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grigory [225]
2 years ago
9

the arm of a crane is 15.0 m long and makes an angle of 70.0 degrees with the horizontal. Assume that the maximum load for the c

rane is limited by the amount of torque the load produce around the base of the arm. What is the maximum torque the crank can withstand if the maximum load is 450 N. What is the Maximum load for this crane at an angle of 40.0°​

Physics
1 answer:
Gnoma [55]2 years ago
3 0

To find max torque you need component of force that is perpendicular to the direction of vector r. For second part just set the equation above equal to 2308 and solve for Fmax. Only difference is the angle

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A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a
UNO [17]

Answer:

P_{avg} = 6.283*10^{-9} \ W

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})

where;

E= \frac{d \phi}{dt}

E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

P_{avg} = 6.283*10^{-9} \ W

6 0
2 years ago
A box with a mass of 12.5 kg sits on the floor. how high would you need to lift it for it to have a gpe of 355 j
grandymaker [24]
Gpe=mgh

355=12.5×10×h
h=355/125= 2.84m
3 0
2 years ago
Wind blows at the speed of 30m/s across a 175m^2 flat roof if a house.
Makovka662 [10]

Answer:

the net force is 101587.5 N

Explanation:

The speed of wind

v = 30 m/s

The area of roof,

A = 175 m 2

The expression for the Bernoulli's theorem.

P = 12 ρv 2 ...... (1)

Here,

P is the pressure difference,

ρ is the density of air and

v is the speed of wind.

The expression for the pressure.

P = F A ..... (2)

Here,

F is the force and

A is the area of roof.

Part (a)

Substitute the values for the pressure difference in equation (1)

P = 12 × 1.29 × (30) 2 P = 580.5 Pa

Thus, the pressure difference at the roof between the inside and outside air is

580.5 Pa

Part (b)

Substitute the values for the net force in equation (2)

580.5 = F 175 F = 101587.5 N

Thus, the net force is 101587.5 N.

8 0
2 years ago
A diffusion couple composed of two silver– gold alloys is formed; these alloys have compositions of 98 wt% Ag–2 wt% Au and 95 wt
jeyben [28]

Answer:

for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)

It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.

we have D= D0exp( -Qd/RT)

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2 years ago
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crimeas [40]

Answer:

Centripetal acceleration of the car is 17.4 m/s²

Explanation:

It is given that,

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We need to find the centripetal acceleration of the race car. The formula for the centripetal acceleration is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(31.5\ m/s)^2}{57\ m}

a=17.4\ m/s^2

So, the centripetal acceleration of the race car is 17.4 m/s². Hence, this is the required solution.

5 0
2 years ago
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