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nordsb [41]
3 years ago
8

to measure the static friction coefficient between a block and a vertical wall, a spring is attached to the block, is pushed on

the end in a direction perpendicular to the wall until the block does not slip downward. If the spring is compressed, what is the coefficient of static friction
Physics
1 answer:
Stolb23 [73]3 years ago
8 0

Answer:

μ = mg/kx

Explanation:

Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.

Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.

N = F' = kx

So, F = μN = μkx

μkx = mg

So, μ = mg/kx

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A 1500-kg car accelerates from 0 to 25 m/s in 7.0s with negligible friction and air resistance. What is the average power delive
NARA [144]

Answer:

90 hp

Explanation:

Power = work / time

P = ½ (1500 kg) (25 m/s)² / 7.0 s

P = 67,000 W

P = 90 hp

4 0
3 years ago
Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te
seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant \gamma =1.4

specific heat is given as =\frac{\gamma R}{\gamma -1}

gas constant =287 J⋅kg−1⋅K−1

Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}

specific heat at constant volume

Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}

change in internal energy = Cv(T_2 -T_1)

                            \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}

change in enthalapy \Delta H = Cp(T_2 -T_1)

                                 \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}

change in entropy

\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})

\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})

\Delta S = 382.79 J kg^{-1} K^{-1}

7 0
3 years ago
Which of the following factors most affect the rate at which waves erode land features along the shore?
OLEGan [10]

Answer:

The biggest factor affecting coastal erosion is the strength of the waves breaking along the coastline. A wave's strength is controlled by its fetch and the wind speed. Longer fetches & stronger winds create bigger, more powerful waves that have more erosive power.

Explanation:

hope it helps !

5 0
2 years ago
Read 2 more answers
A bald eagle is flying to the left with a speed of 34 meters
Shtirlitz [24]

Answer:

the speed after 3 seconds is 10 m/s

Explanation:

The computation of the speed is shown below:

As we know that

V = U  + at

Here,

U = 34 m/s

a =  - 8 m/s²

t = 3 Sec

V = velocity after 3 sec

V  = 34 + (-8)3

 = 34 - 24

 V = 10 m/s

Hence, the speed after 3 seconds is 10 m/s

4 0
2 years ago
A parallel-plate capacitor has a voltage of 391 v applied across its plates, then the voltage source is removed. what is the vol
andrezito [222]

When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling C_0 the capacitance of the capacitor in air, the charge Q, the capacitance C_0 and the voltage (V_0=391 V) are related by

C_0 =\frac{Q}{V_0} (1)


when the source is disconnected the charge Q remains on the capacitor.


When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):

C=k C_0 = 5.4 C_0

this is the new capacitance. Since the charge Q on the plates remains the same, by using eq. (1) we can find the new voltage across the capacitor:

V=\frac{Q}{C}=\frac{Q}{5.4 C_0}

And since Q=C_0 V_0, substituting into the previous equation, we find:

V=\frac{C_0 V_0}{5.4 C_0}=\frac{V_0}{5.4}=\frac{391 V}{5.4}=72.4 V



7 0
3 years ago
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