A star (not Barnard's star) at a distance of 10 pc is observed to have a proper motion of 0.5 arcsec / year. What is its transverse speed in AU / year?

Answer:

Distance between them increase

Explanation:

The position S of the water droplet can be determined  using equation of motion

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

where [tex]u[/tex] is the initial velocity which is zero here

[tex]t[/tex] is time taken, [tex]a[/tex] is acceleration due to gravity

the position of  first drop after time [tex]t[/tex] is given by

[tex]S_{1} =0 \times t+ \frac{1}{2} at^2=\frac{1}{2} at^2............(1)[/tex]

the position of  next drop at same time is

[tex]S_{2} =\frac{1}{2} a(t-1)^2 = \frac{1}{2} a(t^2+1-2t)............(2)[/tex]

distance between them is [tex]S_{1} -S_{2}[/tex]  is [tex]a(t-1)[/tex]

from the above the difference will increase with the time

As the water drops fall, their velocity increases due to the force of gravity, which causes the distance between each subsequent drop to increase.

The response to the student's question deals with the notion of acceleration due to gravity. As the water drops fall, they are accelerated by gravity, which means their velocity (speed) increases over time. If we consider two subsequent droplets, the second drop begins its descent 1.0 seconds after the first. Therefore, when the second drop begins to fall, the first drop has already accelerated for 1.0 seconds. This causes the distance between the two drops to increase as they fall.

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Using the physics equation of motion and the given initial velocity, reaction time, and deceleration, one can determine whether a truck can stop in time to avoid a collision.

The question focuses on stopping distance and acceleration required to avoid a collision, indicating its base in Physics. If we have a truck moving at a constant velocity and it brakes at a certain distance from an obstacle, the minimum acceleration needed to avoid a collision can be calculated using the equation of motion v^2 = u^2 + 2as. Here, 'v' is the final velocity (0 m/s as the truck needs to stop), 'u' is the initial velocity, 'a' is the acceleration, and 's' is the distance over which the truck needs to stop.

To determine if the truck will hit the child, we must account for the driver's reaction time as well. During this reaction time, the truck continues to travel at its initial speed. After the reaction time, the truck will begin decelerating until it comes to a stop. The total stopping distance is the distance covered during the reaction time plus the distance covered during deceleration. The latter can be found using the deceleration rate and the formula mentioned above.

For the given scenario of the truck with an initial velocity of 10 m/s, a braking distance of 50 m, reaction time of 0.5 seconds, and deceleration of -1.25 m/s^2, we can calculate whether or not the truck will be able to stop in time to avoid hitting the child.

Final Answer:

The speed at which the car hits the water is approximately 75.2 feet per second.

Explanation:

To find the speed at which the car hits the water, we can use one of the kinematic equations that relates the initial velocity, acceleration due to gravity, the height it fell from, and the final velocity. The kinematic equation that we need is:

[tex]\[ v^2 = u^2 + 2gh \][/tex]

Where:
-  v  is the final velocity,
-  u  is the initial velocity,
-  g  is the acceleration due to gravity (which we will use  [tex]\( 32.174 \, \text{ft/s}^2 \)[/tex] for since we are dealing with feet),
-  h  is the height.

Here, we are given:
-  [tex]\( u = 8 \, \text{ft/s} \)[/tex] (initial velocity)
- [tex]\( h = 87 \, \text{ft} \)[/tex] (height)
- [tex]\( g = 32.174 \, \text{ft/s}^2 \)[/tex] (acceleration due to gravity)

Let's find the final velocity \( v \) using these values.

[tex]\[ v^2 = u^2 + 2gh \][/tex]

[tex]\[ v^2 = (8 \, \text{ft/s})^2 + 2 \cdot 32.174 \, \text{ft/s}^2 \cdot 87 \, \text{ft} \][/tex]

[tex]\[ v^2 = 64 \, \text{ft}^2/\text{s}^2 + 2 \cdot 32.174 \, \text{ft/s}^2 \cdot 87 \, \text{ft} \][/tex]
[tex]\[ v^2 = 64 \, \text{ft}^2/\text{s}^2 + 5591.148 \, \text{ft}^2/\text{s}^2 \][/tex]
[tex]\[ v^2 = 5655.148 \, \text{ft}^2/\text{s}^2 \][/tex]

Now we take the square root of both sides to solve for the final velocity \( v \):

[tex]\[ v = \sqrt{5655.148} \, \text{ft/s} \][/tex]

Performing the square root calculation, we get:

[tex]\[ v \approx 75.2 \, \text{ft/s} \][/tex]

So, the speed at which the car hits the water is approximately 75.2 feet per second.

Answer:

Option (2)

Explanation:

The Cryosphere refers to the frozen water bodies on earth. This includes the glaciers, icebergs, ice sheets and the frozen water surrounding the Arctic as well as Antarctica.

The Hydrosphere refers to all the water bodies on earth including the rivers, streams, lakes, and ponds.

The given condition is based on the interaction between the cryosphere and the hydrosphere.

The frozen ice in the Antarctic and Arctic is melting rapidly due to the increase in the global warming effect. This declining ice in the polar region results in the rise in the global sea level. This can be catastrophic as many of the big cities will be flooded because of this increasing height of sea level.

Thus, the correct answer is option (2).

The decline of Arctic sea ice and its impact on water levels is an interaction between two Earth system spheres: the cryosphere and hydrosphere.

The interaction of Arctic sea ice decline and increasing water levels involves the cryosphere and hydrosphere spheres. The cryosphere refers to the frozen components of the Earth system, including ice caps, glaciers, and sea ice. The hydrosphere encompasses all the water on Earth, including oceans, lakes, and rivers.

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Answer:

The correct answer is:

d) A and B have the same magnitude flux

Explanation:

Electric flux is the property of electric field that measures the electric field lines, passing through a surface and electric flux is also directly proportional to the number of electric field lines passing through a surface.  

The formula of electric flux is:

Φ = E A Cos θ

(where E is the electric field, A is the area of face and θ is the angle between the face and the electric field).

Since, faces A and B are perpendicular to the electric field and the electric field lines passing through face A also passes through face B therefore, both of these faces have larger and same magnitude of electric flux.

Since, face C is parallel to the electric field so, the electric flux is smaller at face C, because the magnitude of Cos 180 (when face is parallel) is smaller than the magnitude of Cos 90 (when face is perpendicular).

Face A, which is perpendicular to the uniform electric field, has the largest magnitude electric flux through it because the angle between the field lines and the normal to the surface is zero, maximizing the electric flux.

The question revolves around calculating the electric flux through different faces of a prism in a uniform electric field. Electric flux (Φ) is given by the equation Φ = E ⋅ A ⋅ cos(θ), where E is the magnitude of the electric field, A is the area through which the field lines pass, and θ is the angle between the field lines and the normal (perpendicular) to the surface.

Face A is perpendicular to the electric field, which means the angle θ is 0 degrees and cos(θ) is 1. Thus the flux through Face A is maximum. For Face B, the leaning face, θ is greater than 0 degrees but less than 90 degrees, thus cos(θ) will be less than 1. Hence, flux through Face B will be less than through Face A. Face C, being parallel to the electric field, has θ as 90 degrees, and cos(90) is 0, so the flux through Face C is zero. Therefore, in comparison, Face A has the largest magnitude electric flux through it.

Answer:

532.0725 m

102.17270893 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = g

H = Height of cliff

Distance traveled in 3 seconds

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 3^2\\\Rightarrow s=44.145\ m[/tex]

Distance traveled by sound = 2H-44.145 m

[tex]2H-44.145=ut+\dfrac{1}{2}at^2\\\Rightarrow 2H-44.145=340\times 3\\\Rightarrow H=\dfrac{340\times 3+44.145}{2}\\\Rightarrow H=532.0725\ m[/tex]

The height of the cliff is 532.0725 m

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 532.0725+0^2}\\\Rightarrow v=102.17270893\ m/s[/tex]

Her speed just before she hits the ground is 102.17270893 m/s

Answer:

a)  x_{cm} = m₂/ (m₁ + m₂)   d , b)   x_{cm} = 52.97 pm

Explanation:

The expression for the center of mass is

                [tex]x_{cm}[/tex] = 1 / M  ∑ [tex]x_{i}[/tex] [tex]m_{i}[/tex]

Where M is the total masses, mI and xi are the mass and position of each element of the system.

Let's fix our reference system on the oxygen atom and the molecule aligned on the x-axis, let's use index 1 for oxygen and index 2 for carbon

              x_{cm} = 1 / (m₁ + m₂)   (0+ m₂ x₂)

Let's reduce the magnitudes to the SI system

             m₁ = 17 u = 17 1,661 10⁻²⁷ kg = 28,237 10⁻²⁷ kg

             m₂ = 12 u = 12 1,661 10⁻²⁷ kg = 19,932 10⁻²⁷ kg

             d = 128 pm = 128 10⁻¹² m

The equation for the center of mass is

               x_{cm} = m₂/ (m₁ + m₂)   d

b) let's calculate the value

            x_{cm} = 19.932 10⁻²⁷ /(19.932+ 28.237) 10⁻²⁷    128 10-12

            x_{cm} = 52.97 10⁻¹² m

            x_{cm} = 52.97 pm

(a) The expression for the center mass of these two atoms relative to oxygen atom is  [tex]X_{cm} = \frac{m_1 d_0 \ +\ m_2d}{m_1 + m_2}[/tex]

(b) The numeric value for the center of mass of carbon monoxide is 53 pm.

The given parameters;

The center mass of these two atoms relative to oxygen atom is calculated as follows;

[tex]X_{cm} = \frac{m_1 d_0 \ +\ m_2d}{m_1 + m_2}[/tex]

where;

(b)

The numeric value for the center of mass of carbon monoxide in units of pm is calculated as follows;

[tex]X_{cm} = \frac{17u(0) \ +\ 12u(128 \ pm)}{(12u + 17u)}\\\\X_{cm} = \frac{(12 \times 128u) \ pm}{29u} \\\\X_{cm} = 53 \ pm[/tex]

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Answer:

a. [tex]t=19.56 s[/tex]

b.[tex]d=718.34[/tex]

Explanation:

The solution to the differential equation

[tex]\dfrac{dv}{dt}=9.8-\dfrac{v}{5}[/tex]

is the exponential function

[tex]v(t)=ce^{-0.2t}+49[/tex]

and we find [tex]c[/tex] from the initial condition [tex]v(0)=0:[/tex]

[tex]0=ce^{-0.2*0}+49\\\\0=c+49\\\\c=-49[/tex]

Therefore, we have

[tex]v(t)=-49e^{-0.2t}+49[/tex]

[tex]\boxed{ v(t)=49(1-e^{-0.2t})}[/tex]

Part A:

The maximum velocity that the object can reach is 49 (which the maximum value [tex]v(t)[/tex] can have).

Now, 98% of 49 is 48.02; therefore,

[tex]48.02=49(1-e^{-0.2t})[/tex]

[tex]0.98=1-e^{-0.2t}[/tex]

[tex]e^{-0.2t}=0.02[/tex]

[tex]\boxed{t=19.56 s}[/tex]

Part B:

The distance traveled is the integral of the speed:

[tex]d=\int_0^{19.56}v(t)*dt[/tex]

[tex]d=\int^{19.56}_0 {49(1-e^{-0.2t})} \, dt[/tex]

[tex]d=49[t+5e^{-0.2t}]_0^{19.56}[/tex]

[tex]\boxed{d=718.34}[/tex]

To find the time that must elapse for the object to reach 98% of its limiting velocity, we need to solve the differential equation. We can then find the distance the object falls by integrating the velocity function with respect to time.

To find the time it takes for the object to reach 98% of its limiting velocity, we need to solve the differential equation. First, we separate the variables by writing it as:

dv / (9.8 - (v/5)) = dt

Next, we integrate both sides:

∫ (1 / (9.8 - (v/5))) dv = ∫ dt

After evaluating the integrals, we can solve for v:

v = 49 - 49e^(-t/5)

Substituting v with 0.98 times the limiting velocity (which is 49), we can solve for t:

49 - 49e^(-t/5) = 0.98 * 49

Solving this equation will give us the time it takes for the object to reach 98% of its limiting velocity.

To find the distance the object falls, we need to integrate the velocity function, v, with respect to time:

∫ v dt

By evaluating the integral, we can calculate the distance the object falls in the time found in part (a).

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Answer:

maximum width of the doorway = 35.77ft

Explanation:

The detailed calculation and derivation from first principle is as shown in the attachment

Answer:

the maximum width is x= 4√2 ft = 5.656 ft

Explanation:

for the parabola

y= a*x² + b*x + c

where y= height and x= width

an aircraft hangar should be symmetric with respect to the y axis , then

y(-x)=y(x) → a*x² + b*x + c = a*x² - b*x + c →-2*b*x =0 → b=0

it also should be pointing downwards → a is negative

, then the parabola would be

y= c- a*x²

since c= maximum height = 18 ft

then for y=0 , x= 48 ft/2 = 24 ft  →  0 = 18 ft - a*(24 ft)² → a= 1/32 ft⁻¹

then

y= 18 ft- 1/32 ft⁻¹ *x²

since the doorway cannot go beyond the parabola , the maximum possible doorway is obtained when the doorway touches the parabola.

then for a height y= 8 ft

8 ft = 18 ft- 1/32 ft⁻¹ *x²

x= 4√2 ft = 5.656 ft

To produce a second tone or the first overtone in a tube closed at one end, the length of the tube required is three times the length used for the fundamental frequency, resulting in a length of 1.008 m.

To understand the length required to produce a second tone or the first overtone in a tube closed at one end, it's essential to grasp the concept of harmonics in sound resonance. In such a tube, the resonant frequencies occur in odd multiples of the fundamental frequency. The first resonance the students observed, with the fundamental frequency of 256 Hz at a length of 0.336 m, corresponds to a quarter wavelength of the sound wave in the tube.

For the first overtone (second resonance), the air column in the tube must accommodate three-quarters of a wavelength, meaning the effective length will be three times larger than that of the fundamental. Thus, if the fundamental resonance occurs at a length of 0.336 m, the length for the second resonance will be:

This calculation is based on the understanding that the second tone or first overtone in a closed tube happens at three times the length necessary for the fundamental frequency, leading to the described increase in the length of the air column.

To find the length of tube for the second resonance, halve the initial length where the first resonance occurred at a fundamental frequency of 256 Hz.

The length required to produce a second tone under the same experimental conditions can be calculated based on the concept of resonance in a closed tube.

To find the length for the second resonance (first overtone), we know that the first resonance occurs at 0.336m for a fundamental frequency of 256 Hz. The second resonance, in this case, would occur at half the wavelength of the fundamental frequency, so the length would be half of the initial length: 0.168m.

Answer:

Power delivered by the source will be 182.912 watt

Explanation:

We have given a load is consist of a resistor of 70 ohm in parallel with [tex]90\mu F[/tex] capacitance

Voltage source is given [tex]v_s(t)=160cos(2000t)[/tex]

So maximum value of voltage source is 160 volt

So rms value [tex]v_{r}=\frac{v_m}{\sqrt{2}}=\frac{160}{1.414}=113.154volt[/tex]

We know that average power delivered by the source will be equal to average power absorbed by the resistor

So power absorbed by the resistor [tex]P=\frac{v_r^2}{R}=\frac{113.154^2}{70}=182.912watt[/tex]

So power delivered by the source will be 182.912 watt

To calculate the molecular weight of a polyethylene molecule with n=750, multiply the molecular weight of the ethylene unit by n.

To calculate the molecular weight of a polyethylene molecule with n=750, we need to know the molecular formula and the atomic weights of the elements present in the molecule. Polyethylene is made up of repeating ethylene (C2H4) units, so we can calculate the molecular weight of the polyethylene molecule by multiplying the molecular weight of the ethylene unit (28.05 g/mol) by the value of n (750).

Calculation:
Molecular weight of polyethylene = Molecular weight of ethylene unit × n = 28.05 g/mol × 750 = 21,037.5 g/mol

Therefore, the molecular weight of the polyethylene molecule with n=750 is 21,037.5 g/mol, rounded to three significant figures.

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Answer:

After 1.938 sec velocity of rock will be 36 m/sec

Explanation:

We have given initial velocity at which rock is thrown u = 55 m/sec

Final velocity v = 36 m/sec

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

According to first equation of motion we know that [tex]v=u+gt[/tex], here v is final velocity, u is initial velocity, g is acceleration due to gravity and t is time

So [tex]36=55-9.8t[/tex] ( Negative sign is due to rock is thrown upward )

So [tex]9.8t=19[/tex]

t = 1.938 sec

So after 1.938 sec velocity of rock will be 36 m/sec

In a single phase, the addition or removal of energy changes Kinetic Energy not Potential Energy. However, during a phase change, this energy addition or subtraction results in a change in Potential Energy, not Kinetic Energy.

The subject of this question is energy and phase, particularly in the context of Potential Energy (PE) and Kinetic Energy (KE). When only one phase is present, adding or removing energy will mainly change the KE, not the PE. This is because the energy is utilized to speed up or slow down the particles, thus changing their kinetic energy. However, during a phase change, adding or removing energy changes PE but not KE as it alters the state rather than the speed of the particles. Statement B is the one that is accurate while only one phase is present, whereas the correct option for the phase change scenario is option D.

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Answer:

31 m/s

Explanation:

As both the monkey and the darts are subjected to constant gravitational acceleration g = 9.8 m/s2 and both start from rest (vertically speaking). Their vertical position will always be the same. For the dart to hit the monkey, its horizontal position must be the same as the monkey's, which is unchanged before reaching the ground. Therefore, the time it takes for the dart to travel across 70 m must be less than the time it takes for the monkey to drop 25m to the ground. We can find it out using the following equation of motion

[tex]s_m = gt_m^2/2[/tex]

[tex]25 = 9.8t_m^2/2[/tex]

[tex]t_m^2 = 50/9.8 = 5.1[/tex]

[tex]t_m = \sqrt{5.1} = 2.26 s[/tex]

For the dart to takes less that 2.26 s to travel 70m, its horizontal speed must at least be 70 / 2.26 = 31 m/s

Answer:

12.9121148614 m/s

35.9393048982 m

Explanation:

t = Time taken

u = Initial velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 38=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{38\times 2}{9.81}}\\\Rightarrow t=2.78337865516\ s[/tex]

Time taken for the bag to fall is 2.78337865516 seconds

Time she has been jogging for

9+2.78337865516 = 11.78337865516 seconds

Total distance traveled by her

[tex]s=vt\\\Rightarrow s=3.05\times 11.78337865516=35.9393048982\ m[/tex]

Henrietta is 35.9393048982 m away

Velocity of throwing

[tex]\dfrac{35.9393048982}{2.78337865516}=12.9121148614\ m/s[/tex]

The velocity of throwing is 12.9121148614 m/s

Final answer:

Bruce must throw the bagels at an initial speed of 12.92 m/s for Henrietta to catch them, and Henrietta will be 35.93 m from the point directly below Bruce's window when she catches the bagels.

Explanation:

Projectile Motion and Kinematics Problem

To find the initial speed Bruce must throw the bagels, we need to consider two aspects of projectile motion: the horizontal motion, which is constant because air resistance is neglected, and the vertical motion, which is influenced by gravity.

Firstly, we need to calculate the time it takes for the bagels to fall from the window to the ground. Using the equation for free fall h =
1/2 g t², where h is the height (38.0 m), and g is the acceleration due to gravity (9.81 m/s²), we can solve for t, the time to fall:

38.0 m = 1/2 * 9.81 m/s² * t²

t = sqrt(2 * 38.0 m / 9.81 m/s²) = sqrt(7.74) ≈ 2.78 s

Bruce throws the bagels 9.00 s after Henrietta has passed below the window. In this time, Henrietta has jogged a distance of d = speed * time = 3.05 m/s * 9.00 s = 27.45 m horizontally.

Since Henrietta is already past the point directly below the window, we need to add the distance she will cover in the time it takes for the bagels to fall. This distance is additional distance = jogging speed * fall time = 3.05 m/s * 2.78 s ≈ 8.48 m.

Overall, Henrietta will be approximately 27.45 m + 8.48 m = 35.93 m from the point directly below the window when she catches the bagels.

To find the initial speed with which Bruce throws the bagels, we use the horizontal motion formula initial speed = distance / time, which gives us an initial speed of approximately 35.93 m / 2.78 s ≈ 12.92 m/s.

Bruce must throw the bagels horizontally at an initial speed of approximately 12.92 m/s for Henrietta to catch them just before they hit the ground, at a distance of approximately 35.93 m from the point directly below Bruce's window.

To solve this problem we will apply the concepts related to centripetal acceleration, which will be the same - by balance - to the force of gravity on the body. To find this acceleration we must first find the orbital velocity through the Doppler formulas for the given periodic signals. In this way:

[tex]v_{o} = c (\frac{\lambda_{max}-\bar{\lambda}}{\bar{\lambda}}})[/tex]

Here,

[tex]v_{o} =[/tex]  Orbital Velocity

[tex]\lambda_{max} =[/tex] Maximal Wavelength

[tex]\bar{\lambda}} =[/tex] Average Wavelength

c = Speed of light

Replacing with our values we have that,

[tex]v_{o} = (3*10^5) (\frac{3.00036-3}{3})[/tex]

Note that the average signal is 3.000000m

[tex]v_o = 36 km/s[/tex]

Now using the definition about centripetal acceleration we have,

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

Here,

v = Orbit Velocity

r = Radius of Orbit

Replacing with our values,

[tex]a = \frac{(36km/s)^2}{100000km}[/tex]

[tex]a= 0.01296km/s^2[/tex]

[tex]a = 12.96m/s^2[/tex]

Applying Newton's equation for acceleration due to gravity,

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

Here,

G = Universal gravitational constant

M = Mass of the planet

r = Orbit

The acceleration due to gravity is the same as the previous centripetal acceleration by equilibrium, then rearranging to find the mass we have,

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

[tex]M = \frac{(12.96)(100000000)^2}{ 6.67*10^{-11}}[/tex]

[tex]M = 1.943028*10^{27}kg[/tex]

Therefore the mass of the planet is [tex]1.943028*10^{27}kg[/tex]

Answer:

The y-component of the car's position vector is 670m/s.

The x-component of the acceleration vector is -3, and the y-component is 40.

Explanation:

The displacement vector of the car with velocity

[tex]\boldsymbol{v}= (-3t\boldsymbol{i}+2t^2\boldsymbol{j})m/s[/tex]

is the integral of the velocity.

Integrating [tex]\boldsymbol{v}[/tex] we get the displacement vector [tex]\boldsymbol{d}[/tex]:

[tex]\boldsymbol{d}=(-\dfrac{3}{2}t^2\boldsymbol{i}+\dfrac{2}{3}t^3\boldsymbol{j} )[/tex]

Now if the initial position if the car is

[tex]\boldsymbol{r}= (3.0\boldsymbol{i}+2.0\boldsymbol{j})[/tex]

then the displacement of the car at time [tex]t[/tex] is

[tex]\boldsymbol{d(t)}= \boldsymbol{r+d}[/tex]

[tex]\boxed{\boldsymbol{d(t)}=(-\dfrac{3}{2}t^2+3.0\boldsymbol{i}+\dfrac{2}{3}t^3+2.0\boldsymbol{j} )}[/tex]

Now at [tex]t=10s[/tex], we have

[tex]\boxed{\boldsymbol{d(t)}=(-147\boldsymbol{i}+670\boldsymbol{j} )}m[/tex]

The y-component of the car's position vector is 670m/s.

The acceleration vector is the derivative of the velocity vector:

[tex]\boldsymbol{a(t)}=\dfrac{d\boldsymbol{v(t)}}{dt} =(-3\boldsymbol{i}+4t\boldsymbol{j})[/tex]

and at [tex]t=10s[/tex] it is

[tex]\boldsymbol{a(t)}=(-3\boldsymbol{i}+40\boldsymbol{j})m/s^2[/tex]

The x-component of the acceleration vector is -3, and the y-component is 40.

The x and y components of the car's position at 10 s are -147.0 m and 668.67 m, respectively. The x and y components of the car's acceleration at 10 s are -3 m/s² and 40 m/s², respectively.

The problem involves determining the position and acceleration components of a remote-controlled car from given velocity functions over time.

1.) X Component of Car's Position at 10 s:

Given the velocity component, vx = -3t m/s, we need to integrate it with respect to time to find the position (x). The initial position x0 is 3.0 m.

When t = 10 s:

2.) Y Component of Car's Position at 10 s:

Given the velocity component, vy = 2t² m/s, integrating it with respect to time gives the position (y). The initial position y0 is 2.0 m.

When t = 10 s:

3.) X Component of Car's Acceleration at 10 s:

Given the velocity component, vx = -3t m/s, the acceleration is the time derivative of velocity.

Hence, at t = 10 s:

4.) Y Component of Car's Acceleration at 10 s:

Given the velocity component, vy = 2t² m/s, the acceleration is the time derivative of velocity.

Hence, at t = 10 s:

Answer:

Explanation:

False

Electric conductivity in metals is the result of the motion of charged particles called electrons. Atoms of metallic elements are distinguished by the availability of valence electrons. Valence electrons are the electrons in the outer shell of an atom and free to move around.

These free electrons are free to move in the lattice of metal and result in electric current when a voltage is applied.

Answer:

i. +/- 1.43% and +/- 0.71% ii. +/- 0.33%

Explanation:

[tex]% Error = \frac{Error}{Measurement}* 100%\\[/tex]

Answer:

0.0458 kWh

6.5736 cents

Explanation:

The formula for electric energy is given as

E = Pt................. Equation 1

Where E = Electric energy, P = Electric power, t = time.

Given; P = 550 W, t = 5 min = (5/60) h = 0.083 h.

Substituting into equation 1

E = 550(0.083)

E = 45.83 Wh

E = (45.83/1000) kWh

E = 0.0458 kWh.

Hence the kWh = 0.0458 kWh.

If the makes a toast four morning per week, and the are Four weeks in a month.

Total number days he makes toast in a month = 4×4 = 16 days.

t = 16×0.083 h = 1.328 h.

Total energy used in a month = 550(1.328)

E = 730.4 Wh

E = 0.7304 kWh.

If the cost of energy is 9.0 cents per kWh,

Then for 0.7304 kWh  the cost will be 9.0(0.7304) = 6.5736 cents.

Hence this would add 6.5736 cents to his monthly electric bill

A 550-W toaster in operation for 5.0 minutes uses 2.75 kWh of energy. If you make toast four mornings per week, it would add an estimated cost of $4.30 to your monthly electric energy bill.

To calculate the energy used by the toaster, we can use the formula E = Pt, where P is the power and t is the time. In this case, the power of the toaster is 550 watts and the time it is in operation is 5.0 minutes. Plugging these values into the formula, we get E = (550 W)(5.0 min) = 2750 W.min. To convert this to kilowatt-hours (kWh), we need to divide by 1000, so the energy used by the toaster is 2.75 kWh.

To estimate how much this would add to your monthly electric energy bill, we need to know how many times you use the toaster in a month. If you use it four mornings per week, that would be 4 days x 52 weeks / 12 months = 17.33 days per month. Multiplying the energy used by the toaster (2.75 kWh) by the number of days in a month (17.33), we get an estimate of 47.75 kWh per month. Finally, to find the cost, we multiply the energy (47.75 kWh) by the cost per kilowatt-hour (9.0 cents/kWh) and convert it to dollars, giving us an estimated cost of $4.30 per month.

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Answer: at when distance r = infinity.

Explanation: The formulae for the electric potential of an electric charge to an arbitrary point is given by the formulae below

V = q/4πεr

V = electric potential (volts)

q = magnitude of electric charge

ε = permittivity of free space

r = distance between arbitrary point and charge.

In the equation above, it can be seen that only electric potential (v) and distance (r) is a variable, and there is an inverse relationship between them (an increase in one leads to a decrease in the other)

Thus to have zero value of electric potential (v= 0) we have to have the largest value of r ( r = infinity).

Same goes for electric potential energy between two charges, the formulae is given below as

W = q1 *q2/4πεr

W= electric potential energy

q1 = magnitude of first charge.

q2 = magnitude of second charge

ε = permittivity of free space

r = distance between arbitrary point and charge.

Also, all values are constant aside from electric potential energy (w) and distance (r) which have an inverse relationship.

Thus to have zero value of electric potential energy (w =0), we have to get an infinite value of distance ( r =infinity)

Answer:

Reduced by 49 times

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

[tex]F_G = G\frac{M_1M_2}{R^2}[/tex]

where G is the gravitational constant. [tex]M = M_1 = M_2[/tex] are the masses of the 2 objects. and R is the distance between them.

Since R squared is in the denominator of the formula, if we make it 7 times as large with no change in mass, gravitational force would be dropped by 7*7 = 49 times

To solve the problem we should know about Newton's Law of gravity.

According to Newton's law of gravity, there is an attractive force between any two-particle carrying mass, such that the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

[tex]F \propto m_1m_2\\\\F \propto \dfrac{1}{R^2}[/tex]

[tex]F = G\dfrac{m_1m_2}{R^2}[/tex]

Where G is the proportionality constant and the value of G is 6.67 x 10-11 N m² / kg².

The force between the two will be [tex]\dfrac{1}{49}[/tex] time of the force before.

Given to us,

Assumption

Let's assume that the distance between the moon and the planet is d.

Values

As it is given that there is no change in the mass of the moon or the planet, therefore,

Also, it is given that the distance between them changes to 7 times, therefore,

Substitute the value Newton's Law of gravity,

[tex]F = G\dfrac{m_1m_2}{(7d)^2}\\\\\\F = G\dfrac{m_1m_2}{49d^2}[/tex]

Thus, the force between the two will be [tex]\dfrac{1}{49}[/tex] time of the force before.

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Answer:

Explanation:

Given

Initial velocity of ball [tex]u=10\ m/s[/tex]

height of window [tex]h=20\ m[/tex]

Using Equation of motion

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

where u=initial velocity

t=time

a=acceleration

As ball is already is at a height of 20 m so

[tex]Y=ut+\frac{1}{2}at^2+20[/tex]

[tex]Y=10\times t+0.5\times (-9.8)t^2+20[/tex]

[tex]Y=-4.9t^2+10t+20[/tex]

(b)highest point is obtained at v=0

[tex]v^2-u^2=2as[/tex]

where

v=final velocity

u=initial velocity

a=acceleration

s=displacement

[tex](0)-10^2=2\times (-9.8)\times s[/tex]

[tex]s=\frac{100}{19.6}[/tex]

[tex]s=5.102\ m[/tex]

Highest Point will be [tex]s+20=25.102\ m[/tex]

(c)Time taken when the ball hit the ground i.e. at Y=0

[tex]-4.9t^2+10t+20=0[/tex]

[tex]t=3.28\ s[/tex]

impact velocity [tex]v=\sqrt{2\times 9.8\times 25.102}[/tex]

[tex]v=22.181\ m/s[/tex]

(a) The equation be "Y = -4.9t² + 10t + 20".

(b) The highest point be "25.102 m".

(c) The impact velocity be "22.181 m/s"

According to the question,

Ball's initial velocity, u = 10 m/s

Window's height, h = 20 m

(a) By using equation of motion,

Y = ut + [tex]\frac{1}{2}[/tex]at²

By substituting the values,

  = ut + [tex]\frac{1}{2}[/tex]at² + 20

  = 10 × t + 0.5 × (9.8)t² + 20

  = -4.9t² + 10t + 20

(b) We know that,

→ v² - u² = 2as

here, Final velocity, v = 0

0 - (10)² = 2 × (-9.8) × s

          s = [tex]\frac{100}{19.6}[/tex]

             = 5.102 m

(c) Time taken will be:

→ -4.9t² + 10t + 20 = 0

                            t = 3.28 s

hence,

The impact velocity,

v = [tex]\sqrt{2\times 9.8\times 25.102}[/tex]

  = 22.181 m/s

Thus the above response is correct.

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Answer:

83816746.4254 m/s

Explanation:

m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]

q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

V = Voltage = [tex]2\times 10^4\ V[/tex]

The kinetic energy of the electron is

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

Energy is given by

[tex]E=qV[/tex]

Balancing the energy

[tex]qV=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2qV}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}}\\\Rightarrow v=83816746.4254\ m/s[/tex]

The velocity of the electrons is 83816746.4254 m/s

The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.

Through these waves, you can know that there is beyond the clouds.

Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.

Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.

It is indeed possible to learn about the universe beyond the clouds due to other non-visual forms of radiation, mainly radio waves and gamma rays, which can penetrate through the clouds and reach the earth's surface.

Yes, even if Earth were completely blanketed with clouds and we could not see the sky, we could still learn about the universe beyond the clouds. This is because, in addition to visible light which would be blocked by the clouds, the universe also emits various other forms of radiation that can penetrate the clouds and reach the ground.

Two major types of radiation that could penetrate the dense clouds are radio waves and gamma rays. Radio waves are a form of electromagnetic radiation used in many areas of science and technology, while gamma rays are highly energetic forms of radiation and are used in fields such as astronomy to get valuable information about distant celestial bodies.

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A constellation, in astronomy, is a conventional grouping of stars, whose position in the night sky is apparently invariable. The peoples, generally of ancient civilizations, decided to link them through imaginary strokes, thus creating virtual silhouettes on the celestial sphere. From 1928, the International Astronomical Union (UAI) decided to officially regroup the celestial sphere into 88 constellations with precise limits, such that every point in the sky would be within the limits of a figure. When an astronomer says he saw a comet in Capricorn last night, it means that he saw a comet in the direction of the constellation of Capricorn.

Final answer:

To find the car's acceleration, use the kinematic equation v = u + at. The car's acceleration is 2.5 m/s^2. To find the car's speed as it passes the 0.14 km post, plug the values into the kinematic equation v = u + at. The car's speed is 20 m/s.

Explanation:

To find the car's acceleration, we can use the kinematic equation:
v = u + at
where v is the final velocity, u is the initial velocity (which is 0 m/s since the car starts from rest), a is the acceleration, and t is the time. We are given that the car passes the 0.14 km post at 8.0 s after starting, which means the car travels a distance of 0.14 km in 8.0 s. Converting 0.14 km to meters gives us 140 m. Plugging the values into the equation, we have:
20 = 0 + a * 8.0
Simplifying, we find that the car's acceleration is 2.5 m/s^2.

To find the car's speed as it passes the 0.14 km post, we can use the kinematic equation:
v = u + at
Since the car starts from rest (u = 0 m/s) and the car's acceleration is 2.5 m/s^2 (which we just found), we can plug these values into the equation along with the time (8.0 s) to find the car's speed:
v = 0 + 2.5 * 8.0
Simplifying, we find that the car's speed as it passes the 0.14 km post is 20 m/s.

Answer: Asteroids, meteoroids, and comets are remnants of the early solar system. This Statement is TRUE.

Explanation:

METEOROID: these are small rocky or metallic objects found in outer space.

ASTEROIDS: these are also known as minor planets of the inner solar system. They are irregularly shaped object in space that orbits the Sun.

COMETS: these are dusty chunk of ice, that moves in a highly elliptical orbit about the sun.

Asteroids, meteoroids, and comets as remnants of the early solar system was further proved in nebular hypothesis

initially proposed in the eighteenth century by German philosopher Immanuel Kant and French mathematician Pierre-Simon Laplace. (The word nebula means a gaseous cloud.) According to the modern version of the theory, about 4.5 to 5 billion years ago the solar system developed out of a huge cloud of gases and dust floating through space. These materials were at first very thin and highly dispersed.

Answer:

The angle θ between the direction of the current and the direction of the uniform magnetic field is 41° or 139°.

Explanation:

The force on a current carrying wire is given by the following equation:

[tex]\vec{F} = I\vec{L}\times \vec{B}[/tex]

The cross-product can be written with a sine term:

[tex]F = ILB\sin(\theta)\\0.025 = (3.4)(0.28)(0.04)\sin(\theta)\\\sin(\theta) = 0.6565[/tex]

Therefore, the angle θ is 41.03° and 138.97°

The angles can be calculated using the formula sin(θ) = F / (I L B), giving two symmetrical values about 90° in the first and second quadrants because the sine function is periodic.

The question is asking for the angles at which the force on a current-carrying wire in a magnetic field is a specific magnitude. The magnitude of the force exerted on a current-carrying wire placed in a magnetic field is given by the formula F = I L B sin(θ), where F is the force, I is the current, L is the length of the wire, B is the magnetic field strength, and θ is the angle between the direction of the current and the direction of the magnetic field.

By rearranging for θ, we get the equation sin(θ) = F / (I L B). Plugging in the values from the question, we find sin(θ) = 0.0250 N / (3.40 A  imes 0.280 m  imes 0.0400 T). This gives us θ values that correspond to the sine of this ratio.

There are two angles that will produce the same sine value because sine is a periodic function, which are θ and 180°-θ. Therefore, the two angles between the direction of the current and the direction of the uniform magnetic field for which the force on the wire has a magnitude of 0.0250 N will be symmetrical about 90° in the first and second quadrants.