a student practicing for a track meet ran 250 m in 30 seconds. What was her average speed?

Answer:

(1.9756, -2.1951)

Explanation:

The center of mass equation is: [tex]x_{cm}[/tex] = [tex]\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}[/tex], where m represents the masses and x represents the position.

In order to find the coordinates of the center of mass, we need to use this equation for both the x-values and the y-values.

x-values:

[tex]x_{cm}[/tex] = [tex]\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}[/tex] = [tex]\frac{4(-1)+6(4)+7(-3)+10(5)+2(-2)+12(3)}{4+6+7+10+2+12}[/tex] = [tex]\frac{(-4)+(24)+(-21)+(50)+(-4)+(36)}{41}[/tex] = [tex]\frac{81}{41}[/tex] = 1.9756

y-values:

[tex]y_{cm}[/tex] = [tex]\frac{m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3} + m_{4}y_{4} + m_{5}y_{5} + m_{6}y_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}[/tex] = [tex]\frac{4(1)+6(2)+7(-2)+10(-4)+2(4)+12(-5)}{4+6+7+10+2+12}[/tex] = [tex]\frac{(4)+(12)+(-14)+(-40)+(8)+(-60)}{41}[/tex] = [tex]\frac{-90}{41}[/tex] = -2.1951

center of mass:

(1.9756, -2.1951)

Answer:

[tex]1.84[/tex]×[tex]10^{-26}[/tex]

Explanation:

specific charge = [tex]\frac{charge}{mass}[/tex]  so by rearranging for mass we get

mass= [tex]\frac{charge}{sepcifc charge}[/tex]

[tex]\frac{1.6×10^{-26} }{1.7×10^{-27} }[/tex] = answer in kg

Answer:

8

Explanation:

(8√2)² = x² + x²

8² × √2² = 2x²

64 × 2 = 2x²

128 = 2x²

64 = x²

x = 8

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Answer:

i am guessing for reflection but not so sure

Answer:

yes

it does you weigh less on the equator than at the North or South Pole, but the difference is small. Note that your body itself does not change. Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.

Answer:D

Explanation:

Answer:

7.23407 [tex]\frac{m}{s^2}[/tex]

Explanation:

(I will not include units in calculations)

I'm assuming FBD's are already drawn, so I will work from there.

Let the 2.2kg block equal [tex]m_2[/tex], and the 20kg block equal [tex]m_1[/tex].

Summation equation for [tex]m_2[/tex]: [tex]\sum F_x=F_t_2-(F_f+F_g_x)=m_2a[/tex], [tex]\sum F_y=F_n-F_g_y=0[/tex]

Summation equation for [tex]m_1[/tex]: [tex]\sum F_y=F_g-F_t_1=m_1a[/tex]

Torque Summation Equation: [tex]\sum\tau=F_t_1*r-F_t_2*r=I\alpha[/tex]

Do some plugging in with the values given: [tex]\sum\tau=F_t_1*r-F_t_2*r=.5Mr^2\alpha[/tex]

Replace [tex]\alpha[/tex] with [tex]\frac{a}{r}[/tex], and cancel out the r's.

[tex]\sum\tau=F_t_1-F_t_2=.5Ma[/tex]

This step is important: Rearrange the force summation equation to solve for each tension force.

[tex]F_t_2=m_2a+F_f+F_g_x\\F_t_1=m_1g=m_1a[/tex]

Perform Substitution: [tex]\sum\tau=m_1g-m_1a-(m_2a+F_f+F_g_x)=.5Ma[/tex]

Now, we need to find the friction force and the horizontal component of the force of gravity.

Note that [tex]F_f=[/tex]μ[tex]F_n[/tex]

And based on our earlier summation equation: [tex]F_n=F_g_y[/tex]

First, break [tex]F_g[/tex] into x and y components. [tex]F_g_y=F_g\cos(\theta)[/tex], [tex]F_g_x=F_g\sin(\theta)[/tex]

Perform substitution with this and the fact that [tex]F_g=mg[/tex].

[tex]\sum\tau=m_1g-m_1a-(m_2a+\mu*m_2g\cos(\theta)+m_2g\sin(\theta))=.5Ma[/tex]

Solving for a, plugging in numbers yields an answer of 7.23407 [tex]\frac{m}{s^2}[/tex]

Answer:

7.23407

Explanation:

easy

Answer:

D

Explanation:

From the information given:

The angular speed for the block [tex]\omega = 50 \ rad/s[/tex]

Disk radius (r) = 0.2 m

The block Initial velocity is:

[tex]v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s[/tex]

Change in the block's angular speed is:

[tex]\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s[/tex]

However, on the disk, moment of inertIa is:

[tex]I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2[/tex]

The time t = 10s

∴

Frictional torques by the wall on the disk is:

[tex]T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m[/tex]

Finally, the frictional force is calculated as:

[tex]F = \dfrac{T}r{}[/tex]

[tex]F= \dfrac{0.6}{0.2} \\ \\ F = 3N[/tex]

first of all the formula of force is F=ma,so we are searching for m,so we can divide a on both sides F/a=m, after this substitute the values given above 64N/4.51=14.2°Kg

Answer:

The object is moving at constant speed.

Explanation:

The spaces between the dots are equal.

Answer:We have , a relation in frequency f and wavelength  λ of a wave having the velocity v as ,

           v=fλ ,

given f=60Hz , λ=20m ,

therefore velocity of wave , v=60×20=1200m/s

Answer:

Universe

Explanation:

I took the quiz.

Answer:

1/Re= 1/R1 + 1/R2

Explanation:

Two resistors are connected in parallel. If R1 and R2 represent the resistance in Ohms (Ω) of each resistor, then the total resistance R is given by  [tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]. Thus, the rate of R changes when R₁ = 117 Ω and

R₂ = 112 Ω is 0.25 Ω/min

For a given resistor connected in parallel;

[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]

Making R from the left-hand side the subject of the formula, then:

[tex]\mathbf{R = \dfrac{R_1R_2}{R_1+R_2}}[/tex]

Given that:

Now, replacing the values in the above previous equation, we have:

[tex]\mathbf{R = \dfrac{13104}{229}}[/tex]

However, the differentiation of R with respect to time t will give us the rate at which R is changing when R1=117Ω and R2=112Ω.

So, by differentiating the given equation of the resistor in parallel with respect to time t;

[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex],   we have:

[tex]\mathbf{\dfrac{1}{R^2}(\dfrac{dR}{dt})=\dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})}[/tex]

[tex]\mathbf{(\dfrac{dR}{dt})=R^2 \Bigg[ \dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=(\dfrac{13104}{229})^2 \Bigg[ \dfrac{0.4}{117^2}+\dfrac{0.6}{112^2}\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=3274.44 \Bigg[ (7.7052 \times 10^{-5} )\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=0.25\ \Omega /min}[/tex]

Therefore, we can conclude that the rate at which R is changing R1=117Ω and R2=112Ω is 0.25 Ω/min

Learn more about resistors here:

https://brainly.com/question/17390255?referrer=searchResults

Answer:

Force,friction,inertia and momentum

Explanation:

The speed that the marble is moving at can be determined by the amount of force used when pushed or pulled and what kind of surface it's on.Momentum is also a factor because of the mass of the marbles.

Answer:

r = 1,224 10⁻² m

Explanation:

For this exercise let's use Newton's second law

          F = m a

the force is magnetic

         F = q v x B

The bold letters indicate vectors, the module of this expesion is

          F = q v B

The direction of the force is found by the right hand rule

thumb points in the direction of the velicad + x

fingers extended in the direction of B -z

the palm is in the direction of the force + and

the acceleration of the proton is cenripetal

          a = v² / r

we substitute

          q v B = m v² / r

          r = [tex]\frac{m \ v}{q \ B}[/tex]

let's calculate

         r = [tex]\frac{1.7 \ 10^{-27} \ 5.760063 \ 10^7 }{1.6 \ 10^{-19} \ 0.5 }[/tex]

         r = 1,224 10⁻² m

Answer:

It is the force that pulls down an object on the air

Answer: a downward pull on any object

Explanation:

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

The density of air at 1 atm pressure and 20º C is 1.2 kg/m³  

the volume V = 20,000/(1.2×9.8) =  1700 m³

Answer:

Equation of motion is x(t) = [tex]-te^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] sin(4t)

Explanation:

P.S - The exact question is -

Given - A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at t = 0, an external force equal to  [tex]f(t) = 8 cos(4t)[/tex] is applied to the system.

To find - Find the equation of motion if the surrounding medium offers a damping force that is numerically equal to 8 times the instantaneous velocity.

Proof -

Given that,

Mass = 1 slug

We know that, 1 slug = 32 lb

Now,

Force, f = kx

⇒32 = k(2)

⇒k = 16

Now,

Given that, C = 8 ( 8 times the instantaneous velocity)

Now,

The differential equation of motion is equals to

mx'' + Cx' + kx = 8 cos(4t)

⇒x'' + 8x' + 16x = 8 cos(4t)               ...........(1)

Let the General solution of equation (1) be

x(t) = x(c) + x(p)

Now,

The auxiliary equation is

m² + 8m + 16 = 0

m² + 4m + 4m + 16 = 0

m (m+4) + 4 (m+4) = 0

⇒(m+4)(m+4) = 0

⇒m = -4, -4

So,

The Complimentary equation becomes

x(c) = [tex]Ae^{-4t} + Bte^{-4t}[/tex]                  ...........(2)

Now,

Let the particular solution be

x(p) = C cos(4t) + D sin(4t)

x'(p) = -4C sin(4t) + 4D cos(4t)

x''(p) = -16C cos(4t) - 16D sin(4t)

It also satisfy equation (1)

Equation (1) becomes

-16C cos(4t) - 16D sin(4t) + 8 [ -4C sin(4t) + 4D cos(4t) ] + 16 [ C cos(4t) + D sin(4t) ] = 8 cos(4t)

⇒-16C cos(4t) - 16D sin(4t) - 32C sin(4t) + 32D cos(4t) ] + 16C cos(4t) + 16D sin(4t) ] = 8 cos(4t)

⇒-4C sin(4t) + 4B cos(4t) = cos(4t)

By comparing, we get

4B = 1 , A = 0

⇒ B = [tex]\frac{1}{4}[/tex] , A = 0

So, The particular solution becomes

x(p) =  [tex]\frac{1}{4}[/tex] sin(4t)

Now,

The General solution becomes

x(t) = [tex]Ae^{-4t} + Bte^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] sin(4t)       .......(3)

Now,

Given that, At t = 0, initial velocity is zero and the system starts equilibrium

⇒x(0) = 0, x'(0) = 0

By putting t = 0 in equation (3) , we get

A = 0

Now,

Differentiate equation (3), we get

x'(t) = [tex]-4Ae^{-4t} + Be^{-4t} - 4Bte^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] *4 cos(4t)

Put t = 0, we get

0 = -4A + B + 1

⇒B = -1

∴ we get

The general solution becomes

x(t) = [tex]-te^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] sin(4t)

Equation of motion is x(t) = [tex]-te^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] sin(4t)

Answer: yes

Explanation:

Weight is the gravitational force experienced on a body. If you move up to higher altitudes, the distance between you and earth increases. ... Yes, weight drops as you go up in altitude (because of diminishing gravity), though your mass remains the same. However, the effect is not huge.

Answer:  lose weight at high altitudes.

Explanation:

Answer:

Just a week at high altitudes can cause sustained weight loss, suggesting that a mountain retreat could be a viable strategy for slimming down. Overweight, sedentary people who spent a week at an elevation of 8,700 feet lost weight while eating as much as they wanted and doing no exercise

Answer:

option c

Heat is transferred to your hand from a warm cup

conduction is the process of transferring of heat from one material to another when they are in contact

hope it helps

Answer:

C) Mass of the ball

Explanation:

Independent variable is the variable the researcher changes.

Since Martin is testing the mass of the ball, he'll be using different balls and that is the only thing he changes.

The distance traveled by the ball is the dependent variable since it depends on the mass of the ball.

The height and length of the ramp are the constant variables since that's the only ones that remain the same throughout this experiment.

Answer:

Newtons second law

Explanation:

Depends on mass

Answer:

2nd law

Explanation:

:))))))))))))))

Answer:

The atomic mass of an element is affected by the distribution of its isotopes because each isotope of an element has a different number of neutrons in the nuclei of its atoms.

Explanation:

please give me brainlyiest if its right

Answer:

The building was Painted

The building was painted last year by someone.

Answer:

  a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex]  ,       t = [tex]\sqrt{ \frac{2d}{a} }[/tex]

Explanation:

To solve this exercise we must use Newton's second law

For the block

let's set a reference system with the x axis parallel to the plane

X axis

         Wₓ - T = m a

Y axis  

         N- W_y = 0

         N = W_y

for pulley

          ∑τ = I α

           T R = (½ m_p R²) α

let's use trigonometry for the weight components

         sin β = Wₓ / W

         cos β = W_y / W

         Wx = W sin β

angular and linear variables are related

          a = α R

          α = a / R

we substitute and group our equations

         W sin β - T = m a

         T R = ½ m_p R² (a / R)

         W sin β - T = m a

                        T = ½ m_p a

we solve the system of equations

         W sin β = (m + ½ m_p) a

          a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex]

let's find the time to travel the distance (d) through the block

          x = v₀ t + ½ a t²

          d = 0 + ½ a t²

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