Find the equivalent resistance Req between terminals a and b if terminals c and d are open and again if terminals c and d are shorted together. (Round the final answers to two decimal places.) With terminals c-d open, Req = Ω. With terminals c-d shorted, Req = Ω.

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

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 electric flux through the surface is equal to 3.878 x 10³ Nm²/C

The field distance r is equal to half the length of each side of the cube.

From the area the length of each size was calculated and the field distance and charge were used in calculating the magnitude of the electric field vector which was found to be 202 x 10³ N/C

The total flux area available to this electric field is 6x32cm²

Explanation:

The full solution can be found in the attachment below.

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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:

At least 6 N, assuming that the rope has a length of more than 2 meters.

Explanation:

In general, if it takes a force of [tex]F(h)[/tex] newtons to lift an object at height [tex]h[/tex], the work done lifting the object from [tex]h = a[/tex] to [tex]h = b[/tex] can be found using the definite integral about [tex]h[/tex]:

[tex]\displaystyle W = \int \limits_{a}^{b} F(h)\, dh[/tex].

If the value of [tex]F(h)[/tex] is a constant [tex]m \cdot g[/tex] regardless of height [tex]h[/tex], then the result of the integral would be

[tex]\displaystyle \int \limits_{a}^{b} (m \cdot g)\, dh = \left[m \, g\, h \right]_a^b = m\cdot g \, (b - a)[/tex].

However, in this case the value of [tex]F(h)[/tex] does depend on the value of [tex]h[/tex].

In other words, [tex]F(h) = 3\; h[/tex] where [tex]F[/tex] is in Newtons and [tex]h[/tex] is in meters.

Evaluate the integral:

[tex]\begin{aligned} W &= \int \limits_{a}^{b} F(h)\, dh \cr &= \int \limits_{0}^{2}3\, h \, dh && \text{Apply the power rule.}\cr &= \left[\frac{3}{2}\,h^2\right]_{0}^{2}\cr &= \frac{3}{2} \times 2^2 \cr &= 6\end{aligned}[/tex].

The work required to raise one end of the rope to a height of 2 meters with an applied force of 3 N is 6 joules.

We can calculate the work required to raise one end of the rope to a height of 2 meters as follows:

[tex] W = F*d*cos(\theta) [/tex]

Where:

F: is the force exerted = 3 N (newton: unit of force)

d: is the displacement = 2 m

θ: is the angle between the applied force and the displacement

Since the force and the displacement are in the same direction, θ = 0, so:

[tex] W = F*d*cos(\theta) = Fdcos(0) = F*d [/tex]

Hence, the work done is:

[tex] W = 3 N*2m = 6 J [/tex]

Therefore, it is required 6 J of work.

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Answer:

3,3*10^-11 J

Explanation:

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

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.

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:

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 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:

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.

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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Answer:

Part A  B.  Part B C.

Explanation:

A) The elementary charge e (without sign) is equal to the charge of one electron (with negative sign) or to the charge of one proton (with positive sign), so the proton must have a total charge of +e.

if u = +2/3 e and d= -1/3 e, we need a combination which sum gives +3/3 e.

Combination A adds to 6/3e, so it is not possible. C adds to zero, and D gives a negative result.

The only remaining choice is udd:

uud ⇒ +2/3 e + 2/3 e -1/3 e = +3/3 e = +e

So, the statement B is true.

B) As the neutron has no net charge, we need to find a combination which sum adds to zero.

So, A and D are not possible, as they are combinations of the same type of quarks, so the sum is either positive or negative, but not zero.

uud gives +e (we chose it to make a proton in part A), so the only remaining choice is udd:

udd⇒ +2/3 e -1/3 e - 1/3 e = 0

So the statement C is true.

Answer:

0.71 cm

Explanation:

[tex]q_1=75.0nC=75\times 10^{-9}C[/tex]

[tex]1nC=10^{-9}C[/tex]

[tex]q_2=75.0nC=75\times 10^{-9}C[/tex]

Force between two charges=1 N

Coulomb's law of force

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

Where k=[tex]9\times 10^9Nm^2/C^2[/tex]

Using the formula

[tex]1=\frac{9\times 10^9\times 75\times 10^{-9}\times 75\times 10^{-9}}{r^2}[/tex]

[tex]r^2=0.50625\times 10^{-4}[/tex]m

[tex]r=\sqrt{0.50625\times 10^{-4}}=0.71\times 10^{-2} m[/tex]

[tex]r=0.71\times 10^{-2}\times 10^{2}=0.71\times 10^{-2+2}=0.71\times 10^0=0.71 cm[/tex]

Using formula

[tex]1 m=10^2cm, a^x\cdot a^y=a^{x+y}, a^0=1[/tex]

Hence, the distance between two charges =r=0.71 cm

The unit vector that has the same direction as the given vector [tex]\(-3i + 2j - k\) is \(\frac{-3}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j - \frac{1}{\sqrt{14}}k\).[/tex]

To find a unit vector in the same direction as the given vector [tex]\(-3i + 2j - k\)[/tex], we first need to calculate the magnitude of the given vector and then divide each component by that magnitude.

The magnitude of the given vector is given by the formula:

[tex]\[|v| = \sqrt{(-3)^2 + (2)^2 + (-1)^2} = \sqrt{14}\][/tex]

Now, we can find the unit vector [tex]\(\hat{u}\)[/tex] by dividing each component of the given vector by its magnitude:

[tex]\[\hat{u} = \frac{-3}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j - \frac{1}{\sqrt{14}}k\]\[\hat{u} = \frac{-3}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j - \frac{1}{\sqrt{14}}k\][/tex]

So, the unit vector that has the same direction as the given vector [tex]\(-3i + 2j - k\) is \(\frac{-3}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j - \frac{1}{\sqrt{14}}k\).[/tex]

Learn more about unit vector, here:

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To find a unit vector with the same direction as the given vector -3i + 2j - k, divide the given vector by its magnitude.

To find a unit vector with the same direction as the given vector -3i + 2j - k, we need to divide the given vector by its magnitude. The magnitude of the given vector is the square root of the sum of the squares of its components, which is sqrt((-3)^2 + 2^2 + (-1)^2) = sqrt(14).

So, the unit vector with the same direction as the given vector is (-3i + 2j - k) / sqrt(14).

Answer:

Explanation:

Given

For single guitar noise level [tex]SL=70\ dB[/tex]

Intensity of sound

Sound level[tex]=10\log (\frac{I}{I_0})[/tex]

where I=Intensity of sound Produced

[tex]I_0=[/tex]Human threshold frequency [tex](10^{-12}\ W/m^2)[/tex]

[tex]70=10\log (\frac{I}{10^{-12}})[/tex]

[tex]I=10^{-12}\times 10^7[/tex]

[tex]I=10^{-5}\ W/m^2[/tex]

For 2 guitars

[tex]I'=2I=2\times 10^{-5}\ W/m^2[/tex]

[tex]SL=10\log (\frac{I}{I_0})[/tex]

[tex]SL=10\log (\frac{2\times 10^{-5}}{10^{-12}})[/tex]

[tex]SL=10(7+\log (2))[/tex]

[tex]SL=73.01\ dB[/tex]

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.

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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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:

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.

Final answer:

The volume of the bag at the new pressure and temperature is 0.991 L.

Explanation:

To solve this problem, we can use the combined gas law:

P1V1/T1 = P2V2/T2

Where:

P1 = 795 torr (initial pressure)

V1 = 976 mL (initial volume, convert to L: 976 mL / 1000 = 0.976 L)

T1 = 25°C (initial temperature, convert to Kelvin: 25°C + 273 = 298 K)

P2 = 553 torr (final pressure)

T2 = 10°C (final temperature, convert to Kelvin: 10°C + 273 = 283 K)

Now we can plug in these values into the equation:

P1V1/T1 = P2V2/T2

(795 torr)(0.976 L)/(298 K) = (553 torr)(V2)/(283 K)

Solving for V2, we get:

V2 = (795 torr)(0.976 L)(283 K) / (553 torr)(298 K) = 0.991 L

Therefore, the volume of the bag at the new pressure and temperature is approximately 0.991 L.

Answer:

0.0005217 m/s or 5.217×10⁻⁴ m/s

Explanation:

Speed: This can be defined as the ratio of the distance covered by a body to the time taken to cover that distance. The S.I unit of speed is m/s. Speed is a scalar quantity, because it can only be represented by magnitude only.

Mathematically, speed is expressed as

S = d/t ......................................................... Equation 1

Where S = speed, d = total distance, t = time taken to cover the distance

Given: d = 120 mm = (120/1000) m = 0.12 m, t = 230 s.

Substituting into equation 1

S = 0.12/230

S = 0.0005217 m/s

Hence the speed of the security guard = 0.0005217 m/s or 5.217×10⁻⁴ m/s

The speed of the security guard is 0.52 mm/ss.

To find the speed of the security guard, we need to divide the distance traveled by the time taken. In this case, the distance is 120 mm and the time is 230 ss. The formula to calculate speed is:

Speed = Distance / Time

Substituting the given values:

Speed = 120 mm / 230 ss

Simplifying the expression, we get:

Speed = 0.52 mm/ss

Therefore, the speed of the security guard is 0.52 mm/ss.

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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:

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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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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Answer

given,

mass of block 1, m = 0.3 Kg

speed of block 1, v = 4 m/s

mass of second block,M = 2 Kg

initial speed of block = 0 m/s

spring constant, k = 100 N/m

a) for block 1

  linear momentum before collision

   P₁ = m v = 0.3 x 4 = 1.2 Kg.m/s

  Kinetic energy

   [tex]KE_1 = \dfrac{1}{2}mv^2[/tex]

   [tex]KE_1 = \dfrac{1}{2}\times 0.3\times 4^2[/tex]

   [tex]KE_1 =2.4\ J[/tex]

b) After impact

  final velocity calculation

 using conservation of momentum

  m v  = (m + M )v_f

   0.3 x 4 = 2.3 x v_f

    v_f = 0.522 m/s

   Linear momentum

   P₂ = (m+M) v_f

   P₂ = 1.5 x 0.522

   P₂ = 0.783 kg.m/s

   Kinetic energy

   [tex]KE_2= \dfrac{1}{2}(M+m)v^2[/tex]

   [tex]KE_2= \dfrac{1}{2}\times 2.3\times 0.522^2[/tex]

   [tex]KE_2=0.313\ J[/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

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

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:

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