frac{a}{R}[/tex]

Here,

a = Acceleration

R = Radius of the circular path of the object

The expression for the rotational speed of the object is

[tex]\omega = \frac{\Delta \theta}{\Delta t}[/tex]

Here,

[tex]\Delta \theta[/tex] is the angular displacement of the object

The explanation by which when climbing a mountain uphill is changed to a larger pinion, is because it produces a greater torque but it is necessary to make more pedaling to be able to travel the same distance. Basically every turn results in less rotations of the rear wheel. Said energy that was previously used to move the rotation of the wheel is now distributed in more turns of the pedal. Therefore option a and c are correct.

This would indicate that the correct option is D.

Shifting to a larger-diameter gear in a 10-speed bicycle allows for an increased force exerted by the chain on the gear and a greater torque on the wheel, making it easier to ride uphill.

When riding a 10-speed bicycle up a hill, shifting to a larger-diameter gear attached to the back wheel is preferred compared to a smaller gear because it increases the force exerted by the chain on the gear and allows the cyclist to exert a greater torque on the wheel.

By shifting to a larger gear, the chain wraps around a larger portion of the gear's circumference, resulting in a greater force being applied to rotate the wheel. This increased force allows the cyclist to overcome gravity more efficiently and climb the hill with less effort.

The larger gear also allows the cyclist to apply a greater torque to the wheel. Torque is the rotational equivalent of force and represents the ability to turn the wheel. With a larger gear, the cyclist can pedal with more force and generate a larger torque, which is necessary to propel the bike up the hill.

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Answer:

<-16969.7, 0, 0> N

Explanation:

[tex]p_2[/tex] = Final momentum = < 33000, 0, 0> kg·m/s

[tex]p_1[/tex] = Initial momentum = <89000, 0, 0> kg·m/s

t = Time taken = 3.3 seconds

Impulse is given by

[tex]J=p_2-p_1\\\Rightarrow Ft=p_2-p_1\\\Rightarrow F=\dfrac{p_2-p_1}{t}\\\Rightarrow F=\dfrac{< 33000, 0, 0>-<89000, 0, 0>}{3.3}\\\Rightarrow F=\dfrac{<-56000, 0, 0>}{3.3}\\\Rightarrow F=<-16969.7, 0, 0>\ N[/tex]

The net force exerted on the truck is <-16969.7, 0, 0> N

The net force exerted on the truck by the surroundings is calculated using the change in momentum and the time interval. It is found to be approximately <-16970, 0, 0> Newtons.

To find the net force exerted on the truck by the surroundings, we need to use the change in momentum (Δp) and the time (Δt) over which the change occurs. The change in momentum is obtained by subtracting the final momentum vector from the initial momentum vector. In this case:

Δp = <33000, 0, 0> kg·m/s - <89000, 0, 0> kg·m/s = <-56000, 0, 0> kg·m/s

The time interval Δt is given as 3.3 seconds. The net force (F) can be calculated using Newton's second law in its impulse-momentum form:

F = Δp / Δt

The net force vector is:

F = <-56000, 0, 0> kg·m/s / 3.3 s = <-16969.7, 0, 0> N

Therefore, the net force exerted on the truck by the surroundings as a vector is approximately <-16970, 0, 0> Newtons.

Answer:

a) [tex]E=364N/C[/tex]

b) No

Explanation:

A) Because the electron is affected by an acceleration force in this case by the electric field, we can use the formulas of 2-dimension movement.

We will assume the electron missed the upper plate, so we need to calculate the time to travel all the way through the plate, that is:

[tex]x=v_x*t[/tex]

[tex]where:\\x=distance\\v=speed\\t=time[/tex]

so:

[tex]t=\frac{x}{v_x}=\frac{0.02m}{1.6\cdot 10^6m/s}\\\\t=1.25\cdot10^{-8}[/tex]

the electron experiences an accelerated motion in the vertical direction, so we can obtain the acceleration of the electron:

[tex]y=\frac{1}{2}.a.t^2\\\\where:\\y=vertical\_distance\\a=acceleration\\t=time[/tex]

so:

[tex]a=\frac{2.y}{t^2}\\\\a=\frac{2*(\frac{0.01}{2}m)}{(1.25\cdot10^{-8}s)^2}\\\\a=6.4\cdot10^{13} m/s^2[/tex]

now we can use the relation:

[tex]F=m.a=E.q\\so\\E=\frac{m.a}{q}[/tex]

[tex]where:\\\\E=electric\_field\\m=electron\_mass=9.1\cdot10^{-31}kg\\q=Charge=1.6\cdot10^{-19}\\a=acceleration[/tex]

Now we can calculate the electric field:

[tex]E=\frac{9.1\cdot10^{-31}kg\cdot6.4\cdot10^{13}m/s^2}{1.6\cdot10^{-19}C}\\\\E=364N/C[/tex]

B) Because the proton has the same charge but positive it will go in the opposite direction, so because we assume the electron didn't touch the plate, the proton won't.

The magnitude of the electric field is 8.91 x 10^-6 N/C. Both the electron and the proton would just miss the upper plate.

(a) To find the magnitude of the electric field, we can use the equation of motion for a charged particle in an electric field:

F = qE

Where F is the force on the particle, q is the charge of the particle, and E is the electric field strength. In this case, the force on the electron is given by:

F = (9.11 x 10^-31 kg)(1.10 x 10^6 m/s)(E)

Since the electron just misses the upper plate, the force on the electron due to gravity is equal to the force due to the electric field:

(9.11 x 10^-31 kg)(9.8 m/s^2) = (9.11 x 10^-31 kg)(1.10 x 10^6 m/s)(E)

Solving for E, we find:

E = (9.8 m/s^2) / (1.10 x 10^6 m/s)

E = 8.91 x 10^-6 N/C

(b) To determine if the proton would hit one of the plates, we can use the same approach. The force on the proton is given by:

F = (1.67 x 10^-27 kg)(1.10 x 10^6 m/s)(E)

Comparing this force to the force due to gravity:

(1.67 x 10^-27 kg)(9.8 m/s^2) = (1.67 x 10^-27 kg)(1.10 x 10^6 m/s)(E)

Solving for E, we find:

E = (9.8 m/s^2) / (1.10 x 10^6 m/s)

E = 8.91 x 10^-6 N/C

Therefore, the proton would also just miss the upper plate.

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