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2 years ago

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Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu

t Jari left 15 minutes later. Both drove at a constant speed. The following graphs show the distance (in miles) each of them has traveled on I-10 at time t (in minutes), where t = 0 is 7:00 A.M.

1 answer:

Veronika [31]2 years ago

3 0

Answer:

Jari

Explanation:

The question requires to know who is traveling faster. This is done by comparing the gradients. The steeper the slope (high gradient), the faster the speed and vice versa.

From Jari's line, the starting point is (0, 0) and another point is (6, 7)

The gradient being change in y to change in x

Change in y=7-0=7

Change in x=6-0=6

Slope is 7/6

For Jade, first point is (0, 10) then another point is (6, 16)

Change in y=16-10=6

Change in x=6-0=6

Slope is 6/6=1

Clearly, 7/6 is greater than 6/6 or 1 hence Jari is faster than Jade

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An underground tunnel has two openings, with one opening a few meters higher than the other. If air moves past the higher openin

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Answer:

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Explanation:

This is in accordance with bernoulli's law of fluid flow which states that the pressure exerted by a moving fluid is lesser than it would exert if it were at rest.

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2 years ago

A front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is t

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Explanation:

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Explanation:

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2 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

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2 years ago

A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th

zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d) K = 11.8835 J

U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)

⇒ f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒ E = (1.15)(8.40)²(0.650)² / 2

⇒ E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)

and

U = (1/2)*m*ω²*x² (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒ K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒ U = 5.2581 J

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2 years ago

If you drive through water, your brakes may become slippery and ineffective. To dry the brakes off, __________.

GenaCL600 [577]

Answer:

I am not a driver, but I think it's C.

Explanation:

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2 years ago

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