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GREYUIT [131]
2 years ago
12

If a child pulls a sled through the snow with a force of 50 N exerted at an angle of 38° above

Physics
1 answer:
Pavel [41]2 years ago
7 0

Answer: The horizontal component of the force is 47.754 N and the vertical component of the force is 14.818N.

Explanation:

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Describe what a hydrogen bond is. 6th grade answer
r-ruslan [8.4K]

Answer:

<em>Hydrogen bond is the attractive force between the hydrogen attached electronegative atom </em>

Explanation:

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3 years ago
50 mL of soda in a soda can exerts ____ 50 mL of soda in a 1L bottle.
lutik1710 [3]
More pressure than..



4 0
3 years ago
When Simon grows (let’s say he doubles his mass) what happens to his GPE when he is at the balcony?
DerKrebs [107]

Answer:

His gravitational potential energy will increase as well.

Explanation:

Let gpe represent gravitational potential energy.

gpe = mass × gravitational field strength × height

From the formula above, we can conclude that as the mass of a body increases, it's gpe increases too.

5 0
2 years ago
Each plate of an air-filled parallel-plate capacitor has an area of 45.0 cm2, and the separation of the plates is 0.080 mm. A ba
maw [93]

Answer:

Option (e)

Explanation:

A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,

Energy density = 100 J/m

Let Q be the charge on the plates.

Energy density = 1/2 x ε0 x E^2

100 = 0.5 x 8.854 x 10^-12 x E^2

E = 4.75 x 10^6 V/m

V = E x d

V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V

C = ε0 A / d

C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F

Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C

Q =  190 nC

3 0
3 years ago
A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is
Rudik [331]

Answer:

a=16.2m/s^{2}

Explanation:

From the attached file diagram, the total force acting on the charged box is the downward weight and the repulsive force acting in opposite to the weight force . Hence we can write the total force as

F=masin\alpha -\frac{kQq}{r^{2}} \\\alpha =35^{0}, m=0.495kg, r=0.61m, Q=2.5*10^{-6}, q=75.0*10^{-6}\\

When fixed,F=o

Hence

masin\alpha =\frac{kQq}{r^{2}}\\0.495kg*asin35=\frac{9*10^{9}*2.5*10^{-6}*75.0*10^{-6}}{0.61^{2}} \\0.28a=4.5351\\a=\frac{4.5351}{0.28}\\\\ a=16.2m/s^{2}

The value of the acceleration is 16.2m/s^2

7 0
2 years ago
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