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grin007 [14]
3 years ago
10

The block brake consists of a pin-connected lever and friction block at B. The coefficient of static friction between the wheel

and the lever is μs = 0.3, and a torque of 5 N⋅m is applied to the wheel. Determine if the brake can hold the wheel stationary when the force applied to the lever is (a) P = 30 N, (b) P = 70 N.
Physics
2 answers:
erastovalidia [21]3 years ago
8 0

Answer:

B

Explanation:

Anika [276]3 years ago
7 0

Answer:

a) Insufficient

b) Sufficient

Explanation:

To hold lever:

Sum of moments about point O center of wheel = 0

Fb * 0.15 - 5 =0

Fb = 33.33N

Contact force Nb:

Nb = (Fb/u) = (33.333 / 0.3) = 111.1 N

Sum of moments about point A lever pin joint = 0

Preq * 0.6 - 111.1*0.2 - 33.33*0.05 = 0

Preq = 39.8 N

part a

P = 30N < Preq = 39.8N

Hence, insufficient

part b

P = 70N > Preq = 39.8N

Hence, sufficient

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At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
What is the Doppler Effect equation?
MAXImum [283]
Fperson =[( velocity of wind )+ or - (velocity of person)] / [(velocity of wind) + or - (velocity of sounds)] x frekuency of sounds
7 0
3 years ago
In one experiment the electric field is measured for points at distances r from a uniform line of charge that has charge per uni
Yuki888 [10]

Answer:

(a) A. Uniform line of charge and B. Uniformly charged sphere

(b) To three digits of precision:

λ = 1.50 * 10^-10 C/m

p = 2.81 * 10^-4 C/m^3

Explanation:

8 0
3 years ago
What is the speed of a cyclist that rides west 88 km in 32 minutes?
suter [353]

Answer:

The speed of the cyclist is 2.75 km/min.

Explanation:

Given

  • The distance d = 88 km
  • Time t = 32 minutes

To determine

We need to find the speed of a cyclist.

In order to determine the speed of a cyclist, all we need to do is to divide the distance covered by a cyclist by the time taken to cover the distance.

Using the formula involving speed, time, and distance

s=\frac{d}{t}

where

  • s = speed
  • d = distance covered
  • t = time taken

substitute d = 88, and t = 32 in the formula

s=\frac{d}{t}

s=\frac{88}{32}

Cancel the common factor 8

s=\frac{11}{4}

s=2.75 km/min

Therefore, the speed of the cyclist is 2.75 km/min.

8 0
3 years ago
A 2 feet piece of wire is cut into two pieces and once a piece is bent into a square and the other is bent into an equilateral t
SpyIntel [72]

Answer:

Explanation:

Let x ft be used to make square and 2-x ft be used to make equilateral triangle.

each side of square = x/4

area of square = ( x /4 )²

Each side of triangle

= (2-x) /3

Area of triangle = 1/2 (2-x)²/9 sin 60

= √3 / 36 x (2-x)²

Total area

A = ( x /4 )² +√3 / 36 (2-x)²

For maximum area

dA/dx = 0

1/16( 2x ) -√3 / 36 x2(2-x) = 0

x / 8 - √3(2-x)/ 18 = 0

x / 8 - √3/9 + √3/18 x = 0

x ( 1/8 + √3/18 ) = √3/9

x(.125 +.096 ) = .192

x = .868 ft

3 0
3 years ago
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