A graph of v(t) is shown for a world-class track sprinter in a 100 -m race. (See figure below. For each answer, enter a number.)
(a) What is his average velocity (in m / s ) for the first 4 s} ?
______ m/ s
(b) What is his instantaneous velocity (in m / s ) at t=6 s ?
______ m / s
(c) What is his average acceleration (in m / s²) between 0 and 4s?
________m / s²
(d) What is his time (in s) for the race?
______ s

The magnitude of the constant acceleration of the ball is 9.7 m/s².

Acceleration is the rate of change of velocity.

To calculate the constant acceleration of the ball, we use the formula below.

Formula:

From the question,

Given:

From the question,

Given:

Substitute these values into equation 1

Hence, the acceleration is 9.7 m/s².

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The best way for her to use the laptop in her science class  is to take notes as her teacher talks

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The mass of the car moving with a velocity of 17 m/s is 3115.4 kg.

Mass is the quantity of matter a body contains. The S.I unit of mass is kilogram

To calculate the mass of the car, we use the formula below

Formula:

Where:

From the question,

Given:

Substitite these values into equation 1

Hence, the mass of the car is 3115.4 kg.

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Explanation:

it takes the earth 24 hours to spin on it's axis once

Answer:

56.7m

Explanation:

horizontal velocity does not matter in this situation because it stays constant (no unbalanced force acting upon it)

a=9.81m/s^2              

t=3.4s

Vi=0m/s

d=Vi*t+1/2at^2

d=1/2at^2

d=1/2*9.81m/s^2*(3.4s)^2

d=4.905m/s^2*11.56s^2

d=56.7m

The force exerted at other end is 100N.

                              W = (F cos θ) d = F. d

Where,

W is the work done by the force.

F is the force, d is the displacement caused by force

θ is the angle between the force vector and the displacement vector

Let , F' = Force exerted on ladder at other end

       d' = distance moved = d/2 (given)

i.e.  W = F'd' and W = Fd

Using these two equations we get, F'd' = Fd

i.e. F' = Fd/d'

Given, F = 50N and d' = d/2

Putting these values in above equation we get, F' = 100N

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Answer:

the correct answer is up and down

1. Create a new question

2. Provide an image of your question/work so we don't have to download anything.

The independent variable that we have in the experiment is the temperature. Option B

We know that if we are talking about an experiment, what we have to look at is that we must consider the interplay of the variables that is taking place in the process. As we change one of the variables in the experiment, there would be a change in another variable in the experiment also.

Now, the variable that we usually beging to alter is the one that we call the independent variable. As we change the value of this particular variable, there would be a change in the independent variable.

We have been told that; a  student conducts an experiment to test how the temperature affects the amount of salt that can dissolve in water.

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Answer:

the surface temperature of Regulus A is 11724.13 K

Explanation:

Given the data in the question;

Sun's surface temperature T = 5800 K

total radiated power, relative to the sun is; P/P[tex]_{sun[/tex] = [tex]([/tex] 1.5(M/M[tex]_{sun[/tex] [tex])^{3.5[/tex]

The star Regulus A is a bluish main-sequence star with mass 3.8M[tex]_{sun[/tex]  and radius 3.1R[tex]_{sun[/tex] .

First, we determine the value power emitted by the sun or sun as follows;

P = eσAT⁴

where P is the power, e is surface emissivity, σ is Stefan Boltzmann, A is area and T is temperature.

so, lets assume emissivity of star and sun is same;

let p be power related to star and p[tex]_{sun[/tex] be power related sun.

Ratio of power radiated by star and power radiated by sun;

P/P[tex]_{sun[/tex] = eσAT⁴ / eσA[tex]_{sun[/tex]T[tex]_{sun[/tex]⁴

we know that AREA A = πR²

we input the formula for area

P/P[tex]_{sun[/tex] = eσ(πR²)T⁴ / eσ(π(R[tex]_{sun[/tex])²)T[tex]_{sun[/tex]⁴

such that we now have;

P/P[tex]_{sun[/tex] = R²T⁴ / R[tex]_{sun[/tex]²T[tex]_{sun[/tex]⁴

given that P/P[tex]_{sun[/tex] = [tex]([/tex] 1.5(M/M[tex]_{sun[/tex] [tex])^{3.5[/tex], we substitute

[tex]([/tex] 1.5(M/M[tex]_{sun[/tex] [tex])^{3.5[/tex] = R²T⁴ / R[tex]_{sun[/tex]²T[tex]_{sun[/tex]⁴

we find temperature of the star T

T = 5800  × [tex][[/tex] 1.5[tex]([/tex]M/M[tex]_{sun[/tex][tex])^{3.5[/tex] (R[tex]_{sun[/tex]²/R²)[tex]]^{1/4[/tex]

Given that;  mass M is 3.8M[tex]_{sun[/tex] and radius R is 3.1R[tex]_{sun[/tex] .

we substitute

T = 5800  × [tex][[/tex] 1.5[tex]([/tex]3.8M[tex]_{sun[/tex]/M[tex]_{sun[/tex][tex])^{3.5[/tex] (R[tex]_{sun[/tex]²/( 3.1R[tex]_{sun[/tex] )²)[tex]]^{1/4[/tex]

T = 5800  × [tex][[/tex] 1.5[tex]([/tex]3.8[tex])^{3.5[/tex] ( 1/( 3.1)²)[tex]]^{1/4[/tex]

T = 5800  × [tex][[/tex] 1.5( 106.9652 ) ( 1/(9.61) [tex]]^{1/4[/tex]

T = 5800  × [tex][[/tex] 16.69592 [tex]]^{1/4[/tex]

T = 5800  ×  2.02140152

T = 11724.13 K

Therefore, the surface temperature of Regulus A is 11724.13 K

Answer:

  x₁ = 0.62 m

Explanation:

In this exercise the force is electric, given by Coulomb's law

         F =[tex]k \frac{q_1q_2}{r^2}[/tex]

This force is a vector, since the three charges are in a line we can reduce the vector sum to a scalar sum.

For the sense of force let us use that charges of the same sign repel and charges of the opposite sign attract.

     ∑ F = F₁₂ - F₂₃

They ask us to find the point where the summaries of the force is zero.

      F₁₂ - F₂₃ = 0

      F₁₂ = F₂₃

let's fix a reference system located in the first charge (more to the left), the distance between the two charges is d = 1.5 m and x is the distance to the location of the second sphere

      k q₁q₂ / x² = k q₂q₃ / (d-x) ²

      q₁ (d-x) ² = q₃ x²

let's solve

       d² - 2 x d + x² = [tex]\frac{q_3}{q_1}[/tex]  x²

       x² (1 -  [tex]\frac{q_3}{q_1}[/tex]) - 2x d + d² = 0

we substitute the values

       x² (1- 4/2) - 2 1.5 x + 1.5² = 0

       x² (-1) - 3.0 x + 2.25 = 0

       x² + 3 x - 2.25 = 0

let's solve the quadratic equation

       x = [-3 ± [tex]\sqrt{ 3^2 + 4 \ 2.25}[/tex]] / 2

       x = [-3 ± 4.24] / 2

       x₁ = 0.62 m

       x₂ = 3.62 m

since it indicates that the charge q₂ e places between the spheres, the correct solution is

            x₁ = 0.62 m

Answer:

The answer is "Option C"

Explanation:

Ohm's Rule we remember [tex]V = I\times R[/tex]

It looks like the sample is drawn y=mx while m is curve slope

The slope of the graph is its resistance if they draw a graph v vs I.

However, the internal battery presence resistance can be modified as the straight line equation

[tex]y = mx -c[/tex]

where c is y-intercept:

[tex]E = I(R+r) \\\\ E = IR+ Ir\\\\E = V + Ir\\\\V = E- Ir\\\\V = -Ir+E[/tex]

R is really the internal battery resistance to draw graph V vs I, the slope is internal resistance to the negative slope cure.

Answer:

C. 28g of Li2O, because all mass must be conserved

Explanation:

my mom is a 8th grade science teacher lol.

Answer:

Body stuff

Explanation:

:) <3

Answer:

Old = senior. geriatric. senescent. unyoung. over-the-hill.

Stubble= bristles

Answer:

24.132 m/s

Explanation:

Note

[tex]U[/tex] = Initial Velocity

[tex]U_x[/tex] = Initial Horizontal Velocity

[tex]U_y[/tex] = Initial Vertical Velocity

[tex]V[/tex] = Final Velocity

[tex]V_x[/tex] = Final Horizontal Velocity

[tex]V_y[/tex] = Final Vertical Velocity

[tex]B[/tex] = launch angle

[tex]g[/tex] = gravity

[tex]t[/tex] = time

[tex]U_x=U*cos(B)[/tex]

[tex]U_y=U*sin(B)[/tex]

The horizontal component of the velocity is constant throughout the flight. So [tex]U_x=V_x[/tex] It can be defined as

[tex]V_x=U*cos(B)[/tex]

We can use the kinematics equation

[tex]V=U+at[/tex]

Gravity is acting downwards; gravity would be negative

[tex]V_y=U_y-gt[/tex]

The magnitude of the velocity can be defined as

[tex]V=\sqrt{V_x^2+V_y^2}[/tex]

Inserting some of the other equations gives us an equation at a given time (t).

[tex]V=\sqrt{(U*cos(B))^2+(U*sin(B)-gt)^2}[/tex]

[tex]V=\sqrt{(UcosB)^2+(UsinB)^2+(gt)^2-2gtU*sinB[/tex]

[tex]V=\sqrt{U^2+g^2*t^2-2*t*g*U*sinB}[/tex]

[tex]V(t)=\sqrt{U^2+g^2t^2-2tgUsinB}[/tex]

We are given

[tex]U=34.6[/tex]

[tex]B=80.2[/tex]

[tex]t=1.09[/tex]

[tex]g=9.81[/tex]

[tex]V(1.09)=\sqrt{34.6^2+9.81^2*1.09^2-2*1.09*9.81*34.6*sin80.2}[/tex]

[tex]V(1.09)=\sqrt{34.6^2+114.338-2*1.09*9.81*34.6*sin80.2}[/tex]

[tex]V(1.09)=\sqrt{34.6^2+114.338-739.94868*sin80.2}[/tex]

[tex]V(1.09)=\sqrt{34.6^2+114.338-729.151}[/tex]

[tex]V(1.09)=\sqrt{1197.16+114.338-729.151}[/tex]

[tex]V(1.09)=\sqrt{582.347}[/tex]

[tex]V(1.09)=24.132[/tex]

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Let me know if you have any questions about my work

The gravitational force that is acting on the masses is 1.73 * 10^-2 N.

We know that the force of gravity is that kind of force that is going to act on any two of the objects that we have on the earth. This force as we know it is attractive force. The implication of this is that any two masses that we find on the earth are the forces that would have to be attracted to each other.

Let us note that we would have the masses that we have of the asteroid and that of sally as m1 and m2 while we have the force as;

F = G m1m2/r^2

G = gravitational constant

F = magnitude of the force

m1 and m2 = masses

r = distance of separation.

We then have;

F = 6.6 * 10^-11 * 8.7 * 10^20 * 68/(15 * 10^6)^2

F = 3.9 * 10^12/2.25 * 10^14

F = 1.73 * 10^-2 N

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Explanation:

a) ac = r(omega)^2

where omega = 2.5 rev/s

r = 0.75 m

We need to convert rev/s first into rad/s:

2.5 rev/s × (2pi rad/rev) = 5pi rad/s = omega

Therefore, the centripetal acceleration ac is

ac = (0.75 m)×(5pi rad/s)^2

= 185 m/s^2

b) Fc = mac

= (0.5 kg)(185 m/s^2)

= 92.5 N

Answer:

a)The direction of the impulse that the glass imparts to the baseball Is opposite the direction of the balls motion

b)  [tex]I=0.6336Ns[/tex]

Explanation:

From the Question we are told that:

Mass [tex]m=0.144kg[/tex]

Initial Speed [tex]v_1=14.9m/s[/tex]

Final speed  [tex]v_2=10.5[/tex]

a)The direction of the impulse that the glass imparts to the baseball Is opposite the direction of the balls motion

b)

Generally the equation for impluse magnitude is mathematically given by

 [tex]I=m(v_1-v_2)[/tex]

Therefore

 [tex]I=0.144(14.9-10.5)[/tex]

 [tex]I=0.6336Ns[/tex]

By contrasting a fossil with related rocks and fossils of known ages, relative dating can estimate the fossil's age.

By utilising radiometric dating to track the isotope decay inside the fossil or, more frequently, the rocks it is linked with, absolute dating may pinpoint an object's exact age.

It is possible to identify whether one artefact, fossil, or stratigraphic layer is older than another using relative dating, which does not provide particular dates.

Absolute dating techniques offer more precise dates and times of origin, such as a range in age in years.

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I uploaded the answer to[tex]^{}[/tex] a file hosting. Here's link:

bit.[tex]^{}[/tex]ly/3gVQKw3

Answer:

Explanation:

Polar coordinates formula

Summary. To convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) : x = r × cos( θ ) y = r × sin( θ )

The correct answer is D) convection current.

Convection currents are the cycle of warm air rising and cool air sinking that occurs in Earth's atmosphere (as well as in other fluids) as a result of temperature differences. When a fluid (such as air) is heated, it expands and becomes less dense, causing it to rise. When the warmer fluid cools, it contracts and becomes denser, causing it to sink. This cycle of rising and sinking creates a circular flow known as a convection current.

From conservation of linear momentum, the final velocity is 15.7 m/s and the time taken is 1.6 s

Velocity can be defined as a distance travel in a specific direction per time taken. It is a vector quantity.

Given that 30kg metal ball is dropped from a height of 12.5 m.  

a.  Find the final velocity when the ball  hits the ground.

The maximum K.E of the ball at it final velocity will be equal to its maximum P.E at height 12.5 m. That is,

K.E = P.E

1/2mv² = mgh

Where

Substitute all the parameters

1/2 × 30 × v² = 30 × 9.8 × 12.5

v² = 245

v = √245

v = 15.65 m/s

b. The time it takes for the ball to hit the ground can be found through

h = ut + 1/2gt²

but u = 0

h = 1/2gt²

Substitute all the necessary parameters

12.5 = 1/2 × 9.8 × t²

12.5 = 4.9t²

t² = 12.5/4.9

t² = 2.55

t = √2.55

t = 1.6 s

Therefore, the final velocity when the ball hits the ground is 15.65 m/s and the time it takes for the ball to hit the ground is 1.6 s

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Answer:

v = 1.98 m / s

Explanation:

Let's analyze this exercise a bit, initially the hanging box is at a height h, which is why it has gravitational power energy, as the system is removing the kinetic energy is zero and just when the box reaches the floor its potential energy has dropped to zero and the three bodies have kinetic energy, also between the box and the horizontal surface there is friction, so there is work. Let's use the relationship between work and energy

starting point. Before starting the movement

         Em₀ = U = m g h

final point. Just before the block hit the floor

         Em_f = K = ½ M v² + ½ I w² + ½ m v²

the speed of the two blocks must be the same to maintain the tension of the rope

The work of the friction force

the friction force opposes the movement so its work is negative

          W = - fr x

the law of equilibrium is the largest block

          N-W = 0

          N = W = Mg

we substitute

         W = - μ M g x

the relationship between the work of the non-conservative force (friction) and the energy is

         W = Em_f - Em₀

        - μ Mg x = ½ M v² + ½ I w² + ½ m v² - mg h

the moment of inertia of a hollow sphere is

         I = ⅔ m_s r²

angular and linear velocity are related

         v = w r

         w = v / r

the distance the horizontal block travels must be the same as the distance the vertical block travels

          x = h

let's substitute

         - μ M g h = ½ (M + m) v² + ½ (⅔ m_s r²) (v/r) ² - m g h

          (- μ M + m) g h = ½ (M + m + ⅔ m_s) v²

          v² = [tex]\frac{2(\mu \ M + m ) \ g h }{ M +m + \frac{2}{3} m_s}[/tex]

let's calculate

          v² = 2 (-0.15 7.6 +2.4) 9.8 0.61 / (7.6 + 2.4 + 2/3 1.2)

          v = [tex]\sqrt{\frac{42.32}{ 10.8} }[/tex]

          v = 1.98 m / s

A stronger force can cause a greater change in an object's velocity. The correct option is yes

The force exerted by the thrusters caused the pod to change direction. The thrusters on the ACM pod exerted the same strength force as thrusters on other pods on other missions. If the same strength force is exerted on two objects but the objects have different masses the object with less mass will have a greater change in velocity.

Therefore pod would have had to have a lighter mass meaning a smaller amount of force would have had to be exerted onto the pod making it move in the opposite direction.

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Answer:

The magnitude of the gravitational force which the astronaut experience in satellite [tex]= 448[/tex] Newton

Explanation:

As we know

Gravitational force  F = [tex]\frac{GMm}{r^2}[/tex]

Where G is the gravitational constant [tex]6.67 * 10^{-11}[/tex]

M is the mass of earth [tex]= 6* 10^{24}[/tex]

m is not given

r is the radius of earth [tex]6 400*10^3[/tex]

Substituting the given values, we get -

[tex]F = \frac{6.67*10^{−11}*6*10^{24}* m}{(6400*10^3)^2}[/tex]

[tex]m = 71.6[/tex] Kg

Gravitational force in the satellite

[tex]F = \frac{6.67*10^{−11}*6*10^{24}* m}{(6400*10^3 + 1600 *10^3)^2}\\F = 448[/tex]N