A ship sets out to sail to a point 123 km due north. An unexpected storm blows the ship to a point 112 km due east of its starting point. (a) How far and (b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination

Answer:

The distance the car travels is 115500 m in S.I units

Explanation:

Distance d = vt where v = speed of the car and t = time taken to travel

Now v = 99 km/h. We now convert it to S.I units. So

v = 99 km/h = 99 × 1000 m/(1 × 3600 s)

v = 99000 m/3600 s

v = 27.5 m/s

The speed of the car is 27.5 m/s in S.I units

We now convert the time t = 70 minutes to seconds by multiplying it by 60.

So, t = 70 min = 70 × 60 s = 4200 s

The time taken to travel is 4200 s in S.I units

Now the distance, d = vt

d = 27.5 m/s × 4200 s

d = 115500 m

So, the distance the car travels is 115500 m in S.I units

Answer:

The nail exerts an equal force on the hammer in the opposite direction.

Explanation:

The Newtons third law states that there is an equal an opposite reaction for every action. When hammer pushes the nail, the nail will push the hammer back in opposite direction. When the hammer hits a nail then nail will exert the equal and opposite force to the hammer. These both objects will exert force on each other in opposite directions.

Hello, I am BrotherEye

Answer:

Answers are

1. "The nail exerts an equal force on the hammer in the opposite direction."

2. "500 N"

3. "The iron piece exerts a force of 1 N on the magnet in the opposite direction."

4. "When mass moves closer to the point of rotation, rotational inertia decreases."

5. "The skater spins slower because his rotational inertia has increased."

Explanation:

Answer; 4.4cm

Explanation: There is a ruler upon which the cylinder start from 1.4cm and reaches 10.2cm

distance traveled =10.2-1.4=8.8

since this cylinder is small so the linear distance can be approximately taked as rotational distance(as in case of point charge) so

2x2πxr =8.8

so the circumference will be 2πr=4.4cm

Answer:

c

Explanation:

The marginal cost curve image  has been attached from which we can clearly, indicate that

ATC = average total cost

AFC = average fixed cost

AVC = average variable cost.

From the graph we can indicate that the marginal cost curve

(c) Lies above the AVC curve when the AVC curve slopes downward.

Derived Quantities

Answer:

These are quantities calculated from two or more measurements

Explanation:

They can't me measured directly.

They can only be computed.

They are calculated in PHYSICAL SCIENCE.

hope it helps.

Answer:

=33 .3×10^6Ω

=33.3M Ω

Explanation:

We were told to calculate the Resistance value,

Given the heart pacemaker fires as 72 times a minute, which is the time constant

Then we can convert the pacemaker fires of 72 times a minute to seconds for unit consistency.

1 minutes= 60secs

Then ,Time constant τ=60secs/72=0.8333 seconds

Time constant τ can be calculated using the formula below

τ= RC

Where R= resistance

C = capacitance

Then making RESISTANCE subject of formula we have

R=τ/C

But Capacitance=25.0-nF = 25×10^25F

Substitute the values we have

R=0.833/25×10^25

=33 .3×10^6 Ω

But can still be converted to M Ω= 33.3M Ω

Therefore, the resistance is 33 .3×10^6 Ω or 33.3M Ω

NOTE: 1M= 10^6

The value of the resistance will be "3.33×10⁷ Ω".

According to the question,

Capacitor, C = 25.0 nF

60 sec - 72 fires

now,

Time for 1 fire,

[tex]t = \frac{60}{72}[/tex]

 [tex]= \frac{5}{6} \ sec[/tex]

Now,

⇒            [tex]V = V_0 (1-e^{-\frac{t}{RC} })[/tex]

     [tex]0.632 V_0=V_0(1-e^{\frac{-\frac{5}{6} }{R.25 n C} })[/tex]

[tex]e^{-\frac{1}{R\times 30\times 10^{-9}} } = 1-0.632 = 0.368[/tex]

                  [tex]= 2.72[/tex]

By taking "log" both sides,

⇒ [tex]ln \ e^{\frac{1}{R\times 30\times 10^{-9}} } = ln \ 2.72[/tex]

hence,

The Resistance be:

⇒ [tex]R = \frac{10^9}{30}[/tex]

       [tex]= 3.33\times 10^7 \ \Omega[/tex]

Thus the above approach is right.

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Answer:

The hypothesis may or may not be true and needs to be tested. It might be the answer to the problem. Hence, it must be tested thoroughly. When these predictions are tested again and again in independent scientific experiments and gets verified, the hypothesis is converted into a scientific theory.

Answer:

v1 = −2.201946 m/s ( to the left)

v2 = 0.7780534 m/s ( to the right)

Explanation:

Given the following :

Mass of first glider (m1) = 0.149kg

Initial Speed of first glider (u1) = 0.710 m/s

Mass of second glider (m2) = 0.308kg

Initial Speed of second glider (u2) = 2.27m/s

For elastic collision:

m1u1 + mu2u2 = m1v1 + m2v2

Where V1 and v2 = final velocities if the body after collision.

Taking right as positive ; left as negative

u1 = 0.710m/s ; u2 = - 2.27m/s

u1 - u2 = - (v1 - v2)

0.710 - - 2.27 = - v1 + v2

v2 - v1 = 2.98 - - - - (1)

From:

m1u1 + mu2u2 = m1v1 + m2v2

(0.149 * 0.710) + ( 0.308 * - 2.27) = (0.149 * v1) + (0.308 * v2)

0.10579 + (-0.69916) = 0.149 v1 + 0.308v2

−0.59337 = 0.149 v1 + 0.308v2

Dividing both sides by 0.149

v1 + 2.067v2 = −0.59337 - - - - - (2)

From (1)

v2 = 2.98 + v1

v1 + 2.067(2.98 + v1) = −0.59337

v1 + 6.16 + 2.067v1 = −0.59337

3.067v1 = −0.59337 - 6.16

3.067v1 = −6.75337

v1 = −6.75337 / 3.067

v1 = −2.201946 m/s ( to the left)

From v2 = 2.98 + v1

v2 = 2.98 + (-2.201946)

v2 = 0.7780534 m/s ( to the right)

Answer:

First, as you may know, the light travels at a given velocity.

In vaccum, this velocity is c = 3x10^8 m/s.

And we know that:

distance = velocity*time

Now, if some object (like a star ) is really far away, the light that comes from that star may take years to reach the Earth.

This means that the images that the astronomers see today, actually happened years and years ago (So the night sky is like a picture of the "past" of the universe)

Also, for example, if an astronomer sees some particular thing, he can apply a model (a "simplification" of some phenomena that is used to simplify it an explain it) and with the model, the scientist can infer the information of the given thing some time before it was seen.

The astronomers could know what was happening inside galaxies way back then by the fact that;

they examine the spectra of galaxies (or the overall colors of galaxies) with the highest redshifts they can find

Astronomers Measure the wavelength of the light that is stretched, so the light is seen as 'shifted' towards the red part of the spectrum by using spectroscopy. This measure is also called redshift.

This invokes a ray of light through a triangular prism that splits the light into various components known as spectrum.

The way the astronomers could use this concept to know what was happening in the galaxies before is by examining the spectra of galaxies that have the highest redshifts.

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Answer:

5000 N/m².

Explanation:

The following data were obtained from the question:

Density (d) = 1000 kg/m³

Acceleration due to gravity (g) = 10 N/kg

Height (h) = 50 cm = 50/100 = 0.5 m

Pressure (P) =.?

Pressure is related to density and height by the following equation:

P = dgh

Where

P is the pressure.

d is the density.

g is the acceleration due to gravity.

h is the height.

With the above formula, we can obtain the pressure at the bottom of the cup as follow:

P = dgh

P = 1000 x 10 x 0.5

P = 5000 N/m²

Therefore, the pressure at bottom of the cup is 5000 N/m².

Answer:

66.86m

Explanation:

Velocity of ball thrown, u = 23.5 m/s

Initial height of the ball above the water, H = 11.5 m

Angle of projection, θ = 39.5°

Vertical components of veloclty = usinθ

Horizontal components of veloclty = ucosθ

The soft ball hits the water after time 't'

Considering the second equation of motion

S = ut + 1/2at^2........ 1

But since the ball went through motion under gravity ( free fall ) rather than linear motion, then equation 1 can be rewritten as:

H = ut +/- 1/2gt^2

H = - 11.5m

U = usinθ

θ = 39.5°

a = -g = -9.8m/s^2

- 11.5m = 23.5(sin39.5°)t + 1/2(-9.8)t^2

-11.5m = 23.5(0.6360)t - 4.9t^2

-11.5m = 14.946t - 4.9t^2

4.9t^2 -14.946t-11.5m = 0

Since the ball drifted horizontally

D = (Ucosθ)t

Where θ = 39.5°

U = 23.5m/s t=

Alternatively,

horizontal component of the velocity is 23.5 cos 39.5º = 18.1331 m/s

now how long does it take the ball to raise to a peak and fall to the water.

vertical component of velocity = 23.5 sin 39.5º = 14.947m/s

time to reach peak t = v/g = 11.947/9.8 = 1.5252 sec

peak reached above cliff top is

h = ½gt² = ½(9.8)(1.5252)²

= ½×22.797

= 11.3985m

now the ball has to fall 11.3985+ 11.5 = 22.8985m

time to fall from that height is

t = √(2h/g) = √(2• 22.8986/9.8) = 2.1617 sec

add up the two times to get time it is in the air, 2.1617 + 1.5252 = 3.6869

now haw far does the ball travel horizontally in that time

d = vt = 18.1331 ×3.6869= 66.856m

= 66.86m

Answer:

Rounded to three significant figures:

(a) [tex]2 \times 575\; \rm nm = 1150\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].

(b) [tex]\displaystyle \left(1 + \frac{1}{2}\right) \times (575\;\rm nm) \approx 863\; \rm nm = 8.63\times 10^{-7}\; \rm m[/tex].

Explanation:

Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.  

Let [tex]k[/tex] denote a natural number ([tex]k \in \left\lbrace0,\, 1,\, 2,\, \dots\right\rbrace[/tex].) In a double-split experiment of a monochromatic light:

The path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where [tex]k = 0[/tex] in [tex]\text{Path difference} = 0[/tex].

The path difference increases while moving on the screen away from the center. The first order maximum is at [tex]k = 1[/tex] where [tex]\text{Path difference} = \lambda[/tex].

Similarly, the second order maximum is at [tex]k = 2[/tex] where [tex]\text{Path difference} = 2\, \lambda[/tex]. For the light in this question, at the second order maximum: [tex]\text{Path difference} = 2\, \lambda = 2 \times 575\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].

The dark fringe closest to the center of the screen is the first minimum. [tex]\displaystyle \text{Path difference} = \left(0 + \frac{1}{2}\right)\cdot \lambda = \frac{1}{2}\, \lambda[/tex] at that point.

Add one wavelength to that path difference gives another dark fringe- the second minimum. [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex] at that point.

For the light in this question, at the second order minimum: [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda = \left(1 + \frac{1}{2}\right)\times (575\; \rm nm) \approx 8.63\times 10^{-7}\; \rm m[/tex].

Answer:

5.3333 sec

Explanation:

initial speed: u = 108km/hr or 30 m/s

final speed: v = 0m/s

distance travelled: s = 80m

time the car took to stop: = t sec

[tex]v^{2} - u^{2}[/tex] = 2as,

a = ([tex]v^{2} - u^{2}[/tex])/2s

a = (0-900)/160

a = -5.625 [tex]ms^{-2}[/tex]

v = u + at,

t = (v - u)/a

t= (0 - 30)/(-5.625)

t = 5.3333 sec

Answer:

B.

Explanation:

as there is a '-' sign before the magnification value, it forms a real image.

so the last two options get cancelled.

as the value of magnification is 0.87 ie. lesser than 1, we can say that the image is smaller.

The type of image is formed by a lens if m = -0.87 would be smaller than the object and the object is real

A lens is a transmissive optical tool that employs refraction to focus or disperse a light beam. The power of the lens is expressed in the dioptre which is the reciprocal of the focal length of the lens.

The image formation through the lens is calculated with the help of the lens formula given as follows

1/f = 1/v - 1/u

where f is the focal length of the lens

u is the distance of the object from the lens

v is the distance of the image formed from the

magnification (m) = size of image /size of the object

As given in the problem image is formed by a lens if m = -0.87

Thus, the negative sign represents that the image formed by the object is real, and the magnitude of the magnification is less than 1 this represents that size of the image is smaller than the size of the object.

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Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

Answer:1m

Explanation:

2n=0.4m

5n=?

5n×0.4/2n=1m

Answer:

The correct option is;

3. The total mechanical energy is constant as the block moves back and forth

Explanation:

The total mechanical energy is the sum of the potential and kinetic energies of the system

For a system that is isolated from the effects of external forces, but being acted upon by the internal conservative forces within the system, the total mechanical energy is constant

For a black and spring system, we have total mechanical energy, E = 1/2×K×A².

Where;

K = Constant

A = The amplitude of motion

Therefore, where there is no loss to friction, with A, remaining constant, the total mechanical energy will be constant.

Answer:

Explanation:

Energy of gamma ray = 2.2233 MeV

Let mass of neutron be n  amu  

mass defect of deuteron = 2.013553212 - ( 1.007 276 467 + n ) u .

in terms of energy this mass defect will be equal to energy of gamma ray

1 amu = 931 MeV

931 [ 2.013553212 - ( 1.007 276 467 + n ) ] =  2.2233

( 1.007 276 467 + n ) -  2.013553212  = .00238807733

n = 1.008664822 amu

so mass of neutron = 1.008664822 amu

Answer:

12 i. The work done by Wale = 107.910 kJ

The work done by Lekan = 117.720 kJ

Total work done = 225.36 kJ

ii. Wale's power =  4.3164 kW

Lekan's power = 3.924 kW

Wale has more power and is more powerful than Lekan

13. 313.92 N

Explanation:

i. The work done, W = Force, F × Distance moved by the force, D

The given parameters are

The mass of Wale = 55 kg

The mass of Lekan = 60 kg

The acceleration due to gravity, g =9.81 m/s²

The motion force of Wale and Lekan are;

Motion force of Wale = 9.81 × 55 = 539.55 N

Motion force of Lekan = 9.81 × 60 = 588.6 N

The work done by Wale = 539.55 × 200 = 107910 J = 107.910 kJ

The work done by Lekan= 588.6 × 200 = 117720 J = 117.720 kJ

107910 + 117720 =225630 J = 225.36 kJ

ii. Power = Work done/time

Wale finished the race in 25 s, therefore, his power = 107910/25 = 4316.4 W

Lekan finished the race in 30 s, therefore, his power = 117720/30 = 3924 W

Wale has more power and is more powerful than Lekan

13. The velocity ratio = 5

V. R. = Distance moved by effort/(Distance moved by load)

Efficiency = 80%

Work done by effort = x

Work done by machine = Efficiency × Work done by effort  = 0.8 × x

Distance moved by effort, E = V. R. × Distance moved by load, D = 5 × D

Work done by effort = Force × Distance moved = 200×9.81× E

Work done by effort = 1962×E = 1962×E = 1962×5×D

Work done by machine = 1962 × D, when D = 1, we have;

0.8 × 1962×1 = 1569.6 J

Work done by effort = Force × Distance moved

Work done by effort = Force × 5×D = Force × 5 (D = 1)

From the principle of conservation of energy, we have;

Energy is neither created nor destroyed

Therefore

Work done by effort = Force × 5 = 1569.6 J

Force = 1569.6 /5 = 313.92 N.

Answer:

A dark fringe

Answer:

Diffusion depends among many other things also upon the concentration gradient of the diffusing substance.For example if there are two boxes with given particles first isolated from each other and if they are bought in contact, then there is a net diffusion of particles from a box with higher concentration of particles to box with lower concentration. And also diffusion rate depends at any instant directly on the concentration difference between them at that instant.Now the vaccum is equivalent to an empty box which means with this one leads to a maximum diffusion rate when bought in contact with a box with particles because there is maximum concentration difference.That is vaccum is empty ( except for energy perturbations and a small concentration of particles which will be zero for our purpose) and any bunch of particles will find least resistance to diffuse as just outside this bunch there is ‘0' concentration of the particles.

Answer:

10N

Explanation:

Applying the Hooke law:

F = kx

F: Force

k: stiffness coefficient

x: stretched distance

F = 10N/m x 1m = 10N

Answer:

585×10⁸ m

Explanation:

Distance = rate × time

d = (2.998×10⁸ m/s) (3.25 min) (60 s/min)

d = 585×10⁸ m

Answer:

They are:

1) change position

2) distract yourself

3) Get fresh air

4) Face the direction you are going.

5) Drink water.

6) Play music.

7) Put your eyes on horizon.  

Explanation:

Hope it helps.

Explanation:

It is given that,

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

[tex]x=4t^3+3t^2-5t+2[/tex]

Where,

x is in meters and t is in sec

We know that,

Velocity,

[tex]v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5[/tex]

(a) i. t = 2 s

[tex]v=12(2)^2+6(2)-5=55\ m/s[/tex]

At t = 4 s

[tex]v=12(4)^2+6(4)-5=211\ m/s[/tex]

(b) Acceleration,

[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6[/tex]

Pu t = 3 s in above equation

So,

[tex]a=24(3)+6\\\\a=78\ m/s^2[/tex]

Hence, (a) (i) v = 55 m/s (ii) v = 211 m/s and (b) 78 m/s²

Explanation:

SI is the international system of units

It was developed to express magnitudes and quantities

this process is called parellelogram method of resolving vectors.

Answer:

The correct option is energy levels

Explanation:

Rutherford's model of an atom suggests that an atom has a tiny positively charged central mass (now called the nucleus) which is surrounded by electrons (negatively charged) in a cloud-like manner.

Bohr's model went a bit further than the Rutherford's model in describing an atom by suggesting that the electrons which surrounds in the nucleus travel in fixed circular orbits. This description by Bohr was able to describe the energy levels of orbitals which assumes that smallest orbitals have the lowest energy while the largest orbitals have the highest energy.

Answer:

energy levels

hope this helped!

Explanation:

Answer:

Space in the astronomy and cosmology space is 3 dimensional region and earth atmosphere end is called space.

Explanation:

Space is the everything of the top earth atmosphere moon, starts, milky way, black hole and GPS satellites an distant,space also called between stars,moon,planet.

Space is all the extends far in all directions, space is finite unbound space surface of the earth has finite are no beginning or not end.

Space contains there are three dimensions is called 3 D space,an space is to refer an interval during signal transmitted also used by the character, bytes, words and octets in  digital signal.

space is that term can refer to various in science, communications and mathematics,and maintain orbits for responsible time, space is usually to begin at the lowest attitude satellites  can maintain orbits.

space coordinates are uniquely define the location of any particular point  and that continuum requires more than coordinates,and the number of dimensions and conventional space or digital communications during the  signal represents logic is 0 words in a digital signal.  

Answer:

10 seconds

Explanation:

because the cat is moving one m/s slower than the dog, the dog has a relative speed of 1 m/s. 10 meters would take 10 seconds for the dog to cover