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Alex777 [14]
2 years ago
10

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force

of static friction f on the block?
A) f > mg
B) f = mgsin(?)
C) f > mgcos(?)
D) f = mgcos(?)
E) f > mgsin(?)

Physics
1 answer:
nignag [31]2 years ago
8 0

Answer:

Option B

Explanation:

For a system of block on inclined ramp shown in the attached image. From the attached image, the Normal force N, weight mg and frictional force f act on the block.  The sum of vertical forces should be zero just as sum of vertical forces should be zero when the system is in equilibrium condition.

Taking sum of forces along the inclined plane we deduce that  

[tex]f=mgsin \theta [tex]

Therefore, option B is the correct option.

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6CO2 + 6H2O → C6H12O6 + 6O2
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What is the wavelength of a wave if its frequency is 256 Hz and speed of the wave is 350 m/s?
DiKsa [7]

Answer:

Explanation:

The equation for this is

f=\frac{v}{\lambda} where f is the frequency, v is the velocity, and lambda is the wavelength. Filling in:

256=\frac{350}{\lambda} and

\lambda=\frac{350}{256} which means that

the wavelength is 1.37 m, rounded to the correct number of significant digits.

7 0
2 years ago
A car travels 50 km in 25 km /h and then travels 60km with a velocity 20 km/h and then travels 60km with a velocity 20 km/h in t
Butoxors [25]

Answer:

v = 21.25 km/h

The average velocity is 21.25km/h

Explanation:

Average velocity = total displacement/time taken

v = d/t

Given;

A car travels 50 km in 25 km /h

d1 = 50km

v1 = 25km/h

time taken = distance/velocity

t1 = d1/v1

t1 = 50/25 = 2 hours

and then travels 60km with a velocity 20 km/h

d2 = 60km

v2 = 20km/h

t2 = d2/v2 = 60/20

t2 = 3 hours

and then travels 60km with a velocity 20 km/h in the same direction

d3 = 60km

v3 = 20km/h

t3 = d3/v3 = 60/20

t3 = 3 hours

Average velocity = total displacement/total time taken

v = (d1+d2+d3)/(t1+t2+t3)

v = (50+60+60)/(2+3+3)

v = 170/8

v = 21.25 km/h

The average velocity is 21.25km/h

3 0
2 years ago
A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.
SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

3 0
3 years ago
A venturi is constructed of a 10.0 cm pipe with a 2.0 cm diameter throat. Water pressure in the pipe is twice atmospheric pressu
postnew [5]

Answer:

P_t=5066250.0696\ Pa=50\ atm

Explanation:

Given:

  • diameter of pipe, d=0.1\ m
  • diameter of throat, d_t=0.02\ m
  • velocity of flow, v=0.4\ m.s^{-1}

<u>Pressure in the pipe is twice the atmospheric pressure:</u>

P=2\times 101325=202650\ Pa

<u>Now the force of flow of water:</u>

F=P\times A

<u>now we find cross sectional area of the pipe:</u>

A=\frac{\pi.d^2}{4}

A=\pi \times\frac{0.1}{4}

A=0.007854\ m^2

<u>Therefore,</u>

F=202650\times 0.007854

F=1591.6094\ N

<u>Now the area at throat:</u>

A_t=\frac{\pi.d_t^2}{4}

A_t=\frac{\pi\times 0.02^2}{4}

A_t=0.000314\ m^2

<u>Therefore pressure at throat:</u>

P_t=\frac{F}{A_t}

P_t=\frac{1591.6094}{0.000314}

P_t=5066250.0696\ Pa=50\ atm

6 0
3 years ago
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