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Likurg_2 [28]
2 years ago
7

Chef Andy tosses an orange in the air, then catches it again at the same height. The orange is in the air for 0.75\,\text s0.75s

0, point, 75, start text, s, end text. We can ignore air resistance.
What was the orange's velocity at the moment it was tossed into the air?
Physics
1 answer:
Alika [10]2 years ago
4 0

Answer:

90.82%digrash silsho

You might be interested in
A 250 Hz tuning fork is struck and the intensity at the source is I1 at a distance of one meter from the source. (a) What is the
Zina [86]

Answer:

a) 0.0625 I_1

b) 3.16 m

Explanation:

<u>Concepts and Principles  </u>

The intensity at a distance r from a point source that emits waves of power P is given as:  

I=P/4π*r^2                         (1)

<u>Given Data</u>

f (frequency of the tuning fork) = 250 Hz

I_1 is the intensity at the source a distance r_1 = I m from the source.  

<u>Required Data</u>

- In part (a), we are asked to determine the intensity I_2 a distance r_2 = 4 in from the source.

- In part (b), we are asked to determine the distance from the tuning fork at which the intensity is a tenth of the intensity at the source.  

<u>solution:</u>

(a)  

According to Equation (1), the intensity a distance r is inversely proportional to the distance from the source squared:

I∝1/r^2

Set the proportionality:  

I_1/I_2=(r_2/r_1)^2                                 (2)

Solve for I_2 :  

I_2=I_1(r_2/r_1)^2  

I_2=0.0625 I_1

(b)  

Solve Equation (2) for r_2:  

r_2=(√I_1/I_2)*r_1

where I_2 = (1/10)*I_1:

r_2=(√I_1/1/10*I_1)*r_1

     =3.16 m

3 0
2 years ago
Two thin lenses with a focal length of magnitude 12.0cm, the first diverging and the second converging, are located 9.00cm apart
attashe74 [19]

Answer:

Explanation:

b ) First is concave lens with focal length f₁ = - 12 cm .

object distance u = - 20 cm .

Lens formula

1 / v - 1 / u = 1 / f

1 / v + 1 / 20 = -1 / 12

1 / v =  - 1 / 20  -1 / 12

= - .05 - .08333

= - .13333

v = - 1 / .13333

= - 7.5 cm

first image is formed before the first lens on the side of object.

This will become object for second lens

distance from second lens = 7.5 + 9 = 16.5 cm

c )

For second lens

object distance u = - 16.5 cm

focal length f₂ = + 12 cm ( lens is convex )

image distance = v

lens formula ,

1 / v - 1 / u = 1 / f₂

1 / v + 1 / 16.5 = 1 / 12

1 / v =   1 / 12 -  1 / 16.5

= .08333- .0606

= .02273

v = 1 /  .02273

= 44 cm ( approx )

It will be formed on the other side of convex lens

distance from first lens

= 44 + 9 = 53 cm .

magnification by first lens = v / u

= -7.5 / -20 = .375 .

magnification by second lens = v / u

= 44 / - 16.5

= - 2.67

d )

total magnification

= .375 x - 2.67

= - 1.00125

height of final image

= 2.50 mm x 1.00125

= 2.503mm

e )

The final image will be inverted with respect to object  because total magnification is negative .

6 0
1 year ago
Consider a star that is a sphere with a radius of 6.32 108 m and an average surface temperature of 5350 K. Determine the amount
Mariulka [41]

Answer:

The value is  \Delta s  = 8.537 *10^{25 } \ J/K

Explanation:

From the we are told that

   The radius of the sphere is r =  6.32 *10^{8} \  m

   The temperature is T_x  =  5350 \  K

    The average temperature of the rest of the universe is  T_r  =  2.73 \  K

Generally the change in entropy of the entire universe per second is mathematically represented as

         \Delta s  =  s_r - s_x

Here s_r is the entropy of the rest of the universe which is mathematically represented as

          s_r =  \frac{Q}{T_r}

Here Q is the quantity of heat radiated by the star which is mathematically represented as

           Q =  4 \pi *  r^2 *  \sigma * T^4_x

Here \sigma is the Stefan-Boltzmann constant with value  

           \sigma =  5.67 * 10^{-8 }W\cdot  m^{-2} \cdot  K^{-4}.

=>         Q =  4 \pi *  (6.32*10^{8})^2 *  5.67 * 10^{-8 }  * 5350 ^4

=>         Q =  2.332 *10^{26} \  J

So

      s_r =  \frac{2.332 *10^{26}}{2.73}

=>   s_r =  8.5415 *10^{25}\  J/K

Here s_x is the entropy of the rest of the universe which is mathematically represented as

      s_x =  \frac{Q}{T_x}

=>   s_x =  \frac{2.332 *10^{26} }{5350}

=>   s_x =  4.359 *10^{22} \  J/K

So

      \Delta s  = 8.5415 *10^{25}  - 4.359 *10^{22}

=>   \Delta s  = 8.537 *10^{25 } \ J/K

7 0
1 year ago
A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. 1.84 x 103 J
arsen [322]
<span>work will be equal to the potential energy gained by the person in climbing the stairs. work= potential energy gained = mgh W= 75kg*9.8m/s2*2.50m= 1837.5 J</span>
4 0
1 year ago
On cold winter mornings, Ellen's mother will frequently put a space heater in the kitchen. The space heater has coils that use e
Eduardwww [97]

Answer:

C

Explanation:

It seems right.

5 0
2 years ago
Read 2 more answers
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