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Readme [11.4K]
2 years ago
12

An elevator is moving down with an acceleration of 3.36 m/s2.

Physics
1 answer:
sergeinik [125]2 years ago
5 0

Answer : 413.44N

Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .

  • From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
  • The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
  • When a elevator accelerates down , the weight recorded is less than the actual weight .

From the Free body diagram ,

\sf\longrightarrow Weight = mg - ma \\

\sf\longrightarrow Weight = m ( g - a ) \\

  • Mass of the man = 64.2 kg

\sf\longrightarrow Weight = 64.2( 9.8 - 3.36) N\\

\sf\longrightarrow Weight = 64.2 * 6.44 N\\

\sf\longrightarrow \underline{\boxed{\bf Weight_{apparent}= 413.44 N }} \\

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179.1 kg

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A 82-kg fisherman in a 112-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of vi = 4.8 m/s
abruzzese [7]

Answer:

0.37 m/s to the left

Explanation:

Momentum is conserved.  Initial momentum = final momentum.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Initially, both the fisherman/boat and the package are at rest.

0 = m₁ v₁ + m₂ v₂

Plugging in values and solving:

0 = (82 kg + 112 kg) v + (15 kg) (4.8 m/s)

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The boat's velocity is 0.37 m/s to the left.

8 0
3 years ago
A plane is flying 2400 miles from a to
geniusboy [140]
From a to b speed is 600+40 = 640
from b to a speed is 600-40 = 560

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Time left will be (2400-640t)/640. But if they were to return to a it would fly 640t miles at 560mph which will take (640t/560) hrs

(2400-640t) / 640 = 640t / 560
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3 years ago
A risk-free, zero-coupon bond with a $5000 face value has ten years to maturity. The bond currently trades at $3650. What is the
PtichkaEL [24]

Answer:

The Yield to Maturity of the bond is YTM = 3.20%

Explanation:

Mathematically the Yield to Maturity of the bond  YTM is as follows

           YTM = \frac{C + \frac{F-P}{n} }{\frac{F+P}{2} }

Where C is the amount of payment to be made = $0

           P is the price i.e the present value =$3650

           F is the face value of the bond=$5000

          n is the year of maturity of the bond = 10 years

            YTM = \frac{0+\frac{5000-3650}{10} }{\frac{5000+3650}{2} } * \frac{100}{1}

                      =3.20%

                     

                     

     

   

5 0
3 years ago
When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these
Zigmanuir [339]

Answer:

a)  v = 0.7071 v₀, b) v= v₀, c)  v = 0.577 v₀, d)   v = 1.41 v₀, e)  v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

          v = \sqrt{ \frac{T}{ \mu } }

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

          m = 2m₀

let's find the new linear density

          μ = m / l

iinitial density

          μ₀ = m₀ / l

final density

          μ = 2m₀ / lo

          μ = 2 μ₀

we substitute in the equation for the velocity

initial            v₀ = \sqrt{ \frac{T_o}{ \mu_o} }

with the new dough

                    v = \sqrt{ \frac{T_o}{ 2 \mu_o} }

                    v = 1 /√2  \sqrt{ \frac{T_o}{ \mu_o} }

                    v = 1 /√2 v₀

                    v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

              μ = μ₀

   in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

           μ = 3m₀ / l₀

           μ = 3 m₀/l₀

           μ = 3 μ₀

            v = \sqrt{ \frac{T_o}{ 3 \mu_o} }

            v = 1 /√3  v₀

            v = 0.577 v₀

d) l = 2l₀

            μ = m₀ / 2l₀

            μ = μ₀/ 2

           v = \sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }

           v = √2 v₀

            v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

             μ = 10 m₀/2l₀

             μ = 5 μ₀

we look for speed

             v = \sqrt{ \frac{T_o}{5 \mu_o} }

             v = 1 /√5  v₀

             v = 0.447 v₀

5 0
3 years ago
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