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ikadub [295]
2 years ago
7

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air

resistance, so the brick is in free fall.(a) How tall, in meters, is the building?(b) What is the magnitude of the brick’s velocity just before it reaches the ground?(c) Sketch ay−t,vy−t, and y-t graphs for the motion of the brick.

Physics
1 answer:
kozerog [31]2 years ago
5 0

Answer:

h=17m

v=18.6 m/s

Explanation: The question can be solved by applying kinematic equations of motion

Data

u=0

a=g

t=1.9 secs

firstly to calculate the height

s=ut+0.5at^2\\h=ut+0.5at^2\\h=0*1.9+0.5*9.81*1.9^2\\h=17.707 m

to find the final velocity

v=u+at\\v=0+9.81*1.9\\v=18.639

The acceleration graph is straight line of equation y=9.8 as acceleration is constant:

Velocity graph is given by y=9.8x ( y as velocity and x as time):

Displacement graph is given by y=4.9x^2 ( x as time, y as displacement):

These graphs are only applicable from x=0 to x=1.9 ... ignore the other graph sections

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Rank the following situations according to the magnitude of the impulse of the net force, from largest value to smallest value.
wolverine [178]

Answer:

V

I and II

III and IV

Explanation:

The impulse is equal to the change in momentum of the object involved, so we can calculate the change in momentum in each situation and compare them all.

Taking always east as positive direction, and labelling

u the initial velocity

v the final velocity

m = 1000 kg the mass (which is always equal)

We find:

(i)

u = 25 m/s

v = 0

|I|=m(v-u)=(1000)(0-25)=25,000 Ns

(II)

u = 25 m/s

v = 0

|I|=m(v-u)=(1000)(0-25)=25,000 Ns

(III)

In this case,

F = 2000 N is the force

\Delta t = 10 s is the time

So the magnitude of the impulse is

|I| =F\Delta t = (2000N)(10)=20,000 Ns

(IV)

F = 2000 N is the force

\Delta t = 10 s is the time

So the magnitude of the impulse is

|I| =F\Delta t = (2000N)(10)=20,000 Ns

(V)

u = 25 m/s

v = -25 m/s

|I|=m(v-u)=(1000)(-25-25)=50,000 Ns

So the ranking from largest to smallest is:

V

I and II

III and IV

5 0
2 years ago
The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material o
zavuch27 [327]

Answer:

pp

Explanation:

7 0
2 years ago
Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a
garik1379 [7]

Answer:

T=2.94*10^-10  N/m.

Explanation:

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those at which desirable prey might struggle. Orb spider web silk has a typical diameter of 20μm, and spider silk has a density of 1300 kg/m³.

To have a fundamental frequency at 150Hz , to what tension must a spider adjust a 14cm -long strand of silk?

l=length of the spider silk, 14cm

velocity of wave = √(T/μ)          

where T = tension and

μ = mass per unit length)

λ/2=l

for fundamental frequency λ/2 =14cm    

 (λ= wavelength of standing wave;  as there will be no node

   except the endpoints of silk strand)

               λ = 28 cm = 0.28 m

and since frequency * wavelength = speed of wave. we have,

                  150 * 0.28 = √(T/μ)                                        ..................(#)

now μ = mass/length = [volume * density]/length = [(length*area) * density] / length = area * density

         = [π * (10 * 10^(-6))²] * 1300  = 13π * 10^(-8).

now putting this in equation (#) we get

    150 * 0.28 = √(T/[13π * 10^(-8)]).

thus T = [13π * 10^(-8)] * (42)²     =  

2.94*10^-10  N/m.

6 0
2 years ago
Two students, sitting on frictionless carts, push against each other. Both are initially at rest and the mass of student 1 and t
Zepler [3.9K]

Answer:

  v₂ = v/1.5= 0.667 v

Explanation:

For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.

Initial moment before pushing

    p₀ = 0

Final moment after they have been pushed

    p_{f} = m₁ v₁ + m₂ v₂

   p₀ =  p_{f}

   0 = m₁ v₁ + m₂ v₂

   m₁ v₁ = - m₂ v₂

Let's replace

   M (-v) = -1.5M v₂

   v₂ = v / 1.5

  v₂ = 0.667 v

6 0
2 years ago
Two identical light springs with spring constant k3 are now individually hung vertically from the ceiling and attached at each e
Alenkasestr [34]

Answer:

 Keq = 2k₃

Explanation:

We can solve this exercise using Newton's second one

                F = m a

Where F is the eleatic force of the spring F = - k x

Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement

          F + F = m a

         k₃ x + k₃ x = m a

         a = 2k₃  x / m

To find the effective force constant, suppose we change this spring to what creates the cuddly displacement

       Keq = 2k₃

7 0
2 years ago
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