How to magnetic stripes provide evidence of seafloor spreading

The height of the object will be -5.19 cm

A concave mirror's reflecting surface curves inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.

Given an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall

So let,

v =  Image distance from the mirror = -33.5 cm

u = object distance from the mirror (concave) = 24 cm

hi = Image height = 7.25 cm

h = height of the object = ?

Using below formula to find height of the object

-v/u = hi/h

Putting all value in the formula we get

-(-33.5)/(-24) = 7.25/h

h = -5.19 cm

Therefore the height of the object will be -5.19 cm

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The mass in one period travels 'a distance less than 25 cm which will be 0.

The amplitude of a pendulum is the maximum displacement of the pendulum.

If the maximum displacement of the pendulum is 25 cm, the mass in one period travels a distance less that 25 cm.

Thus, then the mass in one period travels 'a distance less than 25 cm which will be 0.

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The magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

u is the initial velocity of canonball= 47.4 m/sec

g is the acceleration of free fall = 9.81 m/sec²

v is the velocity after 1.23 s

According to Newton's first equation of motion,

[tex]\rm v=u +gt \\\\ v= 47.4 \ m/sec + 9.81 \ m/sec^2 \times 1.23 sec \\\\\ v=59.46 \ m/sec[/tex]

Hence, the magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.

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Explanation:

an object's gravitational potential energy Eg is m×g×h where:

m=mass

g=9.8m/s²

h=height relative to the closest object below it (because it cannot potentially fall through it

so Eg = 15×9.8×5=735J

The weight of an object on pluto will be 15.56. Mass multiplied by the gravitational acceleration gives weight.

Gravitation is a natural law by which all things with all matter are attracted towards one another. Gravity is responsible for large-scale structures present in the Universe.

By dividing the object's weight on Earth by 9.8 m/s², as illustrated below, one may calculate the object's mass.

m = 250 N / 9.8 m/s²

m = 25.51 kg

Multiply the acquired mass by Pluto's gravitational acceleration (g);

W = (25.51 kg) x (0.61 m/s²)  

W = 15.56 N

Thus, the item will only be 15.56 N in weight.

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Answer: B.

15.6 newtons

Explanation: edmentum

Two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

When the body cell passes through cell division process, it divides into two daughter cells. In the second cell division, two cells divides into four cells. In the third cell division, the four cells divides into eight cells. In the fourth cell division, the eight cells divides and sixteen cells are produced.

So we can conclude that two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

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(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), the coefficient of friction corresponding to the frictional force is 0.612

(c) The increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands is 7.78 m

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.

Given is a uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring).

(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) angular frequency = velocity / radius

ωf = 10/2 = 5 rad/s and ωi = 20 rad/s

Angular acceleration, α = ωf - ωi /t

Put the values, we get

α = -15/5 = -3 rad/s²

Coefficient of friction, μ = a/g = rα/g

Plug the values, we get

μ = 0.612

Thus, the coefficient of friction corresponding to the frictional force is 0.612.

(c) The energy lost = heat generated

 energy lost = 1/2 Iω² + 1/2 Mv²

 energy lost = 1/2 MR²ω² + 1/2 Mv²

Plug the values, we get

 energy lost = 7500 J

Thus, the  increase in the thermal energy is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands

s= (v- 2gH) sin2θ /g

s = (10 - 2x (-9.81)x5 ) sin (2x π/6) / 9.81

s = 7.78m

Thus, the horizontal distance is  7.78 m

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The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.

Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h

We have to find the magnitude of the angular momentum

Let,

ρ = Density of air = 1.29 kg/m^3

v = Speed of wind = 73.0 mi/h = 0.032 km/s

M = angular momentum of air

Let the volume of air be 1 m^3

Mass = Volume x ρ = 1 x 1.29 = 1.29 kg

Momentum = M = mass x velocity

Momentum = 1.29 x 0.0032

Momentum = 4.128 x 10^(-3) kg·m^2/s

Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

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Answer:

D. s=3x+4y

Explanation:

The line is at the 3rd column in the 4th row.

HIV can be contracted from contact with bloodborne pathogens.

Other bloodborne diseases are HBV, malaria, syphilis and brucellosis

Bloodborne pathogens can be defined as those microorganisms or pathogenic organisms that cause disease and are present in human blood.

Blood borne pathogens can also be contacted through the following means

In conclusion; HIV can be contracted from contact with bloodborne pathogens.

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Answer:

it is much more efficient than any other process to generate power

We can obtain the velocity from the graph and plot is against time to obtain a velocity time graph.

The velocity time graph shows the change of velocity with time. Recall that velocity is the change of position with time.

As such, when we have a position time graph as shown, we can obtain the velocity from the graph and plot is against time to obtain a velocity time graph.

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Answer:

a) 0.2m/s

b) 5m/s

Explanation:

a) acceleration=∆v/∆t

v=10

t=50

10/50=1/5

=0.2m/s

b) time= d/s

d=50m

s=10

50/10

=5m/s

Answer:

See below

Explanation:

Average speed =  (0+ 10)/2 = 5 m/s

  then to cover 50 m     will take    50 m / 5 m/s  = 10 seconds

      change in velocity/ change in time = acceleration = 10/10 = 1 m/s^2

The initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is;

u is the initial velocity of fall = ?  m/sec

h is the distance of fall = 200 m

g is the acceleration of free fall = 9.81 m/sec²

H is the height of the building

t is the time period = 10 second

According to Newton's second equation of motion,

[tex]\rm h= ut+\frac{1}{2} gt^2 \\\\\ h-\frac{1}{2} gt^2 =ut \\\\ 200 - 0.5 \times(9.81) \times 10^2 = 10 u \\\\ u = - 29.05 \ m/sec[/tex]

- ve shows the direction is downward.The magnitude of the initial velocity is found as;

u = 29.05 m/sec

The height of the building

[tex]\rm H= ut+\frac{1}{2} gt^2 \\\\\ H = 29.05 \times 10 + 0.5 \times 9.81 \times 10^2 \\\\ H = 781 \ m[/tex]

Hence the initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.

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This is an exercise of thermal scales.

T = 40 °C

T = °F ¿?

To convert degrees Celsius to degrees Fahrenheit, we have the formula:

[tex]\bf{^{\circ}F=\dfrac{9}{5} \ ^{\circ}C+32 \ \ \to \ \ \ Formula }[/tex]

We substitute our data into the formula.

[tex]\bf{^{\circ}F=\dfrac{9}{5} \times 40+32}[/tex]  

First we divide 9/5 multiplied by 40.

[tex]\bf{=75+32}[/tex]

Add 75 plus 32

[tex]\bf{=104 \ ^{\circ}F}[/tex]

Therefore, the temperature change of 40 °C on the °F scale is equal to 104.

[tex]\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Piscis04}}}}}}[/tex]

Let's see

[tex]\\ \rm\Rrightarrow mgh=\dfrac{1}{2}mv^2[/tex]

[tex]\\ \rm\Rrightarrow 2gh=v^2[/tex]

[tex]\\ \rm\Rrightarrow 2(10)(247)=v^2[/tex]

[tex]\\ \rm\Rrightarrow v²=4940[/tex]

[tex]\\ \rm\Rrightarrow v=70.3ms^{-1}[/tex]

The minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed without sleeping is 1.97.

The minimum coefficient of friction is calculated as follows;

F = μmg = mac

μmg = mac

where;

μg = ac

μg = v²/r

μ = v²/rg

where;

μ = (1.39²)/(0.1 x 9.8)

μ = 1.97

Thus, the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed without sleeping is 1.97.

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If the displacement is tripled then the spring force also is tripled.

The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N).

Froom the hooks law;

F = kx

Where,

F is the applied force

x is the displacement in case 1

Case 2;

x'=3x

F'=kx'

F'=3KX

F'=3x'

Where,

x' is the displacement when the force is tripled

F' is the force in the case2

Hence, if the displacement is tripled then the spring force also be tripled.

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NGC 300 is different from our solar system because of its smaller size.

NGC 300 galaxy is different from our solar system because of its size. NGC 300 is smaller than our solar system which means that it has less number of stars as compared to our solar system.

So we can conclude that NGC 300 is different from our solar system because of its smaller size.

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At pH = 2 the ratio of the concentration of the hydrogen to the hydroxide ion is 10¹⁰. The hydrogen ion concentration at pH, 2 is 10⁻² and the hydroxide ion is 10⁻¹².

The parameter of measuring the concentration of the hydroxide and the hydrogen ion in the solution in order to determine the basic and the acidic nature is known as pH.

The concentration of H⁺ ions is calculated as:

pH = -log [H⁺]

2 = -log [H⁺]

[H⁺] = 10⁻²

The concentration of OH⁻ ions is calculated as:

pH + pOH = 14

pOH = 14 - 2

pOH = 12

Solving further:

pOH = -log [OH⁻]

12 =-log [OH⁻]

[OH⁻] = 10⁻¹²

At a pH = 2, the ratio of hydrogen ions to hydroxide ions:

10⁻² ÷ 10⁻¹² = 10¹⁰

Therefore, at pH = 2 the ratio is 10¹⁰.

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Answer:

Using 3 periods to get an accurate reading:

3T = (6.40 S - 0.90 s) = 5.50 S So, T = 1.83 s

m = 0.250 kg

Using algebra:

k= [tex]\frac{4\pi ^{2}m }{T^{2} }[/tex]=[tex]\frac{4\pi ^{2}0.250 }{1.90^{2} }[/tex]=2.95kg/sec

Answer:

Define universal gravitational constant. Why is newton's law of gravitation called universal law?

The impulse given to the ball by the floor is 0.2865 kg.m/s.

The change in momentum is equal to the product of impact force applied while colliding and time for that impact.

Impulse F. t = m (Vf -Vi)

where, Vf is the final velocity and Vi is the initial velocity.

A ball of mass 500g is dropped from a height 1.5m . It rebounds the floor and reached the height 1.2m.

The initial velocity u = √2x 9.81 x 1.5 = 5.425 m/s

The final velocity v = √2x 9.81 x 1.2 = 4.852 m/s

Substitute the values into the expression, we get

Impulse = m(v- u)

Impulse=0.5 x (4.852- 5.425 )

Impulse = - 0.2865 kg.m/s

Thus, the magnitude of impulse given to the ball by the floor is 0.2865 kg.m/s.

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The final velocity of the projectile when it strikes the ground below is 198.51 m/s.

The time taken for the projectile to fall to the ground is calculated as follows;

h = vt + ¹/₂gt²

where;

265 = (185 x sin45)t + (0.5)(9.8)t²

265 = 130.8t + 4.9t²

4.9t² + 130.8t - 265 = 0

solve the quadratic equation using formula method,

t = 1.89 s

vyf = vyi + gt

where;

vyf = (185 x sin45) + (9.8 x 1.89)

vyf = 149.322 m/s

vxf = vxi

where;

vxf = 185 x cos(45)

vxf = 130.8 m/s

vf = √(vyf² + vxf²)

where;

vf = √(149.322²  +  130.8²)

vf = 198.51 m/s

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Answer:

Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.

Answer:

D

Explanation:

GPE = mgh

GPE / gh = m

m = GPE/gh

m = 915 / 6.5 x 9.8

m = 915 / 63.7

m = 14.4kg

Answer:

Approximately [tex]9.39 \; {\rm m\cdot s^{-1}}[/tex] after the sphere has travelled a distance of [tex]5\; {\rm m}[/tex].

Approximately [tex]16.3\; {\rm m\cdot s^{-1}}[/tex] right before touching the ground (a distance of [tex]15\; {\rm m}[/tex].)

Assumption: [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].

Explanation:

Weight of the sphere: [tex]m\, g = 9.81\; {\rm N \cdot kg^{-1}} \times 10\; {\rm kg} = 98.1\; {\rm N}[/tex], downwards.

Drag on the sphere: [tex]10.0\; {\rm N}[/tex] upwards.

Net force on the sphere: [tex]98.1\; {\rm N} - 10\; {\rm N} = 88.1\; {\rm N}[/tex] downwards.

Acceleration of the sphere: [tex]a = F_\text{net} / m = 88.1\; {\rm N} / (10\; {\rm kg}) = 8.81\; {\rm m\cdot s^{-2}}[/tex].

Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity ([tex]0[/tex] in this case, as the sphere was released from rest,) and [tex]x[/tex] is the distance (displacement) that the sphere has travelled so far.

Rearrange this equation to obtain an expression for [tex]v[/tex]:

[tex]\displaystyle v = \sqrt{2\, a\, x + u^{2}}[/tex].

For example, after the ball travelled a distance of [tex]5.00\; {\rm m}[/tex], [tex]x = 5.00 \; {\rm m}[/tex]:

[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 5.0\; {\rm m} + 0} \\ &\approx 9.39\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Similarly, [tex]x = 15.0\; {\rm m}[/tex] right before landing, such that:

[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 15.0\; {\rm m} + 0} \\ &\approx 16.3\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Answer:

Explanation:

Comment

It's high. It takes quite a bit of heat to get water to change its heat content. Most substances don't reach much more the 0.5 J/(gram*oC)

The stress on the wire is determined as  6.37 x 10⁶ N/m².

The stress on the wire is calculated as follows;

σ = F/A

where;

Let the diameter of the wire = 2 mm

Area = πr²

Area = π(1 x 10⁻³)²

Area = 3.142 x 10⁻⁶ m²

Stress = 20/(3.142 x 10⁻⁶)

Stress = 6.37 x 10⁶ N/m²

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Answer:

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