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mote1985 [20]
2 years ago
9

A rare and valuable antique chest is being moved into a truck using a 4.00 m long ramp. the kj weight of the chest plus packing

material is 1,500 n. if the truck bed is 1.00 m above the ground, find the work done by the movers as they slide the chest up the ramp if the coefficient of friction between the chest and the ramp is 0.200.
Physics
1 answer:
Lisa [10]2 years ago
7 0

First let us calculate for the angle of inclination using the sin function,

sin θ = 1 m / 4 m

θ = 14.48°

 

Then we calculate the work done by the movers using the formula:

W = Fnet * d

 

So we must calculate for the value of Fnet first. Fnet is force due to weight minus the frictional force.

Fnet = m g sinθ – μ m g cosθ

Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48

Fnet = 84.526 N

 

So the work exerted is equal to:

W = 84.526 N * 4 m

<span>W = 338.10 J</span>

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Two speakers, A and B, produce identical sound waves. A listener is 3.2 m away from speaker A. The listener finds the lowest fre
earnstyle [38]

Answer:

  0.83 m or 5.57 m

Explanation:

Destructive interference will occur when the distances from the speakers differ by 1/2 wavelength.

The length of 1 cycle of 72.4 Hz is ...

  λ = v/f = (343 m/s)/(72.4 Hz) ≈ 4.738 m

So, the distance of the listener from speaker B is ...

  3.2 m ± (4.738 m)/2 = {0.83 m, 5.57 m} . . . either of these distances

_____

The location could be at additional multiples of 4.738 m, but we think not. The sound intensity drops off with the square of the distance from the speaker, so identical sound waves from the speakers will sound quite different at different distances from the speakers. For best interference, the distances need to be as close to the same as possible. That will be at 3.2 m and 5.57 m.

_____

<em>Comment on the speed of sound</em>

We don't know what speed you are to use for the speed of sound. We have used 343 m/s. Some sources use 340 m/s, which will give a result different by 2 or 3 cm.

8 0
2 years ago
Often what one expects to see influences what is perceived in the surrounding environment. Please select the best answer from th
Stella [2.4K]
Hello <span>Andijwiltbank 
</span>

Question: <span>Often what one expects to see influences what is perceived in the surrounding environment. True or False?

Answer: True

Reason: What we observe about the environment decides what we believe about it and how we react.


Hope This Helps :-)
-Chris</span>
8 0
2 years ago
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A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in 15 seconds
solong [7]

A. Formula: F=ma or F/m=a

10,000N/1,267kg≈7.9m/s^{2}

B. Formula: a=\frac{V-V_{0} }{t} and s=d/t

speed= 394.6/15

s=26.3m/s

a=\frac{26.3-0}{15}

a=1.75m/s^{2}

C. 7.9-1.75=difference of 6.15m/s^{2}

D. The force that most likely caused this difference is friction forces

3 0
2 years ago
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A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
2 years ago
Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
2 years ago
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