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inysia [295]
3 years ago
10

What minimum speed must the block have at the base of the 70 m hill to pass over the pit at the far (right-hand) side of that hi

ll and reach the other side (at a 50 m level)
Physics
1 answer:
Drupady [299]3 years ago
4 0

Answer:

initial velocity is v = 4.95 m / s

Explanation:

To solve this exercise we use the projectile launch ratios, when the block leaves the hill its speed is horizontal, let's find the time it takes to fall to the other point.

Initial vertical velocity is zero

          y = y₀ + v_{oy} t - ½ g t²

          y-y₀ = 0 -1/2 g t²

          t = \sqrt{ \frac{ 2(y_o -y)}{g} }

calculate

          t = \sqrt{ \frac{2 ( 70-50)}{9.8} }

          t = 2.02 s

with this time we can substitute in the horizontal displacement equation

          x = v₀ₓ t

          v₀ₓ = x / t

suppose that the distance between the two points is x = 10 m

          v₀ₓ = 10 / 2.02

          v₀ₓ = 4.95 m / s

initial velocity is v = 4.95 m / s

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What does the diameter in meters need to be of a round parachute that the designer has determined needs to have an area of 500 s
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Area of a circle is
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5 0
3 years ago
Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher
siniylev [52]

Answer:

v = 4.44 \times 10^5 m/s

Explanation:

By Einstein's Equation of photoelectric effect we know that

h\nu = W + \frac{1}{2}mv^2

here we know that

h\nu = energy of the photons incident on the metal

W = minimum energy required to remove photons from metal

\frac{1}{2}mv^2 = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

W = \frac{hc}{351 nm}

here we know that

hc = 1242 eV-nm

now we have

W = \frac{1242}{351} = 3.54 eV

now by energy equation above when photon of 303 nm incident on the surface

\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2

4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2

(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2

8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2

v = 4.44 \times 10^5 m/s

6 0
3 years ago
A child pulls a wagon with a force F at an angle o with respect to the horizontal. Which of the following free body diagrams is
NeTakaya

Option d) is the correct free body diagram.

Explanation:

A free-body diagram is a diagram that shows all the forces acting on a body. Each force is represented using an arrow, where:

- The length of the arrow is proportional to the magnitude of the force

- The direction of the arrow corresponds to the direction of the force

For the block in this problem, we have the following forces:

- The force F applied from the child, which acts at an angle with respect to the horizontal --> this rules out option a) and c), where the force acts horizontally

- The force of gravity (the weight of the object), labelled with W, which always acts downward --> this rules out option b), since the weight acts downward.

Therefore, the correct option is d).

(in reality, there should be another force: the normal reaction exerted by the floor on the block, N, acting upward).

Learn more about forces and weight:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

5 0
3 years ago
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