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Setler [38]
2 years ago
8

A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that

is perpendicular to the platform at its center. the angular speed of the merry-go-round is constant, and a child at a distance of 1.4 m from the axis has a tangential speed of 2.2 m/s. what is the tangential speed of another child, who is located at a distance of 2.1 m from the axis?
Physics
1 answer:
julsineya [31]2 years ago
3 0

Since angular speed is persistent, v/r is constant.
Where:

 v is tangential speed; and

 r is distance from axis.

 

Then equate v/r in both cases to get v in the second case.

Hence, speed = 2.2 x 2.1 / 1.4 meters

= 3.3 meters / seconds

 

Alternative solution would be:

w = 2.2 / 1.4 = 1.57

v = rw = 2.1 x 1.57 = 3.3 meters / seconds 

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6 0
2 years ago
A woman on a bridge 82.2 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an a
solong [7]

Answer:

0.71 m/s

Explanation:

We find the time it takes the stone to hit the water.

Using y = ut - 1/2gt² where y = height of bridge, u = initial speed of stone = 0 m/s, g = acceleration due to gravity = -9.8 m/s² (negative since it is directed downwards)and t = time it takes the stone to hit the water surface.

So, substituting the values of the variables into the equation, we have

y = ut - 1/2gt²

82.2 m = (0m/s)t - 1/2( -9.8 m/s²)t²

82.2 m = 0 + (4.9 m/s²)t²

82.2 m =  (4.9 m/s²)t²

t²  = 82.2 m/4.9 m/s²

t² = 16.78 s²

t = √16.78 s²

t = 4.1 s

This is also the time it takes the raft to move from 5.04 m before the bridge to 2.13 m before the bridge. So, the distance moved by the raft in time t = 4.1 s is 5.04 m - 2.13 m = 2.91 m.

Since speed = distance/time, the raft's speed v = 2.91 m/4.1 s = 0.71 m/s

5 0
3 years ago
Five locations are marked on the world map. which location is most prone to hurricanes? five locations are marked on the world m
Alex17521 [72]

Five locations are marked on the world map. The spot that is prone to hurricane mostly A.

<h3>What is hurricane?</h3>

A hurricane is a cyclone with winds of 74 miles (119 kilometers) each hour or more prominent that is typically joined by downpour, thunder, and lightning, and that occasionally moves into calm scopes.

The locations on the world map spot A, B, C, D, E are regions with country prone to hurricane.

The SPOT A, is GULF of Mexico, a sea bowl nearer to the Atlantic sea, encompassed with the North American landmass, the Hawaiian island is likewise inside that locale.

The SPOT B, is the southern Pacific sea with nations like BRAZIL, PERU, ECUADOR, CHILE, in that equivalent locale. This is likewise visited by the tropical storm.

The SPOT C, is inside the locales of AFRICA, nations at the edge are likewise impacted by typhoon.

The SPOT D, is Asian landmass with Russia at the edge. They are likewise inclined to typhoon.

The SPOT E, is the North Pacific sea locale situated at the upper left hand side of the world guide, with nations like Canada, Alaska and co. Near the artic circle.

Thus, location A is the most prone to hurricanes.

Learn more about hurricane.

brainly.com/question/18950883

#SPJ1

5 0
2 years ago
Make a claim for the behavior of charged objects with other charged objects. Make a claim for the behavior of charged objects wi
emmainna [20.7K]

charged objects will either attract or repel other charged objects

3 0
2 years ago
Read 2 more answers
Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum
cestrela7 [59]

Answer:

T_{2}=278.80 K

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

We can find a relation between the volumes of the initial and the final state.

\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}

Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40

Finally, T2 will be:

T_{2}=278.80 K

6 0
3 years ago
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