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Answer:

F = 6666.7 N

Explanation:

Given that,

Mass of a chip, m = 0.1 mg

Initial speed, u = 0

Final speed,[tex]v=4\times 10^{3}\ m/s[/tex]

Time of collision,[tex]t=6\times 10^{-8}\ s[/tex]

We know that,

Force, F = ma

Put all the values,

[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.1\times 10^{-6}\times (4\times 10^3-0)}{6\times 10^{-8}}\\\\F=6666.7\ N[/tex]

So, the required force is 6666.7 N.

Answer:

changing the N to S. that's how the error will be corrected

Answer:

C is the correct answer

Explanation:

i took the test

Answer:

a)  [tex]F_p=882N[/tex]

b)  [tex]P=4410W[/tex]

c)  [tex]V_p'=24135[/tex] ,[tex]n=15.2\%[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=1500kg[/tex]

Velocity [tex]v=4.9m/s[/tex]

Coefficient of Rolling Friction [tex]\mu=0.06[/tex]

a)

Generally the equation for The Propulsion Force is mathematically given by

 [tex]F_p=\mu*mg[/tex]

 [tex]F_p=0.06*1500*9.81[/tex]

 [tex]F_p=882N[/tex]

b)

Therefore Power Required at

 [tex]V_p=5.0m/s[/tex]

 [tex]P=F_p*V_p[/tex]

 [tex]P=882*5[/tex]

 [tex]P=4410W[/tex]

c)

 [tex]V_p' =15mpg[/tex]

 [tex]V_p'=15*\frac{1609}[/tex]

 [tex]V_p'=24135[/tex]

Generally the equation for Work-done is mathematically given by

 [tex]W=F_p*V_p'[/tex]

 [tex]W=882*15*1609[/tex]

 [tex]W=2.13*10^7[/tex]

Therefore

Efficiency

 [tex]n=\frac{W}{E}*100\%[/tex]

Since

Energy in one gallon of gas is

 [tex]E=1.4*10^8J[/tex]

Therefore

 [tex]n=\frac{2.1*10^7}{1.4*10^8}*100\%[/tex]

 [tex]n=15.2\%[/tex]

Answer:

A(predicts an organisms ability to adapt to its enviroment, it is not a fact that each organization can adapt)

Explanation:

Answer:

Atmospheric pressure at Badwater is 1.01022 atm

Explanation:

Data given:

1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa

Elevation (h) = 86m

gravity (g) = 9.8 m/s2

Density of air P = 1.225 kg/m3

Therefore pressure at bad water Pb = Pi + Pgh

Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)

Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa

hence:

Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm

Answer:

Use a lighter string of the same length, under the same tension.

Stretch the string tighter to increase the tension

Explanation:

The wave speed depends on propertices of the medium, not on how you generate the wave. For a string

Increasing the tension or decreasing the linear density (lighter string) will increase the wave speed.

A pulse will be sent down a horizontal extended thread if a person flicks the wrist. To raise the tension, pull the string tighter.

Pulse is the same thing as monitoring heartbeat. The pulse rate could also be measured via auscultation, which includes hearing to the heart rhythm using a stethoscope then counting it for one minute.

A pulse will be sent down a horizontal extended thread if a person holds one end of the rope and flicks their wrist.

To raise the tension, pull the string tighter.

Find out more information about pulse here:

https://brainly.com/question/14524756?referrer=searchResults

Answer:

8m/s

Explanation:

Vavg= 16-0/2=8m/s

Answer:

15cm

Explanation:

Since the lens is a diverging lens, the image distance is negative (virtual)

v = -30cm

u = 10cm

Required

focal length f

Using the lens formula;

1/u + 1/v = 1/u

1/10 - 1/30 = 1/f

(3-1)/30 = 1/f

2/30 = 1/f

f = 30/2

f = 15cm

Hence the focal length of the lens is 15cm

Answer:

686.7N

Explanation:

Given data

Mass= 70kg

We know that the buoyant force experienced by the person is equal to the weight of the person

Hence the weight is

Weight = mass* Acceleration

Weight= 70*9.81

Weight= 686.7N

Therefore the weight is 686.7N

Answer:

Hence the amount of heat transfer is 918.75 Btu.

Explanation:

Now,

Answer:

Myopia curvature of the cornea if it is negative the curvatures are positive,

hypermetry,

Explanation:

Myopia is the visual defect that does not allow to see distant objects, which is why it is corrected with a divergent lens so that the image is formed on the retina, therefore, by reforming the curvature of the cornea if it is negative

therefore the curvature must decrease

To correct hypermetry, the curvatures are positive, consequently the curvature of the lens must increase

Answer:

Answer: Think its true

Let the point where A is launched act as the origin, so that the horizontal positions at time t of the respective projectiles are

• A : x = (19.5 m/s) cos(30°) t

• B : x = 62.2 m + (19.5 m/s) cos(60°) t

These positions are the same at the moment one projectile is directly above the other, which happens for time t such that

(19.5 m/s) cos(30°) t = 62.2 m + (19.5 m/s) cos(60°) t

Solve for t :

(19.5 m/s) (cos(30°) - cos(60°)) t = 62.2 m

t = (62.2 m) / ((19.5 m/s) (cos(30°) - cos(60°))

t ≈ 8.71 s

Answer:

(b)

Explanation:

The voltage always lags the current by 90°, regardless of the frequency.

Answer:

stationary waves are transverse waves

Answer:

4.7 GHz

Explanation:

Applying,

v = λf................. Equation 1

Where v = velocity of the radio wave, λ = wavelength, f = frequency

make f the subject of the equation

f = v/λ.............. Equation 2

Note: A radio wave is an electromagnetic wave, as such it moves with a velocity of 3.00 x 10⁸ m/s

From the question,

Given: λ = 0.0644 meters

Constant: v = 3.00 x 10⁸ m/s

Substitute these values into equation 2

f = (3.00 x 10⁸)/0.0644

f = 4.66×10⁹ Hz

f = 4.7 GHz

The image is missing and so i have attached it.

Answer:

A) E = 8740 N/C

B) E = -6536 N/C

Explanation:

The formula for electric field is;

E = kq/r²

Where;

q is charge

k is a constant with value 8.99 x 10^(9) N•m²/C²

A) Now, to find the net electric field at point A, the formula would now be;

E = (kq1/(r1)²) - (kq2/(r2)²)

Where;

r1 is distance from charge q1 to point A

r2 is distance from charge q2 to point A.

q1 = -6.25 nC = -6.25 × 10^(-9) C

q2 = -12.5 nC = -12 5 × 10^(-9) C

From the attached image, r1 = 25 cm - 10 cm = 15 cm = 0.15 m

r2 = 10 cm = 0.1 m

Thus;

E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.15^(2)) - ((-12.5 × 10^(-9))/0.1^(2))

E = 8740 N/C

B) similarly, electric field at point B;

E = (kq1/(r1)²) + (kq2/(r2)²)

Where;

r1 is distance from charge q1 to point B

r2 is distance from charge q2 to point B.

q1 = -6.25 nC = -6.25 × 10^(-9) C

q2 = -12.5 nC = -12 5 × 10^(-9) C

From the attached image, r1 = 10 cm = 0.1 m

r2 = 25cm + 10 cm = 35 cm = 0.35 m

Thus;

E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.1^(2)) + ((-12.5 × 10^(-9))/0.35^(2))

E = -6536 N/C

Answer:

7772.72N

Explanation:

When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page  (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.

Now which direction is the static friction, assume that it is pointing inward so

Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N

Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.

Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

[tex]\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s[/tex]

[tex]N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns[/tex]

Therefore, the number of turns of wire needed is 573.8 turns

Answer:

T = 13.25°C

Explanation:

From the law of conservation of energy:

Heat Lost by Watermelon = Heat Gained by Cans

[tex]m_wC_w\Delta T_w = m_cC_c\Delta T_c[/tex]

where,

[tex]m_w[/tex] = mass of watermelon = 6.48 kg

[tex]m_c[/tex] = mass of cans = (12)(0.35 kg) = 4.2 kg

[tex]C_w[/tex] = specific heat capacity of watermelon = 3800 J/kg.°C

[tex]C_c[/tex]  = specific heat capacity of cans = 4200 J/kg.°C

[tex]\Delta T_w[/tex] = Change in Temprature of watermelon = 29.4°C - T

[tex]\Delta T_c[/tex] = Change in Temperature of cans = T - 3.78°C

T = final temperature = ?

Therefore,

[tex](4.2\ kg)(3800\ J/kg.^oC)(29.4^oC-T)=(6.48\ kg)(4200\ J/kg^oC)(T-3.78^oC)\\469224\ J-(15960\ J/^oC)T = (27216\ J/^oC)T-102876.48\ J\\469224\ J + 102876.48\ J = (27216\ J/^oC)T+(15960\ J/^oC)T\\\\T = \frac{572100.48\ J}{43176\ J/^oC}[/tex]

T = 13.25°C

Answer:

C

Explanation:

20 cm = 0.2m

since uncertainty is 0.1 cm (0.001 m), should be recorded to same number of decimal place as uncertainty

therefore it's 0.200m

Answer:

0.013 m/min

Explanation:

Applying,

dV/dt = (dh/dt)(dV/dh)............. Equation 1

Where

V = πr²h................ Equation 2

Where V = volume of the tank, r = radius, h = height.

dV/dh = πr²............ Equation 3

Substitute equation 3 into equation 1

dV/dt = πr²(dh/dt)

From the question,

Given: dV/dt = 2 m³/min, r = 7 m, π = 3.14

Substitute these values into equation 3

2 = (3.14)(7²)(dh/dt)

dh/dt = 2/(3.14×7²)

dh/dt = 0.013 m/min

Answer:

Nonrenewable energy

Explanation:

Renewable energy is also known as clean energy and it can be defined as a type of energy that are generated through natural sources or technology-based processes that are replenished constantly. Some examples of these natural sources are water (hydropower), wind (wind energy), sun (solar power), geothermal, biomass, waves etc.

Basically, a renewable energy source is sustainable and as such can not be exhausted.

On the other hand, a non-renewable energy refers to an energy source such as fossil fuels that takes a very long time to be created or their creation happened long ago and isn't likely to happen again e.g uranium.

For example, fossil fuels such as coal, oil, and natural gas, come from deep inside the Earth where they formed over millions of years ago.

In this scenario, the kind of energy the newspaper sources are talking about is a nonrenewable energy source because they are capable of being exhausted in the not-so-distant future.

Answer:

Torques must balance

F1 * X1 = F2 * Y2

or M1 g X1 = M2 g X2

X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7

X2 = 22.4 cm

Torque = F1  * X2 =

62.3 gm* 980 cm/sec^2  * 22.4 cm = 137,000 gm cm^2 / sec^2

Normally x cross y   will be out of the page

r X F  for F1 will be into the page so the torque must be negative

(a) As it gets compressed by a distance x, the spring does

W = - 1/2 (52.1 N/m) x ²

of work on the object (negative because the restoring force exerted by the spring points in the opposite direction to the object's displacement). By the work-energy theorem, this work is equal to the change in the object's kinetic energy. At maximum compression x, the object's kinetic energy is zero, so

W = ∆K

- 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²

==>   x ≈ 0.118 m

(b) Taking friction into account, the only difference is that more work is done on the object.

By Newton's second law, the net vertical force on the object is

∑ F = n - mg = 0

where n is the magnitude of the normal force of the track pushing up on the object. Solving for n gives

n = mg = 2.45 N

and from this we get the magnitude of kinetic friction,

f = µn = 0.120 (2.45 N) = 0.294 N

Now as the spring gets compressed, the frictional force points in the same direction as the restoring force, so it also does negative work on the object:

W (friction) = - (0.294 N) x

W (spring) = - 1/2 (52.1 N/m) x ²

==>   W (total) = W (friction) + W (spring)

Solve for x :

- (0.294 N) x - 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²

==>   x ≈ 0.112 m

For the 0.250 kg object moving along a horizontal track and collides with and compresses a light spring, with a spring constant of 52.1 N/m, we have:

a) The spring's maximum compression when the track is frictionless is 0.118 m.

b) The spring's maximum compression when the track is not frictionless, with a coefficient of kinetic friction of 0.120 is 0.112 m.

a) We can calculate the spring's compression when the object collides with it by energy conservation because the track is frictionless:

[tex] E_{i} = E_{f} [/tex]

[tex] \frac{1}{2}m_{o}v_{o}^{2} = \frac{1}{2}kx^{2} [/tex]  (1)

Where:

[tex]m_{o}[/tex]: is the mass of the object = 0.250 kg

[tex]v_{o}[/tex]: is the velocity of the object = 1.70 m/s

k: is the spring constant = 52.1 N/m

x: is the distance of compression

After solving equation (1) for x, we have:

[tex] x = \sqrt{\frac{m_{o}v_{o}^{2}}{k}} = \sqrt{\frac{0.250 kg*(1.70 m/s)^{2}}{52.1 N/m}} = 0.118 m [/tex]

Hence, the spring's maximum compression is 0.118 m.

b) When the track is not frictionless, we can calculate the spring's compression by work definition:

[tex] W = \Delta E = E_{f} - E_{i} [/tex]

[tex] W = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} [/tex]   (2)

Work is also equal to:

[tex] W = F*d = F*x [/tex]     (3)

Where:  

F: is the force

d: is the displacement = x (distance of spring's compression)  

The force acting on the object is given by the friction force:

[tex] F = -\mu N = -\mu m_{o}g [/tex]   (4)

Where:

N: is the normal force = m₀g

μ: is the coefficient of kinetic friction = 0.120

g: is the acceleration due to gravity = 9.81 m/s²

The minus sign is because the friction force is in the opposite direction of motion.

After entering equations (3) and (4) into (2), we have:

[tex]-\mu m_{o}gx = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2}[/tex]

[tex]\frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} + \mu m_{o}gx = 0[/tex]

[tex] \frac{1}{2}52.1 N/m*x^{2} - \frac{1}{2}0.250 kg*(1.70)^{2} + 0.120*0.250 kg*9.81 m/s^{2}*x = 0 [/tex]        

Solving the above quadratic equation for x

[tex] x = 0.112 m [/tex]  

Therefore, the spring's compression is 0.112 m when the track is not frictionless.

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Answer:

See the answer below

Explanation:

After seeing an object on a slide at the low-power objective of the microscope and it disappears on changing to high power, the following can be done to resolve the problem

1. Drop a few drops of immersion oil on the slide and view again under high the power objective.

2. If the object is still not visible after the action above, return the microscope to the low-power objective and make sure the object is refocused and centered. Then carefully change back to the high power objective and use the fine adjustment to bring it into focus.

Answer:

True

Explanation:

A magnetohydrodynamic drive or MHD accelerator is a method which is used for propelling the vehicles using only by applying the electric and magnetic fields. It has no moving parts. It accelerates an electrically conductive propellant (liquid or gas) with magnetohydrodynamics.

Its working principle is same as an electric motor except that in an MHD drive, the moving rotor is replaced by the fluid acting directly as the propellant.

An MHD accelerator is reversible.

So,  the statement is true.

Answer:

It is a example of physical change