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Mkey [24]
3 years ago
12

A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. a) Calculate the m

agnitude of the vertical component of the force that the wall exerts on the left end of the beam if the angle between the cable and horizontal is θ = 47°.
Physics
1 answer:
Rus_ich [418]3 years ago
4 0

Answer:

1004.277N

Explanation:

Given that g = 9.817 ms^-2

If the angle between the cable and horizontal is 47°.

The angle between the cable and vertical is 90-47= 43°.

The weight of the beam is mg

The weight of the beam is

= 150*9.817

The weight of the beam is

= 1472.55N

magnitude of the vertical component of the force is mgsin(tita)

= 1472.55*sin43

= 1004.277N

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What will be the weight of a man of mass 90kg when he is on a planet with acceleration due to gravity of 15ms-2?
Feliz [49]

Answer:

1350N

Explanation:

Weight = Mass x Acceleration Due to Gravity

W=mg

W=90*15=1350N

7 0
3 years ago
4. A businesswoman on a trip flies a total of 23,000 km. The first day she traveled 4000 km, the second day 11,000 km, and on th
lutik1710 [3]

Answer:

14.035087719298246 ≈ 14 hours

Explanation:

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3 0
3 years ago
Suppose that the inverse market demand for silicone replacement tips for Sony earbud headphones is p ​= pN ​- 0.1Q, where p is t
lbvjy [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

   The effect of a change in the price of a new pair of headphones on the equilibrium price of replacement tips​ ( ​dp/dpN​) is

                   \frac{\delta p}{\delta p_N} =1

b

 The value of Q and p at equilibruim is

          Q_e = 250    and    p_d = 5

The consumer surplus is  C= 3125

The producer surplus  is   P = 375

Explanation:

      From the question we are told that

           The inverse market demand is  p_d = p_N -0.1Q

                The inverse supply function is     p_s = 2+ 0.012Q

a

The effect of change in the price  is mathematically given as

                  \frac{\delta p}{\delta p_N}

Now differntiating the inverse market demand function with respect to p_N

We get that  

                   \frac{\delta p}{\delta p_N} =1

b

   We are told that p_N =$30

        Therefore the inverse market demand becomes

                             p_d = 30 -0.1Q

At  equilibrium

                  p_d = p_s

So we have

               30 -0.1Q_e = 2+ 0.012Q_e

Where Q_e is the quantity at equilibrium

                    28 = 0.112Q_e

                     Q_e = \frac{28}{0.112}

                    Q_e = 250

Substituting the value of  Q into the equation for the inverse market demand function

                p_d = 30 - 0.1 (250 )

                    p_d = 5

Looking at the equation for p_d \ and \ p_s we see that

     For  Q =  0

             p_d = 30

             p_s =2

 And  for Q =  250

                 p_d = 5

                 p_s = 5

Hence the consumer surplus is mathematically evaluated as

           C = \frac{1}{2} * Q_e * (30 -5)

Substituting value

        C = \frac{1}{2} * 250 (30-5)

           C= 3125

And

  The  producer surplus is mathematically evaluated as

                    P = \frac{1}{2} *250 * (5-2)

                    P = 375

     

         

           

                     

4 0
3 years ago
If you drop a bowling ball, a tennis ball, and a feather from the top of a tall building at the same time, which one will hit th
olasank [31]

They hit at the same time. Everything falls at the same rate.

8 0
3 years ago
Read 2 more answers
The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically μr=0.002. Suppose a 180
Irina18 [472]

Answer:

t = 0.354 hours

Explanation:

given,

coefficient of rolling friction μr=0.002

mass of locomotive = 180,000 Kg

rolling speed = 25 m/s

The force of friction = μ mg

                                 = (.002) x (180000) x (9.8)

                                 = 3528 N

F = m  a

now,

m a =  3528 N

180000 x a = 3528

a = 0.0196 m/s²

Then apply

v = u + at  

0 = 25 - 0.0196 x t

t = 1275.51 sec

t = 1275.61/3600 hours

t = 0.354 hours

time taken by the locomotive to stop = t = 0.354 hours

7 0
3 years ago
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