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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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What are the difference between conductor and insulator?

Anvisha [2.4K]

Answer:

A conductor allows energy e.g. electric charge or heat to pass through it easily. While an insulator does not let electric current or heat to travel through it.
Insulators have strong molecular bonds. While molecular bonds are very weak in conductors.
Insulators have very low conductivity. While in conductors, it is very high.
Insulators have a very high resistance and therefore the electrons are held together very firmly. The conductors, on the other hand, have a very low resistance.
Insulators do not have any electric field, neither inside nor on the surface. While in conductors, it is found on the surface and continues to be zero in the inner part of the conductor.

5 0

2 years ago

What is #6<br><br> IM GIVING 40 POINTS

frosja888 [35]

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4 0

2 years ago

A force of 30n is applied to a screwdriver to pry the lid off of a can of paint. The screwdriver applies 75n of force to the lid

gizmo_the_mogwai [7]

The mechanical advantage of the screwdriver that is being described above is equal to 75N. This means that for every 30N that is applied on the screwdriver, this simple machine would in turn apply 75N of force to the lid of the can.

3 0

2 years ago

An electron with a speed of 1.7 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts

Alex Ar [27]

Answer:

y = 77.74 10⁻⁵ m

Explanation:

For this exercise we can use Newton's second law

F = m a

a = F / m

a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹

a = 0.538 10¹⁵ m / s

This is the vertical acceleration of the electron.

Now let's use kinematics to find the time it takes to move the

x= 29 mm = 29 10⁻³ m

On the x axis

v = x / t

t = x / v

t = 29 10⁻³ / 1.7 10⁷

t = 17 10⁻¹⁰ s

Now we can look for vertical distance at this time.

y = $v_{oy}$ t + ½ a t²

y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²

y = 77.74 10⁻⁵ m

3 0

3 years ago

Read 2 more answers

Please. Physics is so difficult.

Softa [21]

Answer:

0.010 m

Explanation:

So the equation for a pendulum period is: $y=2\pi\sqrt{\frac{L}{g}}$ where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:

$1.99 s = 2(3.14)\sqrt{\frac{L}{9.8 m\backslash s^2}}\\$

Evaluate the multiplication in front

$1.99 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}$

Divide both sides by 6.28

$0.317 s= \sqrt{\frac{L}{9.8 m\backslash s^2}}$

Square both sides

$0.100 s^2= \frac{L}{9.8 m\backslash s^2}$

Multiply both sides by m/s^2 (the s^2 will cancel out)

$0.984 m = L$

Now now let's find the length when it's two seconds

$2.00 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}}$

Divide both sides by 6.28

$0.318 s = \sqrt{\frac{L}{9.8 m\backslash s^2}$

Square both sides

$0.101 s^2 = \frac{L}{9.8 m\backslash s^2}$

Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)

$0.994 m = L$

So to find the difference you simply subtract

0.984 - 0.994 = 0.010 m

4 0

1 year ago

Read 2 more answers