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maria [59]
3 years ago
11

- The school zone in front of your school has a posted speed limit of 25 mi/h, which is about 11 m/s. Let's

Physics
1 answer:
miv72 [106K]3 years ago
5 0

Answer:

s = 38.7 m

Explanation:

First we calculate the distance covered during uniform motion of reaction time.

s₁ = vt

where,

s₁ = distance covered during uniform motion = ?

v = uniform speed = 11 m/s

t = time = 2.3 s

Therefore,

s₁ = (11 m/s)(2.3 s)

s₁ = 25.3 m

Now, we calculate the distance covered during decelerated motion:

2as₂ = Vf² - Vi²

where,

a = deceleration = -4.5 m/s²

s₂ = distance covered during decelerated motion = ?

Vf = Final Velocity = 0 m/s

Vi = Initial Velocity = 11 m/s

Therefore,

2(-4.5 m/s²)s₂ = (0 m/s)² - (11 m/s)²

s₂ = (-121 m²/s²)/(-9 m/s²)

s₂ = 13.4 m

the total distance will be:

s = s₁ + s₂

s = 25.3 m + 13.4 m

<u>s = 38.7 m</u>

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The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to
DedPeter [7]

Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

r_o is the distance of the new object from the sun (orbital radius)

T_o=180 d is the orbital period of the object

r_e = 1.50\cdot 10^{11} m is the orbital radius of the Earth

T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

3 0
3 years ago
What is the main function of the spongy bone'
storchak [24]
Oooooo there's a spongy bone? that's cool! Lol okay okay, I will research it and help you out. 

Here's what I found:

Cancellous bone<span>, also known as </span>spongy<span> or </span>trabecular bone<span>, is one of the </span>two<span> types of </span>bone<span> tissue found in the human body. ... It is very porous and contains red </span>bone<span>marrow, where blood cells are made.</span>
7 0
3 years ago
Read 2 more answers
Help ME QUICK PLEASE QUESTION 2 AND 4 PLEASE 58 POINTS
monitta

Answer:

For number 4: A vector pointing to the right with a magnitude of 2.0

Explanation:

Very simple- just subtract 6-2

I am not sure how to do #2- sorry!

6 0
3 years ago
A child in a boat throws a 5.80-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 24.6kg and the ma
Sergio [31]

Answer:

-0.912 m/s

Explanation:

When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

(m_c + m_b)v_b + m_pv_p = 0

where m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg are the mass of the child, the boat and the package, respectively. , v_p = 10m/s, v_b are the velocity of the package and the boat after throwing.

(24.6 + 39)v_b + 5.8*10 = 0

63.6v_b + 58 = 0

v_b = -58/63.6 = -0.912 m/s

3 0
3 years ago
a care starting from rest has an acceleration 0.3 m/s square, calculate the velocity and distance travelled by this car after 2
Anna [14]

Answer:

Final velocity (v) = 36 m/s

Distance traveled (s) = 2,160 m

Explanation:

Given:

Initial velocity (u) = 0

Acceleration (a) = 0.3 m/s

Time travel (t) = 2 minutes = 120 seconds

Find:

Final velocity (v) = ?

Distance traveled (s) = ?

Computation:

v = u + at

v = 0 + 0.3(120)

v = 0.3(120)

v = 36 m/s

Final velocity (v) = 36 m/s

Distance traveled (s) = ut + (1/2)at²

Distance traveled (s) = (0.5)(0.3 × 120 × 120)

Distance traveled (s) = 2,160 m

3 0
3 years ago
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