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Dafna11 [192]
3 years ago
7

A woman performs 2000 J of work in order to push a cart full of groceries 50 meters. How much force did she apply to the cart? *

Physics
2 answers:
GaryK [48]3 years ago
5 0
They will give you 4000
ad-work [718]3 years ago
4 0
They will give you 4000
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Nikolas had an idea that he could use the compressed carbon dioxide in a fire extinguisher to propel him on his skateboard.
Vikentia [17]
The Newton’s law Nikolas would use to come up with this idea is the <span>Third law that states:

</span><span>When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
</span>
So, in this case, let's name the first Body A which is the skateboard and the second body B which is <span>the compressed carbon dioxide in a fire extinguisher. Then, as shown in the figure below, according to the Third law:

</span>FA = -FB<span>

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8 0
3 years ago
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Hold a pencil in front of your eye at a position where its blunt end just blocks out the Moon.
Valentin [98]

Answer:

D = 3.55 \times 10^6 m

Explanation:

Light rays coming from moon is blocked by the pencil

so as per figure we know that angle subtended by pencil and angle subtended by moon must be same

so we have

Angle = \frac{Arc}{Radius}

so we have

\frac{D}{3.8 \times 10^8 m} = \frac{0.7 cm}{75 cm}

so we have

D = 3.55 \times 10^6 m

4 0
3 years ago
A cruise started its trip 8:00. Its average speed was 100 km/h.
Aneli [31]

Answer:

120 k/m

Explanation:

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3 years ago
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Why is the weight of an object more on Earth than on the Moon?
JulijaS [17]
<span>Why is the weight of an object more on Earth than on the Moon? i would say c</span>
8 0
3 years ago
The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20
IrinaK [193]

Answer:

The average angular acceleration of the Earth is; α  = 6.152 X 10⁻²⁰ rad/s²

Explanation:

We are given;

The period of 365 revolutions of Earth in 2006, T₁ = 365 days, 0.840 sec

Converting to seconds, we have;

T₁ = (365 × 24 × 60 × 60) + 0.84

T₁ = (3.1536 x 10⁷) + 0.840

T₁ = 31536000.84 s

Now, the period of 365 rotation of Earth in 2006 is; T₀ = 365 days

Converting to seconds, we have;

T₀ = 31536000 s

Hence, time period of one rotation in the year 2006 is;

Tₐ = 31536000.84/365

Tₐ = 86400.0023 s

The time period of rotation is given by the formula;

Tₐ = 2π/ωₐ

Making ωₐ the subject;

ωₐ = 2π/Tₐ

Plugging in the relevant values;

ωₐ = 2π/ 365.046306        

ωₐ = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation in the year 1906 is;

Tₓ = 31536000/365

Tₓ = 86400 s

Time period of rotation,

Tₓ = 2π /ωₓ

ωₓ = 2π / T

Plugging in the relevant values;

ωₓ = 2π/86400

ωₓ = 7.272205217  x 10⁻⁵ rad/s

The average angular acceleration is given by;

α  = (ωₓ -   ωₐ) /  T₁

α = ((7.272205217  × 10⁻⁵) - (7.272205023 × 10⁻⁵)) / 31536000.84

 α  = 6.152 X 10⁻²⁰ rad/s²

Thus, the average angular acceleration of the Earth is; α  = 6.152 X 10⁻²⁰ rad/s²

8 0
3 years ago
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