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Sergeeva-Olga [200]

4 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]4 years ago

4 0

Answer:

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

Explanation:

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How much heat is required to heat the temperature of 338g of aluminum by?

enyata [817]

Specific heat<span> is another physical property of matter. ... we con now ask the following question: by </span>how much<span> will the </span>temperature<span> of an .</span>

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3 years ago

A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod

Anestetic [448]

(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
$L/8=1.3m/8=0.1625 m$

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
$c_6 = 0.0812m+0.1625m=0.2437 m$

(c) The total charge is $Q=-3 \cdot 10^{-8}C$ . To get the charge on each piece, we should divide this value by 8, the number of pieces:
$Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C$

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
$q= -3.75\cdot 10^{-9}C$
If we approximate piece 6 as a single charge, the electric field is given by
$E=k_e \frac{q}{d^2}$
where $k_e=8.99\cdot 10^9Nm^2C^{-2}$ and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
$E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m$
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.

5 0

3 years ago

The pivot point of a ____ is called the fulcrum.

Alexxx [7]

Hello:

The pivot point of a LEVER is called the fulcrum.

8 0

3 years ago

The plane of a rectangular coil, 7.2 cm by 3.7 cm, is perpendicular to the direction of a uniform magnetic field B. If the coil

netineya [11]

Answer:

The rate of change of magnetic field is 2.23 T/s.

Explanation:

Given that,

Dimension of rectangular coil is 7.2 cm by 3.7 cm.

Number of turns in the coil, N = 104

Resistance of the coil, R = 12.4 ohms

Current, I = 0.05 A

We need to find the rate of change of magnetic field in the coil. The induced emf is given by the rate of change of magnetic flux. So,

$\epsilon=-\dfrac{d\phi}{dt}$

Ohm's law is :

$\epsilon=IR$

So,

$IR=-\dfrac{d\phi}{dt}\\\\IR=-\dfrac{d(NBA)}{dt}\\\\IR=-NA\dfrac{dB}{dt}\\\\\dfrac{dB}{dt}=\dfrac{IR}{NA}\\\\\dfrac{dB}{dt}=\dfrac{0.05\times 12.4}{104\times 7.2\times 10^{-2}\times 3.7\times 10^{-2}}\\\\\dfrac{dB}{dt}=2.23\ T/s$

So, the rate of change of magnetic field is 2.23 T/s.

4 0

3 years ago

Please Help with this ASAP:<br> A wave vibrates 45 times in 30 seconds, calculate its frequency

JulijaS [17]

Th answer would be 1.5

4 0

3 years ago

Read 2 more answers