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A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$

$-F_{f} + ma = 0$

$\mu mg = ma$

$\mu = \frac{a}{g}$

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}$

Where:

d: is the distance traveled = 46.1 m

$v_{f}$ : is the final speed of the truck = 0 (it stops)

$v_{0}$ : is the initial speed of the truck = 17.9 m/s

$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

$\mu = \frac{a}{g}$

$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

$\mu = 0.35$

Therefore, the minimum coefficient of friction required is 0.35.

I hope it helps you!

4 0

3 years ago

Explain in your own words how the Doppler Effect is also applicable in our study of light.

arlik [135]

well in my own words, i'd saw the the doppler effect is similar to light because sound has a speed, and light does too.

so my theory is if you go fast enough everything would just become black, or maybe white? idk its hard to explain

but what my point is, is taht the doppler effect works in the same way, like if a car is moving towards you the sound is being emitted from the car and being pushed by the speed of the car making it have a much higher pitch, when the car is going away however it drops to a lower pitch due the the sound waves being DRAGGED by the car.

there hoped this helped I guess

8 0

2 years ago

What charge do electrons have? A. no charge B. a positive charge C. a negative charge

Mandarinka [93]

Electrons have a negative charge.

5 0

2 years ago

If the fundamental frequency of a musical instrument is 42Hz, what is the frequency of the second harmonic?

timurjin [86]

For the majority of instruments f = n f0 where f is the resonating frequency, n is any whole number and f0 is the fundamental.
This applies to trumpets, violins, flutes and a broad range. 
In such a case the first harmonic would be at n=1 and the second harmonic would be at n=2 

which gives a frequency of 84 Hz

7 0

2 years ago

How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00

laiz [17]

Answer:

Work= -7.68×10⁻¹⁴J

Explanation:

Given data

q₁=q₂=1.6×10⁻¹⁹C

r₁=2.00×10⁻¹⁰m

r₂=3.00×10⁻¹⁵m

To find

Work

Solution

The work done on the charge is equal to difference in potential energy

W=ΔU

$Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\ Work=-7.68*10^{-14}J$

4 0

3 years ago