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jeka94
2 years ago
10

A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 30. wha

t is the torque on the bar?
Physics
1 answer:
Alecsey [184]2 years ago
6 0

Answer:

<h2>9.375Nm</h2>

Explanation:

The formula for calculating torque τ = Frsin∅ where;

F = applied force (in newton)

r = radius (in metres)

∅ = angle that the force made with the bar.

Given  F= 25N, r = 0.75m and ∅ = 30°

torque on the bar τ  = 25*0.75*sin30°

τ = 25*0.75*0.5

τ = 9.375Nm

The torque on the bar is 9.375Nm

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Light hits a clear plastic bottle and a granite rock. Which choice most accurately describes the effect of visible light on the
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An x-ray tube is an evacuated glass tube that produces electrons at one end and then accelerates them to very high speeds by the
laila [671]

Answer:

a)  V = 1.866 10² V ,  b)   V = 3.424 10⁵ V , c)   v = 8.1 10⁶ m / s

Explanation:

a) the potential difference is requested to accelerate the electrons up to 2.7% of the speed of light

           v = 0.027 c

           v = 0.027 3 10⁸

           v = 8.1 10⁶ m / s

for this part we can use the conservation of mechanical energy

starting point. When electrons are at rest

           Em₀ = U = q V

final point. Electrons with maximum speed

          Em_f = K = ½ m v2

          Em₀ = Em_{f}

          e V = ½ m v²

          V = ½ m v² / e

let's calculate

          V = ½  9.1 10⁻³¹ (8.1 10⁶)² / 1.6 10⁻¹⁹

          V = 1.866 10² V

           V = 1866 V

         

b) if this acceleration protons is the mass of the proton is m_{p} = 1.67 10-27

          V = ½ 1.67 10⁻²⁷ (8.1 10⁶)² / 1.6 10⁻¹⁹

           V = 3.424 10⁵ V

           V = 342402 V

c)   this potential difference should give the protons the same speed as the electrons

             v = 8.1 10⁶ m / s

5 0
2 years ago
An automobile being tested on a straight road is 400 feet from its starting point when the stopwatch reads 8.0 seconds and is 55
m_a_m_a [10]

Answer:

a)   v = 75 ft / s , b)  v = 55 ft / s , c)   Δx = 1000 ft

Explanation:

We can solve this exercise with the expressions of kinematics

a) average speed is defined as the distance traveled in a given time interval

        v = (x₂-x₁) / (t₂-t₁)

         v = (550 - 400) / (10 -8)

         v = 75 ft / s

b) we repeat the calculations for this interval

   v = (550 - 0) / (10 -0)

   v = 55 ft / s

c)  we clear the distance from the average velocity equation

     Δx = v (t₂ -t₁)

     Δx = 100 (20-10)

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4 0
2 years ago
A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a
Yanka [14]

Answer:

Diameter of the cylinder will be d=2.998\times 10^4m

Explanation:

We have given young's modulus of steel E=207GPa=207\times 10^9Pa  

Change in length \Delta l=0.38mm

Length of rod l=500mm

Load F = 11100 KN

Strain is given by strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}

We know that young's modulus E=\frac{stress}{strain}

So 207\times 10^9=\frac{stress}{7.6\times 10^{-4}}

stress=1573.2\times 10^{-5}N/m^2

We know that stress =\frac{force}{artea }

So 1573.2\times 10^{-5}=\frac{11100\times 1000}{area}

area=7.055\times 10^{8}m^2

So \frac{\pi }{4}d^2=7.055\times 10^{8}

d=2.998\times 10^4m          

6 0
2 years ago
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