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shusha [124]
2 years ago
5

What is specific heat capacity?

Physics
1 answer:
leonid [27]2 years ago
4 0
B is the ans for this question
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A body falls from the top of the tower and during the last second of its fall it fall through 23mvfind height of tower.
DanielleElmas [232]

Answer:

39.7 m

Explanation:

First, we conside only the last second of fall of the body. We can apply the following suvat equation:

s=ut+\frac{1}{2}at^2

where, taking downward as positive direction:

s = 23 m is the displacement of the body

t = 1 s is the time interval considered

a=g=9.8 m/s^2 is the acceleration

u is the velocity of the body at the beginning of that second

Solving for u, we find:

ut=s-\frac{1}{2}at^2\\u=\frac{s}{t}-\frac{1}{2}at=\frac{23}{1}-\frac{1}{2}(9.8)(1)=18.1 m/s

Now we can call this velocity that we found v,

v = 18 m/s

And we can now consider the first part of the fall, where we can apply the following suvat equation:

v^2-u^2 = 2as'

where

v = 18 m/s

u = 0 (the body falls from rest)

s' is the displacement of the body before the last second

Solving for s',

s'=\frac{v^2-u^2}{2a}=\frac{18.1^2-0}{2(9.8)}=16.7 m

Therefore, the total heigth of the building is the sum of s and s':

h = s + s' = 23 m + 16.7 m = 39.7 m

7 0
3 years ago
What do astronomers mean by light pollution?
Margarita [4]

Answer:

a. Light pollution refers to light used for human activities that brightens the sky and hinders astronomical observations.

Explanation:

Light pollution is due to the excessive and misdirected use of artificial light. Light bulbs are often design in an incorrect way, since a great part of its light is not completely directed to the ground and an important percentage is emitted to the sky in where will be scattered and reflected back to ground by the particles in the atmosphere. That brings as an effect a sky glow, therefore the visibility of astronomical objects will be extremely reduce.  

Hence, professional astronomical research and amateur observations will be affected. Light pollution has a negative impact on bird migration at night and in the health of difference species, humans also.          

8 0
3 years ago
Read 2 more answers
La ecuación de la posición de una esferita está dada por: r(t)=(2.Cos(πt) i-3.Sen(πt) j) (m) ¿Cuál es la velocidad de la esferit
katovenus [111]

Answer:

v = (-4.44 i^ + 6.66 j^ )  m/s, a_average =( 0 i^ -2π j^) m/s²

Explanation:

The expression left corresponds to an oscillatory movement (MAS), the speed is defined by

         v = dr / dt

the function of position

         r = 2 cos πt  i^  + 3 sin πt  j^

let us note that it is a movement in two dimensions

let's perform the derivative

          v = -2π sin πt  i^  + 3π cos πt  j^

we evaluate this expression for t = 0.25 s, remember that the angle is in radians

          v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^

           v = (-4.44 i^ + 6.66 j^ )  m/s

To calculate the mean acceleration we use the expression

           a = (v_{f} - v_{o}) / Δt

 

indicates that the time is the first 3 s

       

we look for the initial velocity t = 0 s

           v₀ = 0 i ^ + 3π j ^

         

we look for the fine velocity, t = 3 s

          v_f = - 2π sin (π 3) + 3π cos (π 3) j ^

          v_f = 0 i ^ - 3π j ^

we calculate the average acceleration

            Δt = (3 -0) = 3 s

           a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3

           a_average = (0 i ^ -2π j ^ ) m/s²

6 0
3 years ago
Which type of employee would most likely spend the majority of work time in a vehicle? O a Corrections Officer at a jail O a Dis
dsp73

Answer:

Paramedic

Explanation:

They will be on the move the whole time

8 0
2 years ago
Read 2 more answers
An ammeter with a resistance of 5.0 ohm is connected in series with a 3.0V cell and a lamp rated at 300 mA, 3V. Calculate the cu
Alex

Answer:

I = 0.2 A

Explanation:

Lamp is rated at 300 mA

I_lamp = 0.3 A

Voltage is; V = 3V

Thus; Resistance is given by;

R = V/I

R = 3/0.3

R = 10 ohms

Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;

R_eq = 10 + 5

R_eq = 15 ohms

Ammeter current will be;

I = V/R_eq

I = 3/15

I = 0.2 A

3 0
2 years ago
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