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vredina [299]
3 years ago
7

A child in a tree house uses a rope attached to a basket to lift a 24 N dog upward through a distance of 4.9 m into the house.

Physics
1 answer:
jenyasd209 [6]3 years ago
3 0
<h3>Answer:</h3>

117.6 Joules

<h3>Explanation:</h3>

<u>We are given;</u>

  • Force of the dog is 24 N
  • Distance upward is 4.9 m

We are required to calculate the work done

  • Work done is the product of force and distance
  • That is; Work done = Force × distance
  • It is measured in Joules.

In this case;

Force applied is equivalent to the weight of the dog.

Work done = 24 N × 4.9 m

                  = 117.6 Joules

Hence, the work done in lifting the dog is 117.6 Joules

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he graph shows the distance (x) traveled by an aircraft traveling at constant velocity in corresponding time intervals (t). What
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Answer:

B linear, 825 meters

Explanation:

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3 years ago
Hello please help i’ll give brainliest
ser-zykov [4K]

The correct answer is B.Antartica and Australia were one landmass millions of years ago.

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n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How f
pashok25 [27]

Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

v = \sqrt{\frac{2K.E}{m} } \\\\v =  \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

8 0
3 years ago
A catapult used by medieval armies hurls a stone of mass 32.0 kg with a velocity of 50.0 m/s at a 30.0 degree angle above the ho
NISA [10]

Answer:

The horizontal distance traveled when the stone returns to its original height = 220.81 m

Explanation:

Considering vertical motion of catapult:-

At maximum height,

Initial velocity, u =  50 sin30 = 25 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v = u + at

Substituting

    v = u + at

    0 = 25  - 9.81 x t

    t = 2.55 s

Time of flight = 2 x Time to reach maximum height = 2 x 2.55 = 3.1 s

Considering horizontal motion of catapult:-

Initial velocity, u =  50 cos30 = 43.30 m/s

Acceleration , a = 0 m/s²

Time, t = 5.10 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

   s = 43.30 x 5.10 + 0.5 x 0 x 5.10²

  s = 220.81 m

The horizontal distance traveled when the stone returns to its original height = 220.81 m

5 0
3 years ago
Josie ran at an average speed of 16 m/s. if her mass 10 kg, what was her kinetic energy as she crosses her finish line?
Helen [10]

Answer: 1280 J

KE=\frac{1}{2}mv^{2}  \\\\KE= \frac{1}{2} (10 kg)(16 m/s)^2\\=1280 J

3 0
2 years ago
Read 2 more answers
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