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3 years ago

6

11. (II) What is the linear speed, due to the Earth's rotation,

1 answer:

Maksim231197 [3]3 years ago

7 0

Answer:

A) 464m/s

B)185m/s

C)345m/s

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What does the diagram represent?

solong [7]

Answer:

Atoms in a element on the periodic table I dont know which one tho so try google that sorry I am not much of a help

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3 years ago

Which is an example of observational learning?

vekshin1

Janice is the one who learned something by watching what was going on in the world around her.

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3 years ago

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On a trip to the Colorado Rockies, you notice that when the freeway goes steeply down a hill, there are emergency exits every fe

Zanzabum

Answer:

The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.

Explanation:

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3 years ago

John’s mass is 95.6 kg, and Barbara’s is 55.3 kg. He is standing on the x axis at xJ = +10.9 m, while she is standing on the x a

Anna11 [10]

Answer:Shifted towards Left by distance of 2.243 m

Explanation:

Given

Mass of john $m_1=95.6 kg$

Mass of barbara $m_2=55.3 kg$

John is standing at $x=10.9 m$

Barbara is standing at $x=2.50 m$

$x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

$x_{com}=\frac{95.6\times 10.9+55.3\times 2.5}{95.6+55.3}$

$x_{com}=\frac{1180.29}{150.9}$

$x_{com}=7.821 m$

Now if they change their Position then

$x'_{com}=\frac{95.6\times 2.5+55.3\times 10.9}{95.6+55.3}$

$x'_{com}=\frac{841.77}{150.9}$

$x'_{com}=5.578$

Thus we can see that center of mass shifted towards left by a distance of $2.243 m$ because heavier is shifted towards left

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3 years ago

A lab cart flies off the end of a track on a lab table with a horizontal velocity of 2.08 m/s. If it lands 0.96 meters from the

lana [24]

Answer:

Option C. 1.2 m

Explanation:

The following data were obtained from the question:

horizontal velocity (u) = 2.08 m/s

Horizontal distance (s) = 0.96 m

Height (h) of the table =?

Next, we shall determine the time taken for the lab cart to get to the ground. This can be obtained as follow:

horizontal velocity (u) = 2.08 m/s

Horizontal distance (s) = 0.96 m

Time (t) =?

s = ut

0.96 = 2.08 × t

Divide both side by 2.08

t = 0.96 / 2.08

t = 0.5 s

Finally, we shall determine the height of the table as illustrated below:

Time (t) = 0.5 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) of the table =?

h = ½gt²

h = ½× 9.8 × 0.5²

h = 4.9 × 0.25

h = 1.2 m

Thus, the height of the table is 1.2 m

3 0

2 years ago