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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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Why do scientists not use US customary units when reporting their data?

Marianna [84]

From britaññica it said time and money. They didn’t have either to switch over from the industrial period and never did. Also from my own person reasoning i think most of the world uses not US customary, so to make stuff more accessible. hope this helps!

8 0

3 years ago

A 675 kg car moving at 15.7 m/s hits from behind another car moving at 9.6 m/s in the same direction. If the second car has a ma

Maslowich

Answer:

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Explanation:

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3 0

3 years ago

1. An excited lithium atom emits a red light with wavelength a = 671nm. What is the corresponding photon energy? hc (6.63 x 10-3

Rudiy27

Answer:

E = 2,964 10⁻¹⁹ J

Explanation:

The energy of the photons is given by the Planck relation

E = h f

the speed of light is related to wavelength and frequency

c = λ f

we substitute

E = h c /λ

let's reduce the magnitude to the SI system

λ = 671 nm = 671 10⁻⁹ m

let's calculate

E = 6.63 10⁻³⁴ 3 10⁸ /671 10⁻⁹

E = 2,964 10⁻¹⁹ J

6 0

3 years ago

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a

Burka [1]

Solution :

a). B at the center :

$=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$

So,

$\frac{I_1}{d_1}= \frac{I_2}{d_2}$

$=\frac{16}{21}=\frac{I_2}{32}$

$I_2=24.38$ A

Therefore, the current in the outer wire is 24.38 ampere.

3 0

3 years ago

Read 2 more answers

Doe anyone get this

8_murik_8 [283]

Answer:

we know a = F/ M

Explanation:

2 m/s²
0.19 m/s²
9.25 m/ s²
0.04 m/s²
100.39 m/s²

4 0

3 years ago