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kodGreya [7K]
3 years ago
13

A force of 1 N will cause a mass of 1 kg to have an acceleration of 1 m/s2. Therefore, a force of 7 N applied to a mass of 7 kg

will cause it to accelerate at what rate?
1 m/s2

0.14 m/s2

8 m/s2

49 m/s2

7 m/s2
Physics
2 answers:
Lana71 [14]3 years ago
3 0

1 m/s^2

Using F=ma,

7=7a

a=1

Angelina_Jolie [31]3 years ago
3 0
<h2>Answer:</h2>

<u>The correct choice is </u><u>1 m/s2</u>

<h2>Explanation:</h2>

According to the unitary method if a force of 1 N will cause a mass of 1 kg to have an acceleration of 1 m/s2.

Then obviously  force of 7 N applied to a mass of 7 kg will also cause the acceleration of the same rate which will be 7m/s^2

Or

we can use the Newtons law

F = m a

7 = 7 * 1

= 1

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Many fundamental laws of physics and chemistry can be formulated when doing this.
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2 years ago
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A diver is swimming underneath an oil slick with a thickness of 200 nm and an index of refraction of 1.50. A white light shines
Tcecarenko [31]

Answer:

Explanation:

thickness of oil t = 200 nm

index of refraction μ = 1.5

For transmitted light :---

path difference = 2μ t

For constructive interference

path difference = n λ , λ is wavelength  of light

2μ t = n λ

λ = 2μ t /  n

For longest λ , n = 1

λ = 2μ t

= 2 x 1.5 x 200 nm

= 600 nm

Wavelength in water

= 600 / refractive index of water

= 600 / 1.33

= 451.1 nm Ans

4 0
3 years ago
(a) What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) For a hydrogen atom, determine the
crimeas [40]

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

E=\frac {-13.6}{n^2}\ eV

(a) The energy of the electron in n= 6 excited state is:

E=\frac {-13.6}{6^2}\ eV

E=-0.3778\ eV

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

E=\frac {-13.6}{1^2}\ eV

E=-13.6\ eV

Ratio=\frac {E_6}{E_1}

Ratio=\frac {-0.3778}{-13.6}

Ratio = 0.0278

7 0
3 years ago
A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the
yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

       x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

       x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

        x = ½ a₁ t²

        x = ½ a₂ (t-4)²

Let's solve

       a₁  t² = a₂ (t-4)²

      a₁/a₂ t² = t² -2 4 t + 16

      t² (1- 2.0 / 4.0) - 8 t +16

      t² 0.5 - 8 t +16 = 0

      t² -16 t + 32 = 0

Let's solve the second degree equation

     t = [16 ±√( 16² - 4 32)] / 2  

     t = ½ (16 ± 11,3)

Solutions

     t1 = 13.66 s

     t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

6 0
3 years ago
When a gun is fired at the shooting range, the gun recoils (moves backward). Explain this using the law of conservation of momen
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The total momentum is unchanged according to the law of conservation of momentum. When the gun is fired, the bullet gains a high velocity forward (positive velocity), and that velocity multiplied by its mass is the momentum the bullet gains. Therefore, the gun must gain a momentum backwards to cancel out that momentum forward, so the gun recoils back with a negative velocity.
4 0
3 years ago
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