A drag racer can reach a top speed of 98 m/s. How long will it take the racer to travel 1500 m?

Answer:

The maximum angle does not depend on the mass

Explanation:

This is because In as much as the force acting on the normal which is the the maximum force of static friction increases as the mass increases, the component of the force of gravity parallel to the ramp increases at the same rate. Thus The maximum angle is independent of the mass.

Answer:

F = 10.86 units

Explanation:

The magnitude of a vector in terms of the magnitude of its rectangular components is given by the following formula:

F = √(Fₓ² + Fy²)

where,

F = Magnitude of the Vector = ?

Fₓ = magnitude of the x-component of vector = 8.7 units

Fy = magnitude of y component of vector = - 6.5 units

Therefore, using these values in the equation, we get:

F = √[(8.7 units)² + (- 6.5 units)²]

F = √(117.94 units²)

F = 10.86 units

Answer:

The different types of corn used.

Explanation:

Independent variable: In research methods, the term "independent variable" is determined as a variable that is being manipulated, changed, or altered in an experiment by the experimenter in order to see its effect on the dependent variable. The changes in the dependent variable in an experiment depends on the independent variable directly.

The independent variable in the popcorn experiment is brand of popcorn used.

However, the brand of popcorn was changed in this experiment, hence, the brand of the popcorn is the independent variable.

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Answer:

Given that

V2/V1= 0.25

And we know that in adiabatic process

TV^န-1= constant

So

T1/T2=( V1 /V2)^ န-1

So = ( 1/0.25)^ 0.66= 2.5

Also PV^န= constant

So P1/P2= (V2/V1)^န

= (1/0.25)^1.66 = 9.98

A. RMS speed is

Vrms= √ 3RT/M

But this is also

Vrms 2/Vrms1= (√T2/T1)

Vrms2=√2.5= 1.6vrms1

B.

Lambda=V/4π√2πr²N

So

Lambda 2/lambda 1= V2/V1 = 0.25

So the mean free path can be inferred to be 0.25 times the first mean free path

C. Using

Eth= 3/2KT

So Eth2/Eth1= T2/T1

So

Eth2= 2.5Eth1

D.

Using CV= 3/2R

Cvf= Cvi

So molar specific heat constant does not change

The rms speed of the system is 1.6 while Mean free path is 0.25 times the first mean free path.

Given that  

[tex]\bold{\dfrac {V_2}{V_1}= 0.25}[/tex]

In adiabatic process  

[tex]\bold {TV^\gamma^-^1}[/tex]= constant  

So  

[tex]\bold {\dfrac {T1}{T2}=(\dfrac {V1 }{V2})^ \gamma^-^1}}\\[/tex]  

So,

[tex]\bold {\dfrac {T_1} {T_2} = (\dfrac { 1}{0.25})^ 0^.^6^6= 2.5}[/tex]  

for pressure,

[tex]\bold {PV^\gamma = \ Constant }[/tex]  

So

[tex]\bold {\dfrac {P1}{P2}=(\dfrac {V1 }{V2})^ \gamma}}\\\\\bold {\dfrac {P1}{P2}=(\dfrac {1}{0.25})^ 0^.^6^6} = 9.98 } }[/tex]  

A. rms speed can be calculated as

[tex]\bold {\dfrac {Vrms 2}{Vrms1}= \sqrt {T2T1})}\\\\\bold {Vrms2 =\sqrt {2.5} = 1.6\ Vrms1 }[/tex]

B.  The mean free path can be calculated as

[tex]\bold {\dfrac {\lambda_1 }{\lambda_2} = \dfrac {V_1}{V_2} = 0.25 \\ }[/tex]

Mean free path is 0.25 times the first mean free path.

C.

[tex]\bold {Eth= \dfrac {3}{2}kT}}\\\\\bold {\dfrac {E_t2}{E_t1} = \dfrac {T_2}{T_1}}\\\\\bold {E_t2= 2.5\ E_t1}[/tex]  

D.  the molar specific at a constant volume can be calculated a using,

[tex]\bold {CV= \dfrac 3{2}R }[/tex]  

 [tex]\bold {Cv_f= Cv_i }[/tex]

So, molar specific heat constant will not change.

Therefore, the rms speed of the system is 1.6 while Mean free path is 0.25 times the first mean free path.

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Answer:

a)   P' = P_original, b)  P ’= P_original + ρ  g Δh

Explanation:

The expression for nanometric pressure is

          P = ρ g h

where ρ  is the density of the liquid and h is the height

a) we change the radius of the barrel, but keeping the same height

as the pressure does not depend on the radius it remains the same

        P' = P_original

b) We change the barrel height

         h ’≠ h

we substitute in the equation

      P ’= ρ  g h’

      h ’= h + Δh

      P ’= ρ  g (h + Δh)

      P ’= (ρ  g h) + ρ  g Δh

      P ’= P_original + ΔP

In this case, the pressure changes due to the new height,

*if it is higher than the initial one, the pressure increases

*if the height is less than the initial one, the pressure is less

Answer: 1798

Explanation:

Given that there is no dielectric material between the plates,

the permittivity of free space = 9 × 10^-12 f/m

The frequency F = 10 MHz

The ratio of conduction current JC to the displacement current Jd is also known as loss tangent.

Please find the attached file for the solution

Answer:

The velocity of any point on a rigid body is ___Always perpendicular______ to the relative position vector extending from the IC to the point

Explanation:

This is because The instantaneous center (IC) of zero velocity for the rigid body is at the point in contact with ground. The velocity direction at any point on the body is always perpendicular to the line connecting the point to the IC.

The velocity of any point on a rigid body is : Perpendicular   to the relative position vector extending from the IC to the point.

The IC ( instantaneous center ) is the point of the rigid body in contact with the ground when the rigid body is at zero velocity.

The velocity direction from any point of the rigid body will always be perpendicular to the ( IC ) since the IC is the reference point of the rigid body when the body is at rest ( zero velocity ).

Hence we can conclude that the velocity of any point on a rigid body will be  perpendicular.

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Although the options related to your question is missing an accurate answer is provided within the scope of your question

Answer:

d poor accuracy,poor precision

The diagram represents how much accurate and precise the result data is. It is clear from the image that, the data points have poor accuracy and good precision.

Accuracy of a result is the measure of the closeness of the experimental or calculated value to the true value or absolute value of a measurement.  For a reproducible experiment, the results for each trials can be differ or close.

The closeness between values of a set of experiments is called precision of the results. Not all accurate value be can be precise. Similarly precision of values does not need to meet accuracy.

Here, the central dot indicates the true value. The green points are a set of experimental values. They are not so close to the true value and hence it is less accurate. However, all the points and closer together. Hence, they are precise. Therefore, option B is correct.

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Answer:

The distance covered by the cheetah during the motion is 77.75 m

Explanation:

Given;

initial velocity of cheetah, u = 0

final velocity of cheetah, v = 25 m/s

time of acceleration, t = 6.22 s

Apply kinematic equation;

[tex]s = (\frac{v+u}{2} )t[/tex]

where;

s is the distance covered by the cheetah during the motion

[tex]s = (\frac{v+u}{2} )t\\\\s = (\frac{25+0}{2} )6.22\\\\s = 77.75 \ m[/tex]

Therefore, the distance covered by the cheetah during the motion is 77.75 m

Answer:

A stop watch measures time and the number of seconds making an osculation is the period T

Which is T = 2π√(L/g)........solving for L (the length of the pendulum),

Then the L = T²x g/( 4πr²) substitute and then u find length or height of the swing

Answer:

Explanation:

d )

power in an electrical loop = volt x current

= 5.067sin(140πt)  x 2.03sin(140πt)

= 10.286 sin²(140πt) .

10.286 ( 1 - cos 280πt ) / 2

= 5.143 ( 1 - cos 280πt )

e )

Torque on a current carrying loop in a magnetic field

= M B sinθ

M is magnetic moment of coil , B is magnetic field , θ is angle between area vector of coil and direction of  B .

Here magnetic moment of coil

= A i where A is area of coil and i is current .

= .12² x 2.03sin(140πt)

= .029 sin(140πt)

B = .8 T

MB = .8 x .029 sin(140πt)

= .0232  sin(140πt) x sin(140πt)

=.0232 sin²(140πt).

Answer:

4km

Explanation:

d = st

d=0.25*16

d=4km

Answer:

126 g of water, H2O.

Explanation:

First, we'll begin by calculating the number of atoms in 7 mole oxygen gas.

This is illustrated below:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This means that 1 mole of O2 also contains 6.02×10²³ atoms.

Now, if 1 mole of O2 contains 6.02×10²³ atoms, then 7 moles of O2 will contain = 7 × 6.02×10²³ = 4.214×10²⁴ atoms.

Finally, we shall determine the mass of H2O that will contain 4.214×10²⁴ atoms.

This is illustrated below:

6.02×10²³ atoms is present in 1 mole of any substance contains

1 mole of H2O = (2x1) + 16 = 18 g

6.02×10²³ atoms is present in 18 g of H2O.

Therefore, 4.214×10²⁴ atoms will be present in = (4.214×10²⁴ × 18)/6.02×10²³

= 126 g of water, H2O.

Therefore, 126 g of water, H2O will contain the same number of oxygen atoms as 7.0 mol of oxygen gas.

7.0 mol of oxygen gas contain the same number of oxygen atoms as 250 g of water.

We want to know the mass of water (H₂O) that has the same number of oxygen atoms as 7.0 mol of oxygen gas (O₂).

First, we will calculate the number of oxygen atoms in 7.0 mol of O₂ considering the following relations.

[tex]7.0 mol O_2 \times \frac{6.02 \times 10^{23}molecule O_2 }{1molO_2} \times \frac{2atomO}{1molecule O_2} = 8.4 \times 10^{24} atomO[/tex]

Now, we want to calculate the mass of H₂O that contains 8.4 × 10²⁴ atoms of O. We will consider the following relations.

[tex]8.4 \times 10^{24} atomO \times \frac{1moleculeH_2O}{1atomO} \times \frac{1molH_2O}{6.02 \times 10^{23}moleculeH_2O } \times \frac{18.02gH_2O}{1molH_2O} = 250 gH_2O[/tex]

7.0 mol of oxygen gas contain the same number of oxygen atoms as 250 g of water.

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Explanation:

average speed = total distance travelled / total time travelled

time to travel the first 6km: 6 / 50 = 3/25 (h)

time to travel the next 6km: 6 / 90 = 1/15 (h)

[I think there's problem in the question 'cause 900km/h sounds impossible for normal person to travel in normal condition]

The total time: 3/25 + 1/15 = 14/75 (h)

Average speed over the 12 km drive will be:

[tex] \frac{12}{ \frac{14}{75} } = \frac{450}{7} = 64.3 \: km{h}^{ - 1} [/tex]

Answer:

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

I would appreciate it!

Answer:

[tex]R\approx12.43 \,\, \Omega[/tex]

Explanation:

We can use Ohm's Law to find the resistance R of a wire that carries a current I under a given potential difference:

[tex]V=I\,\,R\\R = \frac{V}{I} \\R=\frac{87}{7} \\R\approx12.43 \,\, \Omega[/tex]

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂   -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C   (OR)   q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC   (OR)   q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Answer:

b) The electric field inside is zero.

Explanation:

This is simply because the charges in the field will be in motion, thus this charges kind of spread around to make the conductor field zero.

Answer:

Time Period of Oscillation = 3.04 x 10⁻⁵ s

Explanation:

The time period of oscillation and the the frequency of oscillation are two inter linked quantities. They are actually the reciprocals of each other. So, the time period of oscillation of quartz crystal is given by the following formula:

Time Period of Oscillation = 1/Frequency of Oscillation

where,

Frequency of Oscillation = 32.8 KHz

Frequency of Oscillation = 32800 Hz

Therefore,

Time Period of Oscillation = 1/32800 Hz

Time Period of Oscillation = 3.04 x 10⁻⁵ s

Answer:

a

   [tex]f = 0.4848 \ m[/tex]

b

  [tex]p = 2.063 D[/tex]

Explanation:

From the question we are told that

   The  near point distance is  [tex]k = 45.5 \ cm[/tex]

    The  distance of the glasses from the eye is  [tex]y = 2.1 \ cm[/tex]

    The  distance of an object she can focus with the glass is  [tex]i= 25 \ cm[/tex]

Generally the image distance is mathematically evaluated  as

     [tex]v = -(45.5 - 2.1)[/tex]

     [tex]v = -43.4 \ cm[/tex]

Generally the object distance is mathematically represented as

     [tex]u = (25 -2.1)[/tex]

     [tex]u = 22.9 \ cm[/tex]

The  negative sign tells us that the image was formed behind the eye  

Generally the lens formula  is mathematically represented as  

        [tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]

=>   [tex]\frac{1}{f} = \frac{1}{22.9} + \frac{1}{ - 43.4}[/tex]

=>  [tex]f = 48.48 \ cm[/tex]

 converting to meters  

   [tex]f = 0.4848 \ m[/tex]

Thus the refractive power is mathematically represented as

    [tex]p = \frac{1}{f}[/tex]

=>  [tex]p = \frac{1}{0.4848 }[/tex]

=>   [tex]p = 2.063 D[/tex]

Answer:

1)  speed of a cyclist = 100 m/min.

2) speed of a car moving = 3 km/min.

Explanation:

1)  speed of a cyclist = distance over time

                                  = 2000 m / 20 min

                                  = 100 m/min.

2) speed of car moving = distance over time

                                       = 60 km / 20 min

                                        = 3 km/min.

Answer:

Mass of flake = 67.716 kg

Explanation:

Given:

Length of lawn = 21 ft

Width of lawn = 20 ft

Note:

Each snow flake mass = 1.90 mg

Find:

Weight of total flake in a minute

Computation:

Area of lawn = 21 × 22

Area of lawn = 440 ft²

Amount of flake per minute = 440 × 1350

Amount of flake per minute = 594000 flake/ minutes

Mass of flake = 594000 × 1.90 mg × 60 minutes

Mass of flake = 67,716,000 mg

Mass of flake = 67.716 kg

Answer:

Explanation:

The volume rate of flow = a x v where a is cross sectional area of pipe and v is velocity of flow

putting the values

π x .2945² x 12.1

= 3.3  m³ /s

To know the pipe's diameter at the refinery we shall apply the following formula

a₁ v₁ = a₂ v₂

a₁ v₁ and  a₂ v₂ are volume rate of flow of liquid which will be constant .

3.3 = a₂ x 6.29

a₂ = .5246 m³

π x r² = .5246

r = .4087 m

= 40.87 cm

diameter

= 81.74 cm

Answer:

Given that K.E is

1/2mv²

So to find speed v,

Make it subject

K.E= 1.2mv²

However radial force = magnetic force

So mv²/r= qvB

So v subject

V= 2K.E/ qBr that is speed

To find mass

K.E = 1/2mv²

Puy value of v

So KE= 1/2m(2K.E/qBr)

m= (qBr)/2K.E

That is mass

Answer:

m = qbr/v

v = 2k/qbr

Explanation:

When a charged particle enters a magnetic field, it experiences a force that is always perpendicular to the velocity. This force provides a centripetal force, and thus, we have

qvb = mv²/r

if we make m the subject of the formula, we will have

m = qbr/v

Recall that the kinetic energy, KE = ½mv²

Now, let's make v² the subject of formula, we have

v² = 2K/m

now, we substitute for m from the equation we got earlier

v² = 2K / (qbr/v)

v² = 2Kv / qbr, if we simplify further, we have

v = 2k / qbr

Therefore, we can say that the expression for the mass and speed is respectively,

m = qbr/v

v = 2k/qbr

Answer:

It is higher.

Explanation:

Hello,

In this case, the quantity of energy that a wave is able to transfer is directly related its amplitude exhibiting that the higher the carried energy, the higher its amplitude and on the flip side, the lower the energy the lower the amplitude, therefore, the answer is it is higher. You can verify this on the attached picture.

Regards.

Answer:

It is higher.

Explanation:

Answer:

11 mm

Explanation:

Original length L' = 26 m

Increase in temperature dT = 36.0°C

The gap l =?

The coefficient of linear expansion for steel & = 12 × 10^−6 per °C

The gap will be gotten from the equation of linear expansion.

l = L' x & x dT

substituting, we have

L = 26 x 12 × 10^−6 x 36

L = 0.011 m = 11 mm

Answer:

No, life would not be to develop

Explanation:

Stars less massive than about 2 solar masses can only produce natural element up to carbon and oxygen, which are the basic elements for building life. However, other more massive elements are needed by life to thrive and function properly, more massive elements like phosphorus, iron necessary for oxygen circulation, calcium for a strong support system, and silicon are essential for for life to form and be sustained here on Earth.

Answer:

The most reactive elements are at the bottom left corner of the periodic table.  Those are the elements that have the most active/most reactive.  For example, lithium, sodium, and potassium all react with water!

Explanation:

Don't know what to explain :v

Answer:

18808.7 m/s^2

Explanation:

Given

Length of the pendulum L = 1.44 m

Number of complete cycles of oscillation n = 1.10 x 10^2

total time of oscillation t = 2.00 x 10^2 s

The period of the T = n/t

T = (1.10 x 10^2)/(2.00 x 10^2) = 0.55 ^-s

The period of a pendulum is gotten as

T = [tex]2\pi \sqrt{\frac{L}{g} }[/tex]

where g is the acceleration due to gravity

substituting values, we have

0.55 = [tex]2\pi \sqrt{\frac{1.44}{g} }[/tex]

0.0875 = [tex]\sqrt{\frac{1.44}{g} }[/tex]

squaring both sides of the equation, we have

7.656 x 10^-3 = 144/g

g = 144/(7.656 x 10^-3) = 18808.7 m/s^2

Answer:

6

Explanation:

Given that

dsinθ = mλ,

now, if sinθ = 1, then

m = d / λ, where

m = order of interference

d = distance between the slits

λ = wavelength of light

this is the formula we would use to solve the question

d = 1 / 320 lines/mm

d = 1 / 320*10^3

d = 3.125*10^-6 m

At λ = 551 nm, we have

m = 3.125*10^-6 / 551*10^-9

m = 5.67

5.67 ~ 6

thus, we can say that the orders of visible wavelength 551 nm, can produce is 6