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Vikentia [17]
2 years ago
11

How does acceleration affect speed and velocity

Physics
1 answer:
Nikitich [7]2 years ago
5 0

Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.

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Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an
castortr0y [4]

Answer:

  • Fx = -9.15 N
  • Fy = 1.72 N
  • F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

  F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

  f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

  = -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

  ≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

  ≈ (-9.15309, 1.71982)

The resultant component forces are ...

  • Fx = -9.15 N
  • Fy = 1.72 N

Then the magnitude and direction of the resultant are

  F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

  F∠γ ≈ 9.31∠-10.6°

4 0
3 years ago
Directions: Read each question carefully and answer
Lady bird [3.3K]
I have all the answers here so take this

8 0
2 years ago
An atom’s emission of light with a specific amount of energy confirms that
laila [671]
<span>An atom’s emission of light with a specific amount of energy confirms that </span><span>electrons emit and absorb energy based on their position around the nucleus.

The light emitted from an electron is a result of the electron's quantum jumps/leaps ( atomic electron transitions ) to and from different energy levels.</span>
3 0
3 years ago
Read 2 more answers
The engine starter and a headlight of a car are connected in parallel to the 12.0-V car battery. In this situation, the headligh
stepladder [879]

Answer:

The total power they will consume in series is approximately 2.257 W

Explanation:

The connection arrangement of the headlight and the engine starter = Parallel to the battery

The voltage of the battery, V = 12.0 V

The power at which the headlight operates in parallel, P_{headlight} = 38 W

The power at which the kick starter operates in parallel, P_{kick \ starter} = 2.40 kW

We have;

P = V²/R

Where;

R = The resistance

V = The voltage = 12 V (The voltage is the same in parallel circuit)

For the headlight, we have;

R₁ = V²/P_{headlight}  = 12²/38 = 72/19

R₁ = 72/19 Ω

For the kick starter, we have;

R₂ = V²/P_{kick \ starter} = 12²/2.4 = 60

R₂ = 60 Ω

When the headlight and kick starter are rewired to be in series, we have;

Total resistance, R = R₁ + R₂

Therefore;

R = ((72/19) + 60) Ω = (1212/19) Ω

The current flowing, I = V/R

∴ I = 12 V/(1212/19) Ω = (19/101) A

We note that power, P = I²R

In the series connection, we have;

P_{headlight} = I² × R₁

∴ P_{headlight} = ((19/101) A)² × 72/19 Ω = 1368/10201 W ≈ 0.134 W

The power at which the headlight operates in series, P_{headlight, S} ≈ 0.134 W

P_{kick \ starter} = ((19/101) A)² × 60 Ω = 21660/10201 W ≈ 2.123 W

The power at which the kick starter operates in series, P_{kick \ starter, S} ≈ 2.123 W

The total power they will consume, P_{Total} = P_{headlight, S} + P_{kick \ starter, S}

Therefore;

P_{Total} ≈ 0.134 W + 2.123 W = 2.257 W

4 0
3 years ago
How much power does it take to do 500J of work in 10 sec
tatuchka [14]
P=w/t
so the answer is 500/10=50
8 0
2 years ago
Read 2 more answers
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