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sveta [45]
3 years ago
14

Suppose an elephant has a mass of 2850 kg. show answer No Attempt How fast, in meters per second, does the elephant need to move

to have the same kinetic energy as a 77-kg sprinter running at 7.5 m/s
Physics
1 answer:
Bad White [126]3 years ago
6 0

Answer:

1.23 m/s

Explanation:

The kinetic energy of the sprinter is:

KE = 0.5 * m(s) * v²

KE = 0.5 * 77 * 7.5²

KE = 2165.63 J

If the KE of the sprinter and the KE of the elephant are equal, hence:

2165.63 = 0.5 * m(e) * v²

2165.63 = 0.5 * 2850 * v²

=> v² = 1.52

v = √(1.52)

v = 1.23 m/s

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An object is 3.0 cm from a concave mirror, with a focal length of 1.5 cm. Calculate the image distance. Remember to include your
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Answer:

Construct a quadrilateral ABCD, where

Construct a quadrilateral ABCD, whereAB = 4 cm, BC = 5 cm, CD = 6.5 cm and angle B = 105° and angle C = 80°

7 0
3 years ago
A charge of -8.5 µC is traveling at a speed of 9.0 106 m/s in a region of space where there is a magnetic field. The angle betwe
Sindrei [870]

Answer:

The magnitude of the magnetic field is 9.3\times 10^{-5}\ T.

Explanation:

Given that,

Charge, q=-8.5\ \mu C=-8.5\times 10^{-6}\ C

Speed of the charged particle, v=9\times 10^6\ m/s

The angle between the velocity of the charge and the field is 56°.

The magnitude of force, F=5.9\times 10^{-3}\ N

We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

F=qvB\ \sin\theta

B is the magnetic field.

B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T

So, the magnitude of the magnetic field is 9.3\times 10^{-5}\ T. Hence, this is the required solution.

4 0
3 years ago
An intergalactic rock star bangs his drum every 1.50 s. A person on earth measures that the time between beats is 2.70 s. How fa
sineoko [7]

Answer:

v = 83.1 % of speed of light

Explanation:

given,

T_e is the earth time = 2.7 s

T_s is the ship time = 1.5 s

we know,

T_s = T_e \times \gamma

where c is the speed of light

v is the speed of the rock star moving

T_s = T_e\times \sqrt{1-\dfrac{v^2}{c^2}}

1.5= 2.7\times \sqrt{1-\dfrac{v^2}{c^2}}

\sqrt{1-\dfrac{v^2}{c^2}} =0.556

squaring both side

1-\dfrac{v^2}{c^2}=0.3086

v^2=0.6914c^2

v = 0.831 c

v = 83.1 % of speed of light

7 0
3 years ago
Consider the reaction.
Free_Kalibri [48]

2.1 x 102

Is the correct solution for this problem

5 0
3 years ago
Read 2 more answers
In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The separation between
kap26 [50]

The separation between the slits is d = 8.96

What is fringe width?

  • Fringe width is the distance between two consecutive bright spots (maximas, where constructive interference take place)
  • Or two consecutive dark spots (minimas, where destructive interference take place).

Fringe width is given by β = λL/d

In the first case fringe width is β1 = λLA /d   = 625 x 10-9 x 0.36 / ( 1.4 x 10-5 )  = 0.016071428 m

The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m

No. of fringes in this 0.1m = 0.1 / 0.016071428  = 6.222  

So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 6.

In the second case fringe width is β1 = λLAB /d   = 625 x 10-9 x 0.25 / ( 1.4 x 10-5 )  = 0.011160714 m

The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m

No. of fringes in this 0.1m = 0.1 / 0.011160714  = 8.96

So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 8.  The ninth one will not be seen since the screen is less a little less in width.

Learn more about fringe width

brainly.com/question/14438105

#SPJ4

<u>The complete question is -</u>

In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The separation between the slits is d = 1.4 × 10-3 m. The total width of the screen is 0.20 m. In one version of the setup, the separation between the double slit and the screen is LA = 0.36 m, whereas in another version it is LB = 0.25 m. On one side of the central bright fringe, how many bright fringes lie on the screen in the two versions of the setup? Do not include the central bright fringe in your counting. --Tm = 3 (Bright fringe) ++m = 0 (Bright fringe) -m = 3 (Bright fringe) Figure 27.7

5 0
1 year ago
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