Which of the following does not support the idea of continental drift
A) seafloor spreading
B) magnetic reversals
C) Pangaea
D) Play tectonics

[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\ [/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}[/tex]

[tex]\\[/tex]

☯ As we know that,

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }[/tex]

[tex]\\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\ [/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = 0.04 \: m }[/tex]

[tex]\\[/tex]

☯ For left penetration (s₂)

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}[/tex]

[tex]\\[/tex]

[tex]\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\ [/tex]

it would be the 3rd one. so C

Answer:

the placement of the ammeter in the circuit

Explanation:

Answer: The correct answer is C

Explanation:

Answer:

c. 317 m

Explanation:

Vertical Launch Upwards

It occurs when an object is launched vertically up without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

[tex]\displaystyle h_m=\frac{v_o^2}{2g}[/tex]

We'll assume the acceleration of gravity as [tex]g=10\ m/s^2[/tex].The pellet is vertically upward launched with vo=80 m/s. The maximum height is:

[tex]\displaystyle h_m=\frac{80^2}{2*10}=320[/tex]

[tex]h_m = 320\ m[/tex]

That height is 3 meters above the edge of the cliff, thus the cliff is 320-3=317 m hight

c. 317 m

Answer:

Knowing that these metals are infact good conductors of electricity we can infer that metals are able to hold and conduct certain temperatures. Another thing we can infer is that these good conductors can be used in connection to transferring energy or electricity.

Answer:

The apparent depth d = 19.8495 cm

Explanation:

The equation for apparent depth can be expressed as:

[tex]d = \dfrac{d_1} {\mu_1}+\dfrac {d_2}{\mu_2}[/tex]

here;

[tex]d_1 = d_2 = 15 \ cm[/tex]

[tex]\mu_1[/tex] = refractive index in the first liquid = 1.75

[tex]\mu_2[/tex] = refractive index in the second liqquid= 1.33

∴

[tex]d = \dfrac{15}{1.75}+\dfrac{15}{1.33}[/tex]

[tex]d = 15( \dfrac{1}{1.75}+\dfrac{1}{1.33})[/tex]

[tex]d = 15( 0.5714 +0.7519)[/tex]

d = 15(1.3233 ) cm

d = 19.8495 cm

Answer:

gimmie

Explanation:

Answer:

Explanation:

This is a common question on Khan Academy's "Calculating change in kinetic energy from a force" practice exercises. (AP Physics 1)

Here's the real question without all the formatting:

A truck pushes a pile of dirt in the horizontal direction with a force of 20 N to a distance of 15 m then the kinetic energy of the dirt is 300 J.

In physics, the term "work" refers to the measurement of energy transfer that takes place when an item is moved over a distance by an externally applied, at least a portion of which is applied inside the direction of the displacement.

The duration of the path is multiplied by the component of a force acting along the path to calculate work if the forces are constant. The work W is theoretically equivalent to the pressure f times the range d, or W = fd, to represent this idea. Work is done when a force is applied at an angle of to a displacement, or W = fd cos.

The work done, [tex]W = Force * Displacement[/tex]

W = 15× 20

W = 300 J

To know more about Work:

https://brainly.com/question/18094932

#SPJ3

Answer:

The answer is "1557 meters".

Explanation:

speed of sound in ([tex]\frac{m}{s}[/tex]) [tex]= 331.5 + 0.60 \ T^{\circ}\ C\\\\[/tex]

[tex]\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters[/tex]

Answer:

2.57 seconds  (rounded to 2.6 Seconds)

Step-by-step explanation:

Great question, it is always good to ask away and get rid of any doubts that you may be having.

Before we can solve this question we need to create a formula that calculates the final speed. The formula will be the following,

Where:

Vf is the final Velocity

Vi is the initial velocity

A is the acceleration

t is the time in seconds

Now that we have the formula we can plug in the values given to us in the question and solve for the amount of time (t).

Finally, we can see that it would take the cyclist 2.57 seconds (approximately) to reach a speed of 18 m/s

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Angle \ of \ projection = 30^{\circ} }[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ projectile = 28 \: m/s^{-1} }[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Height_{\:(max)}\: reached\: by \:the \:projectile }[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]

☯ As we know that,

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{u^2\;sin^2\theta}{2\;g} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{H = \dfrac{(28)^2\;sin^2 30^{\circ}}{2\;(9.8)} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{H = \dfrac{784 \times \;sin^230^{\circ}}{19.6} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times sin^2 30^{\circ}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \bigg(\dfrac{1}{2}\bigg)^2 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \dfrac{1}{4} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: {\boxed{\sf{H=10\:m }}}[/tex]

Answer:

Δx = 39.1 m

Explanation:

        [tex]v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)[/tex]

        where  vf is the final velocity (0 in our case), v₀ is the initial velocity

        (25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance

        traveled since the brakes are applied.

        [tex]\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)[/tex]        

The car will travel a distance of 39.1 m before its stops.

To solve the problem above, use the equations of motion below.

Equation:

Where:

From the question,

Given:

Substitute these values into equation 1

Solve for s

Hence, The car will travel a distance of 39.1 m before its stops.

Learn more about acceleration here: https://brainly.com/question/605631

Answer:

123.48J

Explanation:

Given parameters:

Mass of the ball  = 2.8kg

Height  = 4.5m

Unknown:

Potential energy  = ?

Solution:

The potential energy is the energy due to the position of a body. It is mathematically given as;

      P.E   =  mgh

m is the mass

g is the acceleration due to gravity

h is the height

 Now insert the parameters and solve;

          P.E  = 2.8 x 4.5 x 9.8  = 123.48J

Answer:

0

Explanation:

There is 0 PE when its on the ground

Answer:

add answer +5 so so so so so

Answer:

3 trust me

Explanation:

Answer:

bus momentum

p_bus= m_bus x v_bus

=18,200 x 16.5

basball momentum

pball=mball x vball

=0.142 x v

p_bus = pball

18200 x 16.5 = 0.142 x v

v=(18200 x 16.5)/0.142

v is the answer for baseball

Explanation:

⚠️not my answer tryna be honest here⚠️

The momentum of the bus of 18200 kg and velocity of 13.5 m/s is 245700 Kg m/s. To have equal momentum the base ball with 145 g have to throw in a speed of 1.7 × 10 ⁶  m/s.

Momentum of a moving body is the product of mass and velocity. Thus it have the unit of g m/s or Kg m/s. Momentum is a vector quantity and thus having magnitude and direction.

Given that one bus is having a mass of  18200 Kg and 13.2  m/s speed. The momentum is:

p = mv

 =18200 kg × 13.5  m/s

 = 245700 Kg m/s

To have a momentum of 245700 Kg m/s   the base ball with 0. 142 g have to have a velocity  = 245700 Kg m/s   / 0.142 g

                          =1.7 × 10 ⁶  m/s

Hence, the baseball weighs0. 142 g have to move in 1.7 × 10 ⁶  m/s

To find more on momentum, refer here:

http://brainly.com/question/24030570

#SPJ2

Answer: 2

Explanation:

:}

Answer:

Nose has mucous glands with hairs which helps the body in trapping pollutants and infectants from entering inside the body. On the other hand,our feet is composed of millions of sweat pores when dirt and other things accumulate,it smells because of sweat mixed with the dirt and other dirty things of the ground.

Explanation:

Hope this helps

Given :

Two satellites orbit the same planet. if a satellite A has an orbit radius r and satellite B has an orbit radius 2r.

To Find :

The ratio of the period of satellite B to the period of satellite A.

Solution :

We know, square of time period (T) is directly proportional to cube root of radius (r) .

So,

[tex]\dfrac{T_B^2}{T_A^2}=\dfrac{(2r)^3}{r^3}\\\\\dfrac{T_B}{T_A}=2\sqrt{2}[/tex]

Therefore, the ratio of the period of satellite B to the period of satellite A is 2√2 .

Answer:

Its pulling down on the book

Explanation: if it was pushing up the book would be floating and the other choices don't make sense

Answer:

the period of the pendulum will be twice as long.

Explanation:

because i looked it up

Answer:

600 KPa.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V1) = 0.075 m³

Final volume (V2) = 0.45 m³

Final pressure (P2) = 100 KPa

Initial pressure (P1) =?

Temperature = constant

The initial pressure can be obtained by using the Boyle's law equation as shown below:

P1V1 = P2V2

P1 × 0.075 = 100 × 0.45

P1 × 0.075 = 45

Divide both side by 0.075

P1 = 45 / 0.075

P1 = 600 KPa.

Thus, the initial pressure in the balloon is 600 KPa.

Compute the car's weight:

W = m g = (950 kg) (9.8 m/s²) = 9310 N

The net vertical force on the car is

∑ F = N - W = 0

so the normal force has magnitude

N = W = 9310 N

Then the friction force that keeps the car from skidding has magnitude F = µ N, where µ is the coefficient of friction, and it's friction that makes up the net horizontal force on the car. By Newton's second law, we have

F = m a

µ N = m v ² / R

µ (9310 N) = (950 kg) (25 m/s)² / (72 m)

µ ≈ 0.89

Answer: mass = 48.47 kg.

Explanation:

Formula : Weight = mg  , where m = mass of body , g= acceleration due to gravity .

Given: Weight  = 475 N

[tex]g= 9.8\ m/s^2[/tex]

Substitute all values in formula ,  we get

[tex]475= m \times9.8\\\\\Rightarrow\ m = \dfrac{475}{9.8}\\\\\Rightarrow\ m \approx 48.47\ kg[/tex]

Hence, his mass = 48.47 kg.

Answer: The reason for the differences in density is the composition of rock in the plates. When two plates come in contact with each other through plate tectonics, scientists can use the density of the plates to predict what will happen. Whichever plate is more dense will sink, and the less dense plate will float over it.

Explanation:

Hope this helps ( not copied and pasted, this answer was done by me so I don't know if it's good or not)

Answer: 54.88 meters/s

Explanation:

The final velocity will be calculated by using the formula:

v = u + at

where,

v = final velocity

u = initial velocity = 0

a = 9.8

t = 5.6

Therefore, we slot the value back into the formula. This will be:

v = u + at

v = 0 + (9.8 × 5.6)

v = 0 + 54.88

v = 54.88 meters per second

Therefore, the final velocity is 54.88m/s

Answer:

The maximum angular velocity is 20 rad/s

Explanation:

Given;

torque, τ = 10 N

maximum mechanical power, P = 200 J/s

The output power of the pmdc is given as;

P = τω

where;

P is the maximum mechanical power

ω is the maximum angular velocity

ω = P / τ

ω = (200) / (10)

ω = 20 rad/s

Therefore, the maximum angular velocity is 20 rad/s

Answer:

The value is [tex]T_2 =416.9 \ K[/tex]

Explanation:

From the question we are told that

     The mass of the aluminum baking sheet is  [tex]m = 225 \ g = 0.225 \ kg[/tex]

      The energy absorbed is [tex]E = 2.4 *10^{4} \ J[/tex]

       The initial  temperature is  [tex]T_1 = 25 ^oC = 25 + 273 = 298 \ K[/tex]

Generally the heat absorbed is mathematically represented as

         [tex]Q = m * c_a * [T_2 - T_1][/tex]

Here  [tex]c_a[/tex] is the specific heat capacity of  aluminum with value  [tex]c_a = 897 \ J / kg \cdot K[/tex]

So

           [tex]2.4 *10^{4 } =0.225 * 897 * [ T_ 2- 298][/tex]

=>         [tex]T_2 - 298 = 118.915[/tex]

=>          [tex]T_2 =416.9 \ K[/tex]

Answer:

[tex]\theta = 16.70 ^{\circ}[/tex]

Explanation:

The coefficient of static friction is equal to the tangent of the minimum angle at which an object will begin to start sliding down a ramp.  

Since we are given the coefficient of static friction we can solve for the minimum angle that the block will begin to slide.

Let's solve for the force of gravity that is acting on the block. The force of gravity is also known as the weight force, which can be calculated by using w = mg.

We are given the mass of the block (kg) and we know that g = 9.8 m/s².

Now we can use this force in the equation:

Plug [tex]\displaystyle u_s = 0.30[/tex] and 49 N into the equation.

Notice that the gravitational force cancels out in the end, so we can actually start with [tex]0.30=\text{tan} \theta[/tex].

Evaluate this equation by taking the inverse tangent of both sides of the equation.

The minimum angle at which the block will begin to slide is about 16.70 degrees.