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r-ruslan [8.4K]
3 years ago
6

If a force of 250N is applied to a piano being pushed up a ramp at an angle of 18degrees above the horizontal, what force is bei

ng applied to lift the piano?

Physics
1 answer:
klasskru [66]3 years ago
4 0

answer

77.25 N

explanation

(look at picture for free body diagram)

the force on the piano has both vertical and horizontal components, and since the questions asks about the force to lift the piano, we want to find the vertical component

we can use trigonometry where sinA = opposite/hypotenuse

A = 18 degrees

hypotenuse = 250 N

sin18 = opposite (vertical component) / 250

vertical component = sin18 * 250

= 77.25 N

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Describe and name the different types of collision. In which are the linear momentum and kinetic energy conserved
ipn [44]

Answer:

1. Elastic collision

2. Inelastic collision    

Explanation:

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Which of the following provides evidence that geologic activity occurred in the past on Earth's Moon?
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The surface features of moon such as Mares, Craters, mountains, Rays and rills are the proof of some geological activity on the Moon. Mares are the dark patches on the moon's surface formed of solidified lava. Due to negligible atmosphere on the moon, the meteors strike its surface and cause craters to form. Thus, the correct answer is d.

6 0
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Who was the first person referred to as a psychologist?
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3 years ago
How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume
Svetach [21]

Answer:

9.56\cdot 10^{-7} C

Explanation:

A parallel-plate capacitors consist of two parallel plates charged with opposite charge.

Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.

The electric field between two infinite sheets with opposite charge is:

E=\frac{\sigma}{\epsilon_0}

where

\sigma=\frac{Q}{A} is the surface charge density, where

Q is the charge on the plate

A is the area of the plate

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

In this problem:

- The side of one plate is

L = 19 cm = 0.19 m

So the area is

A=L^2=(0.19)^2=0.036m^2

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

E=3\cdot 10^6 N/C

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

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7 0
3 years ago
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