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3 years ago

13

Calculate the speed of a car that traveled 200 kilometers in 3 hours.

2 answers:

nignag [31]3 years ago

8 0

Answer:

The units (km/h) tell you how to do this! 200km/3h = 66.66666666…. BUT technically you only have ONE significant digit: 3 so 66.666… rounded to ONE digit is 70km/h but that is probably not important in this intro class so V = 66.67 or 67 km/h

Kamila [148]3 years ago

7 0

Answer:

66.7km/h

Explanation:

s=d/t

200/3

=66.6666666

round off: 66.7km/h

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A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp

amm1812

Answer:

the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = $-\beta\frac{dx}{dt} = \frac{-10dx}{dt}$

$\beta = 10$

By Newtons second law ,

The diffrential equation of motion with damping is given by

$m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}$

substitute the value of m =1kg, k = 21N/M, and $\beta = 10$

$1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}$

$\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 21x = 0$

suppose the equation of the form $x =e^m^t$ ,

and the auxilliary equation is given by

$m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3$

The general solution for the above differential equation is

$x(t) =C_1e^{-3t}+C_2e^{-7t}$

Derivate with respect to t

$x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}$

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

$x'(0) = 0$

substitute the initial condition

$C_1 +C_2 = 1$

$-3C_1-7C_2=0$

Solve the above equation to get C₁ and C₂

$C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}$

substitute for C₁ and C₂ in general solution

$x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Thus the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

3 0

3 years ago

Read 2 more answers

In noisy factory environments, it's possible to use a loudspeaker to cancel persistent low-frequency machine noise at the positi

Vaselesa [24]

Answer:

7.39 m or 3.61 m

Explanation:

$\lambda$ = Wavelength

f = Frequency = 90 Hz

v = Speed of sound = 340 m/s

Path difference of the two waves is given by

$s_1-s_2=\frac{\lambda}{2}$

Velocity of wave

$v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{340}{90}\\\Rightarrow \lambda=3.78\ m$

$s_1=s_2\pm\frac{\lambda}{2}\\\Rightarrow s_1=5.5\pm \frac{3.78}{2}\\\Rightarrow s_1=7.39\ m, 3.61\ m$

So, the location from the worker is 7.39 m or 3.61 m

7 0

3 years ago

Where are you likely to find the oldest oceanic crust?

Stells [14]

Answer:

it would be letter E. near oceanic ridges

Explanation:

new ocean crust is formed at the mid ocean ridges

4 0

3 years ago

19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,

sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

$t=15.2 s$ is the time it takes to the child to reach the bottom of the slope

$V_{o}=0$ is the initial velocity (the child started from rest)

$\theta=15\°$ is the angle of the slope

$d$ is the length of the slope

Now, the Force exerted on the sled along the ramp is:

$F=ma$ (1)

Where $m$ is the mass of the sled and $a$ its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

$F=mg sin \theta$ (2)

Where $g=9.8 m/s^{2}$ is the acceleration due gravity

Then:

$ma=mg sin \theta$ (3)

Finding $a$ :

$a=g sin \theta$ (4)

$a=9.8 m/s^{2} sin(15\°)$ (5)

$a=2.5 m/s^{2}$ (6)

Now, we will use the following kinematic equations to find $d$ :

$V=V_{o}+at$ (7)

$V^{2}=V_{o}^{2}+2ad$ (8)

Where $V$ is the final velocity

Finding $V$ from (7):

$V=at=(2.5 m/s^{2})(15.2 s)$ (9)

$V=38 m/s$ (10)

Substituting (10) in (8):

$(38 m/s)^{2}=2(2.5 m/s^{2})d$ (11)

Finding $d$ :

$d=288.8 m$

6 0

3 years ago

A rock thrown straight up takes 4.2 s to reach its maximum height what was its initial velocity

liubo4ka [24]

Consider the upward direction of motion as positive and downward direction of motion as negative.

a = acceleration due to gravity in downward direction = - 9.8 $\frac{m }{s^{2}}$

v₀ = initial velocity of rock in upward direction = ?

v = final velocity of rock at the highest point = 0 $\frac{m }{s}$

t = time to reach the maximum height = 4.2 sec

Using the kinematics equation

v = v₀ + a t

inserting the values

0 = v₀ + (- 9.8) (4.2)

v₀ = 41.2 $\frac{m }{s}$

8 0

3 years ago