A skateboarder flies horizontally off a cement planter. After 3 seconds the skateboarder lands on the ground with a final velocity of −4.5 m/s. Which kinematic equation would be most useful for finding the skateboarder’s initial velocity? (Assume a=−9.8 ms^2)

B
C or E
A
D

[tex]\\ \sf\longmapsto Force=ma=250(10)=2500N[/tex]

[tex]\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}[/tex]

[tex]\\ \sf\longmapsto Pressure=\dfrac{2500}{0.015}=1.6666.6Pa[/tex]

Answer:

The moon is being pulled to the earth by gravity, so the net force makes a circular pattern.

Explanation:

The force of gravity is a force that pulls objects toward the centre of the earth. The acceleration due to gravity of the moon is about 1.6 metre per second square which is comparatively smaller than the earth's 10metre per second. Therefore, the moon isn't completely pulled to the centre of the earth, but orbits around the earth

Answer:

firts divide then multiply divide again and your and your answer should d

Answer:

(Δa/a +Δb/b) × 100

Explanation:

Y= a/b

So, maximum percentage error

Δy/y = ( Δa/a + Δ b/b) × 100

Plz, Mark me as the brainlest

Answer:

3.13m/s²

Explanation:

90×1000/3600 note: 1h- 3600s

=25m/s

a=v/t

a= 25m/s ÷ 8s

a= 3.125 / 3.13m/s²

Answer:

wavelength = 2.5 m

Explanation:

λ=. v/f

= [tex]\frac{150m/s}{60Hz} = 2.5m[/tex]

Answer::27 km per hour West + 17 km per hour North

Explanation:

:27 km per hour West + 17 km per hour North

he or she is right good jod

The velocity for the trip would be 31 km per hour, therefore the correct answer is option B.

An object's position changes if it moves in relation to a reference frame, such as when a passenger moves to the back of an airplane or a professor moves to the right in relation to a whiteboard. Displacement describes this shift in location.

As given in the problem, on a trip you travel at a constant speed of 50 kilometers per hour North for 1 hour. Then you turn West and travel at a constant speed of 40 kilometers per hour for 2 hours.

The final displacement of the trip = √(50² + 80²)

                                                         = 94.34 kilometers

The velocity of the trip = total displacement of the trip / time period

                                      =94.34 kilometers / 3

                                      = 31 kilometers per hour

Thus, the velocity of the trip would be 31 kilometers per hour.

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when two charge particles interact

Answer:

0.4 N

Explanation:

We know ,

Answer:

I think its false cause with a slope you have acceleration

Answer:

I am pretty sure that it is A. The Sun and all the planets revolve around Earth. Hope this helps!!!!!

Explanation:

Answer:

fL = fs(V/v-vs) 400000002= 40000010(331/331- vs). then find vs so easy

Answer:

[tex]momentum =mv \\ = 8 \times 3 = 24kg.m. {s}^{ - 1} [/tex]

Answer:

C

Explanation:

Newton's first Law of Motion is also known as law of inertia, which states that "an object at rest remains at rest, and an object that is moving will continue to move straight and with constant velocity, if and only if there is no net force acting on that object". As such, the correct option should be C.

Option A is Newton's third Law of Motion.

Option B is Newton's second Law of Motion

Option D is Newton's Law of Universal Gravity

The height of the ball after the given time of motion is 3.275 m.

The given parameters;

The distance traveled by the ball is calculated as follows;

[tex]h =h_0 + ut - \frac{1}{2}gt^2 \\\\h = 2.5 + 4(0.5) - (0.5\times 9.8\times 0.5^2)\\\\h = 3.275 \ m[/tex]

Thus, the height of the ball after the given time of motion is 3.275 m.

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The kinematic relations allow finding the results for the different questions are:

     1.  The acceleraations is a = 375 m/s²

     2. The time is t = 62.5 s

     3. The velocity at the ground is  v = -14.7 m / s

     4.  The time to the bottoncliff is    1.7 s

     5.  The height isy₀ = 122.5 m

     6.

         A) The velocity is v = 25.2 m / s

         B)  The height  for t=1 s is  y = 18.1 m

     7.

         A)  Time in the air is t = 2 s

         B)  The final veloicty is   v = - 19.6 m / s

Kinematics analyzes the movement of bodies, finding relationships between the position, velocity and acceleration of bodies.

Let's look for the answers to a series of questions:

1. They indicate the initial and final velocities of the body and the time to reach it is 20 s, ask how much the acceleration is worth

Let's use the relationship

             v = v₀ + a t

             a = [tex]\frac{v-v_o}{t}[/tex]

             a = [tex]\frac{17500-10000}{20}[/tex]

             a = 375 m/s²

2. What is the time it takes to travel a distance of 125 m with an acceleration of 4 m / s²

             x = v₀ t + ½ a t²

As it leaves the direction its initial velocity is zero

             x = ½ to t²

             t = [tex]\sqrt{\frac{2x}{a} }[/tex]

             t = [tex]\sqrt{\frac{2 \ 125}{4} }[/tex]

             t = 62.5 s

3. A pencil falls from a height and reaches the floor in a time of 1.5 s what is the speed when reaching the floor

            v = v₀ - g t

As it is released its initial velocity is zero

            v = -g t

            v = - 9.8 1.5

            v = -14.7 m / s

The negative sign indicates that the velocity is directed downwards.

4. From the edge of a cliff 40 m high, an apple is thrown downward with an initial velocity of 15 m / s. How long does it take to get to the bottom?

             y = y₀ + v₀ t - ½ g t²

In this case the initial velocity is negative because it is directed downwards and when it reaches the floor its height is zero, let us substitute

Suppose the acceleration is g = 10 m / s²

             0 = 40 - 15 t - ½ 10 t²

             0 = 40 - 15 t - 5 t²

let's solve the quadratic equation

             t² + 3 t - 8 = 0

             t = -3 + Ts 9 + 4 8/2

            t = -3 + 6. 4/2

             t₁ = - 4.9 s

            t₂ = 1.7 s

The time must be a positive quantity, so the correct answer is 1.7 s

5. A stone is dropped from the ceiling and it takes a time 5 s to reach the floor, which is the height of the ceiling

           y = y₀ +v₀ t - ½ g t²

as it is released from the ceiling its initial velocity is zero and the height upon reaching the floor is zero

          0 = y₀ + 0 - ½ g t²

          y₀ = ½ g t²

          y₀ = ½ 9.8 5²

          y₀ = 122.5 m

6. A man at a height of 30 m throws a ball downward at 7 m / s

    A) The speed when reaching the ground

    B) The height of the ball when a time of 1 s has fallen.

  A) Let's use the kinematics relation

              v² = v₀² - 2 g (y -y₀)

              v = 7² - 2 9.8 (0 -30)

              v = 25.2 m / s

 B)   y = y₀ + v₀ t - ½ g t²

        y = 30  7 1 - ½ 9.8 1²

        y = 18.1 m

7.  A watermelon is dropped from rest from the top of a 20m cliff.  

 A) time in the air

 B) ground speed

  A) Let's use the relation

       y = y₀ + v₀ t - ½ g t²

It is released its initial velocity is zero and when it reaches the ground its height is zero (y = 0)

       0 = y₀ + 0 - ½ g t²

       t = [tex]\sqrt{\frac{2y_o}{g} }[/tex]

       t = [tex]\sqrt{\frac{2 \ 20}{9.8} }[/tex]

       t = 2 s

B) the speed when reaching the ground

       v = v₀ - g t

       v = 0 - 9.8 2

       v = - 19.6 m / s

The negative sign indicates that the speed is down

In conclusion using the kinematic relations we can find the results for the different questions are:

   1.  The acceleraations is a = 375 m/s²

   2. The time is t = 62.5 s

   3. The velocity at the ground is  v = -14.7 m / s

   4.  The time to the bottoncliff is    1.7 s

   5.  The height isy₀ = 122.5 m

   6.

      A) The velocity is v = 25.2 m / s

      B)  The height  for t=1 s is  y = 18.1 m

  7)

      A)  Time in the air is t = 2 s

      B)  Te final veloicty is   v = - 19.6 m / s

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Answer:

Pedometers are designed to detect vertical movement at the hip and so measure the number of steps and provide an estimate of distance walked. They cannot provide information on the temporal pattern of physical activity or the time spent in different activity at different intensity.

Answer:hey can you help me answer my question please.

Explanation:

Answer:

the average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two points approaches zero.

Explanation:

Answer:

Astronomers estimate the distance of nearby objects in space by using a method called stellar parallax, or trigonometric parallax. Simply put, they measure a star's apparent movement against the background of more distant stars as Earth revolves around the sun.

Answer:

it References the position

Explanation:

Displacement is defined to be the change in position of an object.

And the treasure can't move by itself because its an object (and objects don't move).

I'm going to assume you mean

[tex]y = 14t^2 - \dfrac{t^2}{2+5t} - 3[/tex]

Taking the derivative with respect to t on both sides gives

[tex]\dfrac{\mathrm dy}{\mathrm dt} = 2\cdot14t - \dfrac{(2+5t)(2t)-t^2(5)}{(2+5t)^2} - 0 \\\\ \dfrac{\mathrm dy}{\mathrm dt} = \boxed{28t - \dfrac{4t+5t^2}{(2+5t)^2}}[/tex]

You could go on to expand the rational expression into partial fractions. One way to do this:

[tex]\dfrac{5t^2 + 4t}{(2 + 5t)^2} = \dfrac{t(5t+2) + 2t}{(2+5t)^2} \\\\= \dfrac{t}{2+5t} + \dfrac{2t}{(2+5t)^2} \\\\= \dfrac15 \cdot \dfrac{t}{\frac25+t} + \dfrac{2t}{(2+5t)^2} \\\\ = \dfrac15 \cdot \dfrac{t+\frac25-\frac25}{\frac25+t} + \dfrac{2t}{(2+5t)^2} \\\\ = \dfrac15 \left(1 - \dfrac{2}{2+5t}\right) + \dfrac{2t}{(2+5t)^2} \\\\ = \dfrac15 - \dfrac{2}{10+25t}+ \dfrac{2t}{(2+5t)^2}[/tex]

and this would give

[tex]\dfrac{\mathrm dy}{\mathrm dt} = 28t - \dfrac15 + \dfrac2{10+25t} - \dfrac{2t}{(2+5t)^2}[/tex]

Answer:

The force formula is defined by Newton's second law of motion: Force exerted by an object equals mass times acceleration of that object: F = m×a( force = mass×area)

The transmission of light waves is usually done through cornea of the eyes, then move through another opening which is regarded as pupil before it will get to the retina.

Therefore, light wave are form of tiny microscopic particles.

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Explanation:

We know ,

Volcanic Island arcs form when an oceanic tectonic plate is subducted below a tectonic plate  taking seawater with it.

Volcanic island arcs occur offshore during volcanoes. They occur due to the subduction of an oceanic plate under another tectonic plate. Volcanic Island arcs occur mostly in ocean basins.

Due to the subduction of oceanic plates under tectonic plates, Island Arcs are found, mostly along the margins of continent.

Hence we can conclude that Volcanic Island arcs Form when an oceanic tectonic plate is subducted below a tectonic plate  taking seawater with it.

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Answer: 4 times

Explanation:

KE1 = (1/2)mv^2

KE2 = (1/2)m(2v)^2 or (1/2)m*4v^2

KE2/KE1 = ((1/2)m*4v^2)/((1/2)mv^2) = 4 times

Friction is one of the forces working against a moving object, it makes it harder to move the more friction is exerted on the object. So, by reducing friction, you are making it easier for you to be able to move objects.