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Tatiana [17]
4 years ago
5

A person rides a Ferris wheel that turns with constant angular velocity. Her weight is 549.0 N. At the top of the ride her appar

ent weight is 1.500 N different from her true weight. What is her apparent weight at the top of the ride?
Physics
1 answer:
pashok25 [27]4 years ago
8 0

Answer:

Explanation:

Given that the angular velocity is constant.

The actual weight is W = 549N

The ride apparent weight is

W' = 1.5N

Apparent weight at the top of the ride?

Using newton second law of motion

ΣF = m•ar

ar is the radial acceleration

N — W = —m•ar

N = W —m•ar

N = mg —m•ar

N = m(g—ar)

The apparent weight is equal to the normal

W'(top) = N = m(g—ar)

W'(top) = m(g—ar)

W'(top) = mg —m•ar

We know that the actual weight is

W=mg

Also, the apparent weight is

W' = m•ar

Then

W'(top) = Actual weight - apparent

W'(top) = 549—1.5

W'(top) = 547.5 N

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Helen [10]

This Question is not complete

Complete Question:

Part B - Permeability of the lipid bilayer Some solutes are able to pass directly through the lipid bilayer of a plasma membrane, whereas other solutes require a transport protein or other mechanism to cross between the inside and the outside of a cell. The fact that the plasma membrane is permeable to some solutes but not others is what is referred to as selective permeability. Which of the following molecules can cross the lipid bilayer of a membrane directly, without a transport protein or other mechanism?

a) proteins

b) water

c) ions

d) sucrose

e) lipids

f) carbon dioxide

g) oxygen

Answer:

b) water

e) lipids

f) carbon dioxide

g) oxygen

Explanation:

Diffusion is the means by which molecules like water, lipids, carbon dioxide and oxygen easily move across the lipid bilayer of a membrane directly without the use of a transport protein or any another mechanism.

Diffusion is the way or means through which molecules( such as gases or liquids) move across a membrane from and region of higher concentration to a region of lower concentration.

Before a molecule can pass through the semi permeable membrane without the use of any mechanism or transport protein, it has to be:

a) a small molecule

b) an uncharged molecule

c) an hydrophobic molecule

7 0
3 years ago
A car is traveling 20 m/s and slows down at a uniform rate. It stops in 6 seconds. How far has it traveled in this interval?
Scilla [17]

Answer:

Car travel a distance of 60.06 m in 6 sec

Explanation:

We have given initial velocity v = 20 m/sec

Time = 6 sec

As the car stops finally so final velocity v = 0

From the first equation of motion

v = u+at (as the car velocity is slows down means it is a case of deceleration)

So v = u-at

0=20-a\times 6

a=3.33m/sec^2

Now from second equation of motion s=ut-\frac{1}{2}at^2=20\times 6-\frac{1}{2}\times 3.33\times 6^2=60.06m

6 0
3 years ago
2
Xelga [282]

Answer:

About 7.67 m/s.

Explanation:

Mechanical energy is always conserved. Hence:

\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}

Where <em>U</em> is potential energy and <em>K</em> is kinetic energy.

Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:

\displaystyle U_i = K_f

Substitute and solve for final velocity:
\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\  2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ &  =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}

In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.

8 0
2 years ago
Use the drop down menu to complete the sentence. Resistance is measured in units called...
liubo4ka [24]
Ohms is the correct answer
7 0
4 years ago
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A woman is 1.6 m tall and has a mass of 50 kg. She moves past an observer with the direction of the motion parallel to her heigh
eduard

To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.

By definition we know that the relative velocity of an object with reference to the Light, is defined by

V_0 = \frac{V}{\sqrt{1-\frac{V^2}{c^2}}}

Where,

V = Speed from relative point

c = Speed of light

On the other hand we have that the linear momentum is defined as

P = mv

Replacing the relative velocity equation here we have to

P = \frac{mV}{\sqrt{1-\frac{V^2}{c^2}}}

P^2 = \frac{m^2V^2}{1-\frac{V^2}{c^2}}

P^2 = \frac{P^2V^2}{c^2}+m^2V^2

P^2 = V^2 (\frac{P^2}{c^2}+m^2)

V^2 = \frac{P^2}{\frac{P^2}{c^2}+m^2}

V^2 = \frac{(2.1*10^10)^2}{\frac{(2.1*10^10)^2}{(3.8*10^8)^2}+50^2}

V = 2.81784*10^8m/s

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l = l_0*\sqrt{1-\frac{V^2}{c^2}}

l = 1.6*\sqrt{1-\frac{(2.81*10^8)^2}{(3*10^8)^2}}

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Therefore the height which the observerd measure for her is 0.56m

8 0
3 years ago
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