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Sergeeva-Olga [200]

3 years ago

9

A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2. What is the minimum tractive effort required for

this vehicle to maintain a 70 mi/h speed on a 5% upgrade through an air density of 0.002045 slugs/ft3?

1 answer:

emmasim [6.3K]3 years ago

4 0

Answer:

i put this in the calculator and my answer is 600. hope this helps

Explanation:

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Substitution solve for x & y<br>-4x - 2y equals 14<br>-10x+7y=-25

den301095 [7]

Answer:

-4×-2y=14 (1)

-10×+7y=-25 (2)

multiplying eq 1 by 7 and eq 2 by 2 and add eq. 1 and 2

-28×-14y=98

-20×+14y=-50

___________

-28×=48

×=48/-28

×=-12/7

now

-4×-2y=14

-4*-12/7-2y=14

48/7-2y=14

-2y=14-48/7

-2y=(98-48)/7

-2y=50/7

y=-50/14

y=-25/7

8 0

3 years ago

A 12-kg piece of metal displaces 1.6 L of water when submerged. Part A Find its density. Express your answer to two significant

Tatiana [17]

Answer:

ρ = 7500 kg/m³

Explanation:

Given that

mass ,m = 12 kg

Displace volume ,V= 1.6 L

We know that

1000 m ³ = 1 L

Therefore V= 0.0016 m ³

When metal piece is fully submerged

We know that

mass = Density x volume

$m=\rho \times V$

Now by putting the values in the above equation

$\rho=\dfrac{12}{0.0016}\ kg/m^3$

ρ = 7500 kg/m³

Therefore the density of the metal piece will be 7500 kg/m³.

6 0

3 years ago

What is a key phrase for remembering the order of the planets?

horsena [70]

My very eager mother just served us nine pizzas

7 0

3 years ago

A football punter accelerates a .55 kg football

Ronch [10]

Answer:

17.6 N

Explanation:

The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

$F=\frac{\Delta p}{\Delta t}$

where

$\Delta p$ is the change in momentum of the football

$\Delta t=0.25 s$ is the time elapsed

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m = 0.55 kg is the mass of the football

u = 0 is the initial velocity (the ball starts from rest)

v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we find the force exerted on the ball:

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N$

5 0

3 years ago

An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force

sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = $10ms^{-2}$ , F = ma = 15*10 = 150N

2) m = 3kg, v = $4ms^{-1}$ , r = 4m, F = $\frac{mv^{2} }{r}$

$\frac{3*4^{2} }{4}$ = 12N

3) a = $10ms^{-2}$ , r = 10m, v=?

F = $\frac{mv^{2} }{r}$ and F = ma

equating the two equations and cancelling a, we have:

$\frac{v^{2} }{r}$ = a

making v the subject of formula, we have:

v = $\sqrt{ar}$

= $\sqrt{100}$

= 10 $ms^{-1}$

4) r = 10m, v = $5ms^{-1}$ , a = ?

F = $\frac{mv^{2} }{r}$

F = ma

equating the above equations and making a subject of formula, we get:

a = $\frac{v^{2} }{r}$

a = 25/10 = 2.5 $ms^{-2}$

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is tripled

F = $(3v)^{2}$

F = 9 $v^{2}$

We can see that the force will be 9X greater than it was.

9) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is doubled

F = $(2v)^{2}$

F = 4 $v^{2}$

We can see that the force will be 4X greater than it was.

10) F = $\frac{mv^{2} }{r}$

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

7 0

3 years ago