In 2014, a space probe approached the rocky core of the comet Churyumov–Gerasimenko, which is only a few km in diameter. The probe then entered orbit around the comet at a distance of 30 km. The comet was found to have a mass of 1.0 * 10^13 kg. What was the orbital period of the probe around the comet, in earth days?

Answer:

a. Clouds at night prevent the heat radiation of the earth from escaping back into the space by continuously reflecting them back.

b. If during the day the earth has been heated well by the sun and then if the night is cloudy then that night is not so cold.

c. Day time clouds prevent the heating of the earth by reflecting back the radiations of the sun back into the space.

Explanation:

During the day if there is sunshine and then if the night is cloudy then that night will not be so cold because during the day the earth has absorbed the heat of sun's radiation which it radiates at night but when there is cloud at night it acts as a reflecting screen and reflects the heat radiations back to the earth.

But  if there is cloud during the day time it prevents the radiations from the sun to fall on the earth by reflecting them back into the space preventing the rise in temperature of the earth.

Answer:

Visibility conditions.

Explanation:

Safe speed is a speed at which the operator of the boat can take effective action to avoid stops within the distance. To calculate the safe speed visibility conditions (fog, rain, mist, and darkness) should be included. We should also include some other factors given below:

i) Traffic density

ii) Type of vessels in the area

iii) Wind

iv) Water conditions  

The Navigation Rules dictate that a safe speed is determined by factors such as visibility, traffic density, vessel maneuverability, weather conditions, and proximity to navigational hazards.

According to the Navigation Rules, several factors should be taken into account in determining a safe speed. These include the visibility, the traffic density, the maneuverability of the vessel in immediate circumstances, the state of wind, sea, and current, and the proximity of navigational hazards. For example, on a clear day with sparse traffic and calm seas, higher speeds might be safe. However, in heavy traffic or poor visibility, the navigation rules would call for lower speeds to ensure safety.

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Answer : The original activity will be, 303 millicuries.

Explanation :

Half-life = 15 hr

First we have to calculate the rate constant, we use the formula :

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

[tex]k=\frac{0.693}{15hr}[/tex]

[tex]k=4.62\times 10^{-2}\text{ hr}^{-1}[/tex]

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  = [tex]4.62\times 10^{-2}\text{ hr}^{-1}[/tex]

t = time passed by the sample  = 24 hr

a = initial amount of the reactant  = ?

a - x = amount left after decay process = 100 millicuries

Now put all the given values in above equation, we get

[tex]24=\frac{2.303}{4.62\times 10^{-2}}\log\frac{a}{100}[/tex]

[tex]a=302.97\text{ millicuries}\approx 303\text{ millicuries}[/tex]

Therefore, the original activity will be, 303 millicuries.

Answer:

P=133.71mmHg

Explanation:

the standard free energy of formation for liquid ethanol is -174/9 kj/mol and that for gaseous ethanol is -168.6 kj/mol. calculate the vapour pressure of ethanol at

assumption:

is that temperature is at 25C, at standard pressure of 1bar(750mmHg)

ethanol is an ideal gas

The free energy of ethanol liquid does not vary with pressure,

C2H5OH(l)⟶C2H5OH(g)

free energy of formation on the reactant side is  -174.9 kj/mol

fro the product side is  -168.6 kj

∅Gvap-∅G(l)=-168.6kj/mol-(-174.9kj/mol)

+6.3kj/mol

∅G=∅Gvap+RTlnK-∅Gliq

∅G=0

0=+6.3kj/mol+8.314Jk/mol/k(298)InK

-6.3/(RT)=Lnk

taking the exponential of both sides

[tex]e^{-6300/(8.314*298)} =K[/tex]

0.178=k

k=p/[tex]p^{0}[/tex]

P^0=refers to the pressure of ethanol vapour at its standard state

partial pressure , which is 750 mmHg

P=0.178*750

P=133.71mmHg

Final answer:

The vapor pressure of ethanol at a given temperature can be calculated using the Clausius-Clapeyron equation.

Explanation:

The vapor pressure of a substance is related to its standard free energy of formation and temperature. To calculate the vapor pressure of ethanol at a given temperature, we can use the Clausius-Clapeyron equation:

ln(P₂/P₁) = △Hvap/R × (1/T₁ - 1/T₂)

Where P₁ and P₂ are the vapor pressures at temperatures T₁ and T₂ respectively, △Hvap is the enthalpy of vaporization, R is the gas constant, and ln denotes the natural logarithm.

By substituting the given values for △Hvap = -174/9 kJ/mol, T₁ = 20.0 °C (293 K), and T₂ = desired temperature, we can solve for P₂.

Answer:

110 cm

Explanation:

Gradpoint

Answer:

B) The cosmic background radiation is expected to contain spectral lines of hydrogen and helium, and it does.

Explanation:

Answer:

a) [tex]t=2.6\ s[/tex]

b) [tex]s=43.7747\ m[/tex]

Explanation:

Given:

length of inclined roof, [tex]l=54\ m[/tex]

inclination of roof below horizontal, [tex]\theta=17^{\circ}C[/tex]

acceleration of hammer on the roof, [tex]a_r=2.87\ m.s^{-2}[/tex]

height from the lower edge of the roof, [tex]h=46.5\ m[/tex]

Now, we find the final velocity when leaving the edge of the roof:

Using the equation of motion:

[tex]v^2=u^2+2.a_r.l[/tex]

[tex]v^2=0^2+2\times 2.87\times 54[/tex]

[tex]v=17.6057\ m.s^{-1}[/tex]

The direction of this velocity is 17° below the horizontal.

∴Vertical component of velocity:

[tex]v_y=v.sin\ \theta[/tex]

[tex]v_y=17.6057\times sin\ 17^{\circ}[/tex]

[tex]v_y=5.1474\ m.s^{-1}[/tex]

a.

So, the time taken to fall on the ground:

[tex]h=ut+\frac{1}{2} g.t^2[/tex]

here:

initial velocity, [tex]u=v_y=5.1474\ m.s^{-1}[/tex]

putting respective values

[tex]46.5=5.1474\times t+0.5\times 9.8\times t^2[/tex]

[tex]t=2.6\ s[/tex]

b.

Horizontal component of velocity, [tex]v_x=v.cos\ \theta=17.6057\ cos\ 17^{\circ}=16.8364\ m.s^{-1}[/tex]

Since there is no air resistance so the horizontal velocity component remains constant.

∴Horizontal distance from the edge of the roof where the hammer falls is given by:

[tex]s=v_x.t[/tex]

[tex]s=16.8364\times 2.6[/tex]

[tex]s=43.7747\ m[/tex]

Using equations of motion, it's calculated that it takes about 3.07 seconds for the hammer to fall to the ground from the roof. The hammer also travels approximately 89.2 meters horizontally from the roof to the ground.

The question is related to physics concepts of motion under gravity and kinematics. For simplicity, let's ignore air resistance.

The time it takes for the hammer to fall to the ground from the edge of the roof can be calculated using the equation of motion: h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s²), and t is the time. Solving the equation for time (t): t = sqrt(2h/g). Substituting the given values, we get t = sqrt((2*46.5)/9.8) ≈ 3.07 s.

The horizontal distance travelled by the hammer can be calculated using the formula: distance = speed × time. The horizontal speed of the hammer when it falls off the roof will be the same speed it had just as it left the roof due to the roof slope acting on it with constant acceleration. This can be gotten from the equation v = u + at, where u is the initial velocity (0 m/s), a is the acceleration (2.87 m/s²) and t is the time. The time here is the time it takes for the hammer to slide down the roof, gotten by time = distance/speed = 54.0/v. Solving all these gives a distance of approximately 89.2 m.

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Answer:

a.large particles such as gravel

Explanation:

High energy environments are typically associated with larger particles such as gravel due to the ability of these environments to move larger matter. Smaller particles are generally found in lower energy environments.

High energy environments, such as fast-flowing rivers or coastal areas with strong waves, are most capable of moving larger particles. So, among these options, the environments with high energy are most likely to contain large particles such as gravel (a). The other particles listed (silt, manganese nodules, cosmogenous sediments, clay) require less energy to be moved, so they are likely to be found in lower energy environments.

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Answer:

1.0 M HNO3 at 40°C

Explanation:

Rate of chemical reaction: This can be defined as the number of moles of reactant, converted or product formed per unit time.

Factors that affect rate of chemical reaction:

(a) Temperature:  Generally, an increase in temperature increase the rate of chemical reaction by (1) increasing the number of particles with energy equal to or greater than the activation energy, (2) Increasing the average speed of all the reactant particles, due to greater kinetic energy, leading to higher frequency of collision.

(b) Concentration: An increase or decrease in the concentration of the reactant will  result to a corresponding increase or decrease in the effective collision of the reactant and hence in the reaction rate.

other factors that affect the rate of chemical reaction are

(i) Nature of the reactant

(ii) Surface area of reactant

(iii) presence of light

(iv) presence of catalyst.

From the question above,

The condition with the highest temperature and concentration will produce the GREATEST reaction rate.

And that is  1.0 M HNO3 at 40 °C

Answer:

Answer Choose “Faster response” in the sampling pull-down menu

Explanation:

At the top of the report, below the date range selector, select faster response,

This option uses a smaller sampling size to give you faster results. In order to get a better understanding lets define sampling, according to Google analytics website  

In data analysis, sampling is the practice of analyzing a subset of all data in order to uncover the meaningful information in the larger data set.  

For example, if you wanted to estimate the number of trees in a 100-acre area where the distribution of trees was fairly uniform, you could count the number of trees in 1 acre and multiply by 100, or count the trees in a half acre and multiply by 200 to get an accurate representation of the entire 100 acres.

To speed up Go_ogle Analytics reports, you can reduce the date range and complexity of the report, or increase server resources.

To increase the speed at which Go_ogle Analytics compiles reports, several actions could be taken. First, one can reduce the amount of data that is being processed by adjusting the date range of the reports. By analyzing a smaller chunk of data, the report can be generated much faster.

Secondly, one can adjust the complexity of the report. The more complex the report, the more data needs to be processed, hence it will take longer. Reducing the complexity and only focusing on the key metrics can help speed up the report compilation.

Lastly, one can increase the capacity of their server resources. If a company has the means, it can invest in dedicated server resources for Go_ogle Analytics, allowing it to process reports faster.

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Answer:

Explanation:

Given

mass of gas [tex]m=8000 gm=8 kg[/tex]

Area of Piston [tex]A=5 cm^2[/tex]

[tex]T_1=15^{\circ}C[/tex]

[tex]T_2=27^{\circ}C[/tex]

no of moles [tex]n=0.28[/tex]

Work done on the gas is given by

[tex]W=-P\Delta V[/tex]

[tex]P\Delta V[/tex] can also be written as [tex]nR\Delta T[/tex]

as PV=nRT

[tex]W=-nR\Delta T[/tex]

[tex]W=-0.28\times 8.314\times (27-15)[/tex]

[tex]W=-27.93 J[/tex]

negative sign indicates that work is done on the system i.e. surrounding done a positive work on the gas

Explanation:

Given that

             [tex]rA(t)=(3.00 m/s)t\hat{i}+ (1.00 m/s^2)t^2\hat{j}\texttt{ and }rB(t)=(4.00 m/s)t\hat{i}+ (-1.00 m/s^2)t^2\hat{j}[/tex]

We need to find distance when t = 3 s

Substituting t = 3 s

             [tex]rA(t)=(3.00 m/s)\times 3\hat{i}+ (1.00 m/s^2)\times 3^2\hat{j}=9\hat{i}+9\hat{j}\\\\rB(t)=(4.00 m/s)\times 3\hat{i}+ (-1.00 m/s^2)\times 3^2\hat{j}=12\hat{i}-9\hat{j}[/tex]

[tex]\texttt{Displacement = }12\hat{i}-9\hat{j}-(9\hat{i}+9\hat{j})=3\hat{i}-18\hat{j}[/tex]

[tex]\texttt{Magnitude = }\sqrt{3^2+(-18)^2}=18.3m[/tex]

Option C is the correct answer.

The distance between object A and object B is approximately 18.248 meters. (Choice C)

In this question we must apply the concepts of vector difference, dot product and norm to determine the distance between objects A and B, in meters:

[tex]r_{B/A} = \sqrt{(\vec r_{B}-\vec r_{A})\,\bullet\,(\vec r_{B}-\vec r_{A})}[/tex] (1)

Where:

If we know that [tex]\vec r_{A} = (3\cdot t, t^{2})\,\left[m\right][/tex], [tex]\vec r_{B} = (4\cdot t,-t^{2})\,\left[m\right][/tex] and [tex]t = 3\,s[/tex], then the distance of B relative to A is:

[tex]r_{B/A}=\sqrt{t^{2}+4\cdot t^{4}}[/tex]

[tex]r_{B/A} = t\cdot \sqrt{1+4\cdot t^{2}}[/tex]

[tex]r_{B/A} = 3\cdot \sqrt{1+4\cdot 3^{2}}[/tex]

[tex]r_{B/A} \approx 18.248\,m[/tex]

The distance between object A and object B is approximately 18.248 meters. (Choice C) [tex]\blacksquare[/tex]

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Answer:

7

Explanation:

The balanced reaction is as follows:

[tex]2CO+2NO \rightarrow 2CO_2 + N_2[/tex]

Coefficient of CO = 2

Coefficient of NO = 2

Coefficient of CO2 = 2

Coefficient of N2 = 1

Sum of all coefficient = 2 + 2+ 2+ 1

                                    = 7

Answer:

39.4 g

39.4 cm³

1.09137 g/cm³

Explanation:

[tex]\rho[/tex] = Density of water = 1 g/cm³

Mass of water displaced will be the difference of the

[tex]m=43-3.6\\\Rightarrow m=39.4\ g[/tex]

Mass of water displaced is 39.4 g

Density is given by

[tex]\rho=\dfrac{m}{v}\\\Rightarrow v=\dfrac{m}{\rho}\\\Rightarrow v=\dfrac{39.4}{1}\\\Rightarrow v=39.4\ cm^3[/tex]

So, volume of bone is 39.4 cm³

Average density of the bird is given by

[tex]\rho=\dfrac{43}{39.4}\\\Rightarrow \rho=1.09137\ g/cm^3[/tex]

The average density is 1.09137 g/cm³

The spin quantum number (ms) describes the orientation of the spin of the electron: TRUE

The magnetic quantum number (ml) describes the size and energy associated with an orbital. An orbital is the path that an electron follows during its movement in an atom: FALSE

The angular momentum quantum number (l) describes the orientation of the orbital: FALSE

The principal quantum number (n) describes the shape of an orbital: FALSE

Explanation:

The statement contains mixed truths and falsehoods. The spin quantum number describes the orientation of electron spin, the magnetic quantum number pertains to orbital orientation, and the angular momentum quantum number concerns orbital shape, while the principal quantum number describes orbital size and energy.

The statement is false. The spin quantum number (ms) does describe the orientation of the spin of the electron, either up or down. The magnetic quantum number (ml) describes the orientation of the orbital in space. An orbital is defined as a region in space where there's a high probability of finding an electron, not their path. The angular momentum quantum number (l) determines the shape of the orbital. The principal quantum number (n) describes the size and energy level of an orbital.

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Answer:

Time taken to cover the distance 11.45 s.

Explanation:

The given parameters:

Initial Velocity(u)=24 m/s

Acceleration (a)=8 [tex]m/s^{2}[/tex]

Distance or Displacement(s)=800 m

Displacement or Distance is equal in the above because it travels in a straight line.

We have to apply Second Equation of Motion,

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

800=24t + 4[tex]t^{2}[/tex]

200=6t + [tex]t^{2}[/tex]

[tex]t^{2}[/tex] + 6t - 200=0

Solving, the quadratic equation to find out the roots, we get that the possible values of t will be 11.45 s and a negative value.

The negative value will be neglected as time cannot be negative.

Hence, the time taken is 11.45 s.

Answer:

b. activity interrelations are not considered

Explanation:

A bar chart schedule or Gantt chart is used to visualize task that scheduled over time.

Answer:

If the stone is thrown from the ground, the correct answer is A. If it is thrown from a height h, the correct answer is D.

Explanation:

Hi there!

I can´t see the drawing but let´s assume that initially, the stone is on the ground level. If that is the case, initially, the potential energy will be zero and when it returns to Earth it will also be zero. The potential energy depends on the height of the stone. If the final and initial height of the stone is zero, then the change in potential energy will also be zero:

ΔPE = final PE - initial PE

ΔPE = m · g · hf - m · g · hi (where hf and hi are the final and initial height respectively)

ΔPE = m · g (hf - hi)

ΔPE = m · g (0)

ΔPE = 0

Initially, the kinetic energy (KE) of the stone is the following:

KE = 1/2 · m · v0²

As the stone goes up, the kinetic energy is transformed into potential energy; but as the stone starts to fall, the acquired potential energy is transformed again into kinetic energy, so that the final and initial kinetic energy of the stone is the same.

Then:

ΔKE = final KE - initial KE = 0 (because final KE = initial KE).

Then, the correct answer is A.

Always ΔKE = -ΔPE due to the conservation of energy. Potential energy can´t be acquired by the stone if there is no loss of kinetic energy and vice-versa.

Let´s assume now that the stone is thrown from a height hi to the ground.

The final potential energy will be zero (becuase h = 0) but the initial PE will be:

PE = m · g · h1

Then:

ΔPE = final PE - initial PE = 0 - m · g · h1

Then ΔPE will be negative.

The initial kinetic energy will be:

KE = 1/2 · m · v0²

But the final kinetic energy will be equal to the initial kinetic energy plus the loss of potential energy (remember: if potential energy decreases, another type of energy has to increase, in this case, kinetic energy and vice-versa):

ΔKE = final KE - initial KE

ΔKE = 1/2 · m · v0² + m · g · h1 - 1/2 · m · v0²

ΔKE = m · g · h1

Then ΔKE will be positive and the correct answer would be D.

Answer:

The count rate at the end of four days will be 7.5 counts per minute.

Explanation:

First it is important to know that half-life is the time to a piece of radioactive material to decay 50%. So if we know we start with 120 counts per minute and we already know the half-life of the isotope is 1 hour we expect that past 1 hour the material decays 50% (it's halved) so we will count 60 counts per minute, now if we wait another hour 60 counts will decay in to 30 counts per minute and so on. That should be translate to a math equation as:

final counts = initial counts * [tex](\frac{1}{2})^{\#half\,life\,periods}[/tex]

After 4 days we have 4 half-life periods passed so:

final counts= 120 counts per minute * [tex](\frac{1}{2})^{4}[/tex]

final material = 7.5 counts per minute

Answer:

C- ±37.6°

Explanation:

This is the case of a single slit diffraction. In this case, the dark fringe occurs at

sin θ = ± mλ/a

where m = 1

θ = ± 30°

Hence we have sin ± 30° = ± λ/a

±0.5 = ± λ/a

Hence, we have a = 2λ

For a circular aperture, the condition for the first dark fringe is

D = diameter of circle = a = 2λ

so we have sin θ1 = 1.22λ/a

sin θ1 = 1.22λ/2λ

hence sin θ1 = 0.61

θ1 = sin⁻¹0.61 = ±37.6°

Answer:

D. 6 m/s²

Explanation:

Acceleration is change in velocity over time.

a = Δv / Δt

a = (60 m/s − 0 m/s) / 10 s

a = 6 m/s²

The acceleration of the car is 6m/s².

To find the correct option among all the options, we need to know about the acceleration.

Acceleration = ∆V/∆t = 60/10= 6 m/s².

Thus, we can conclude that the option (D) is correct.

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Answer:255.255.255.224

Explanation:

/20 has a subnet mask of 255.255.240.0

We have 20 on bits for the network and 16 off bits for the host. Since the designer wants 20 to 30 hosts in each subnet we will be having 5 off bits for the host. Therefore we will be having a /27..

Which will be 255.255.255.224

Final answer:

The suitable subnet mask for creating subnets with 20-30 hosts within the 10.0.240.0/20 network is 255.255.255.224 or /27, which allows for 32 IP addresses, 30 of which can be used for hosts.

Explanation:

The correct subnet mask to meet the requirement for new subnets that support between a minimum of 20 hosts and a maximum of 30 hosts within the 10.0.240.0/20 network is 255.255.255.224 or /27. This subnet mask allows for 32 IP addresses per subnet, with 30 usable addresses for hosts when excluding the network and broadcast addresses. Subnet masks 255.255.224.0 and 255.255.240.0 provide too many hosts per subnet, and subnet mask 255.255.255.240 or /28 provides only 16 IP addresses, which would be insufficient for the required minimum of 20 hosts.

Answer:

240 cm

Explanation:

Gradpoint

Answer:

Approximately [tex]\rm 11.2 \; N \cdot kg^{-1}[/tex] at that distance from the center of the planet.

Option A) The low value of [tex]g[/tex] near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of [tex]g[/tex] on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

[tex]\displaystyle \frac{G \cdot M \cdot m}{R^2}[/tex],

where

To find an equation for [tex]g[/tex], divide the equation for gravity by the mass of the object:

[tex]\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}[/tex].

In this case,

Calculate [tex]g[/tex] based on these values:

[tex]\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}[/tex].

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for [tex]g[/tex]:

[tex]\displaystyle g = \frac{G \cdot M}{R^2}[/tex].

The mass of the planet is in the numerator. If two planets are of the same size, [tex]g[/tex] would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since [tex]R[/tex] is in the denominator of [tex]g[/tex], increasing the value of [tex]R[/tex] while keeping [tex]M[/tex] constant would reduce the value of [tex]g[/tex]. That explains why the value of [tex]g[/tex] near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately [tex]9.81\; \rm N \cdot kg^{-1}[/tex].

As a side note, [tex]5.82\times 10^7\rm \; m[/tex] likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of [tex]g[/tex] near the cloud tops of Saturn is approximately [tex]\rm 11.2 \; N \cdot kg^{-1}[/tex].

concept is gravitational force. Saturn's low value of g on its surface is possible because of its low average density, which is less than the density of water.

The value of g on the surface of Saturn is determined by its mass and radius. The low value of g on Saturn is possible because of its low average density. Saturn's mass is much larger than Earth's, but its density is much lower, resulting in a lower value of g on its surface.

The low density of Saturn, which is only 0.7 g/cm³, is less than the density of water. This means that Saturn would be light enough to float if placed in water. Therefore, despite its large mass, the low density of Saturn allows for a low value of g on its surface.

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Answer:

40 ft/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = -32 ft/s² = a (downward is taken a negative)

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -32\times -25+0^2}\\\Rightarrow v=40\ ft/s[/tex]

The speed of Isaac as he reached the ground is 40 ft/s

"The final velocity of Isaac when he hits the ground can be calculated using the kinematic equation for an object in free fall under the influence of gravity with an initial velocity of zero:

[tex]\[ v_f^2 = v_i^2 + 2a\Delta y \][/tex]

Plugging in the known values:

[tex]\[ v_f^2 = 0^2 + 2(-32 \text{ ft/sec}^2)(25 \text{ ft}) \] \[ v_f^2 = 0 + (-2)(32)(25) \] \[ v_f^2 = -1600 \][/tex]

Since velocity cannot be negative, we take the positive square root of the absolute value of [tex]\( v_f^2 \)[/tex] to find [tex]\( v_f \)[/tex]:

[tex]\[ v_f = \sqrt{1600} \] \[ v_f = 40 \text{ ft/sec} \][/tex]

Therefore, Isaac is traveling at 40 feet per second when he hits the ground.

The answer is: [tex]40 \text{ ft/sec}.[/tex]

Final answer:

About 8,960 atoms of carbon-14 were present in the object when it was made 18,000 years ago, based on its current carbon-14 content and the known half-life of carbon-14.

Explanation:

To calculate how many atoms of carbon-14 (14C) were present in an object that was created 18,000 years ago, we use the half-life of 14C, which is 5,730 years. Given that there are 1,000 atoms of 14C present now, we need to find out how many half-lives have passed to work backwards and determine the initial quantity of 14C atoms.

First, we divide the age of the object by the half-life of 14C:

18,000 years / 5,730 years per half-life ≈ 3.14 half-lives

Now, we apply the formula for exponential decay, taking into account the number of half-lives:

Initial Quantity = Present Quantity × (2⁰ of half-lives)

Initial Quantity = 1000 atoms × (2≈ 3.14)

Initial Quantity ≈ 1000 atoms × 8.96

Initial Quantity ≈ 8960 atoms of 14C were present when the object was made.

Answer:

Explanation:

A) Initial speed, u = 12 m/s

   Acceleration, a = -7 m/s²

   Final velocity, v = 0 m/s

   We have equation of motion v² = u² + 2as

                0² = 12² + 2 x -7 x s

                s = 10.29 m

  He need 10.29 m to stop.

  Remaining distance = 43 - 10.29 = 32.71 m

  We have

            Remaining distance = Reaction Time x Initial velocity

            32.71 = Reaction Time x 12

            Reaction Time = 2.73 seconds.

B) Distance traveled in 1.1333 s = 1.1333 x 12 = 13.60 m

   Remaining distance = 41 - 13.6 = 27.4 m

    Initial speed, u = 12 m/s

   Acceleration, a = -7 m/s²

   Displacement, s = 27.4 m

                      v² = 12² + 2 x -7 x 27.4

                      v² =  -239.6

   Not possible.

   Motorist will not hit deer.

  He will not reach the deer.

Answer:

The magnitude of the flux is  [tex]2.00 N m^2/C[/tex]

Explanation:

The electric flux through a planar area is defined as the product of electric field and the  component of the area perpendicular to the field.

Electric flux = Electric field * Area * (angle between the planar area and the electric flux)

The equation is

[tex]\phi = E A cos(\theta)[/tex]

Where:

[tex]\phi[/tex]is the Electric Flux

A is the  Area

E is the Electric field

[tex]\theta[/tex] is  angle between a perpendicular vector to the area and the electric field

Now substituting the values,

[tex]\phi = 5.00 \times 4.00 \times cos(0)[/tex]

[tex]\phi = 5.00 \times 4.00 \times 1[/tex]

[tex]\phi = 2.00 N m^2/C[/tex]

The flux of a constant electric field in the z-direction through a rectangle in the xy-plane is zero, because the angle between the direction of the electric field and the direction of the normal to the area is 90 degrees, which makes the dot product zero.

To calculate the magnitude of the flux of an electric field, we use the equation: Φ = E . A where Φ is the electric flux, E is the electric field, and A is the area of the surface. The dot (.) represents a dot product, which means we consider the angle between the field and the area. In this problem, the electric field (E) is given as 5.00 N/C and the area of the rectangle (A) is 4.00 m². Also, because the electric field is in the z-direction (up and down), and the rectangle is in the xy-plane (flat), the angle between the field and the area is 90 degrees.

However, the dot product for angles of 90 degrees is zero because cos(90°) = 0. So, regardless of the magnitudes of the electric field and the area, the flux is zero because Φ = E . A = EAcos(90°) = 0. Therefore, the correct answer is (a) 0.

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Answer:

The Correct Option is C (coarse lithogenic sediment)

Explanation:

This is because all other options are found on sea floors, except for coarse lithogenic sediments that are as a result of erosion on land. Both calcareous and siliceous ooze are common sea sediments. Abyssal clay and manganese nodule are red clay and rock concretion respectively found in the bottom of the sea.

Answer:

A. The horizontal velocity vector points to the right & equals v cos θ.

Explanation:

The motion describes a parabolic path, where the horizontal speed is constant and the horizontal velocity vector always points to the right and equals v*cos θ.