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3 years ago

Which of the following is an obstacle to creating computer-based models for tracking a hurricane?

Physics

1 answer:

iren [92.7K]3 years ago

5 0

Answer:

4. All of the above I think, not to sure about 1. but the rest are right so im like 90.99999 percent sure good luck

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Terry can ride 30 miles in 2 hours. If his riding speed is

marishachu [46]

Answer:

Explanation:

Terry can ride at a speed of

V = 30 miles in 2hours

Speed = distance / time

V = 30 /2

V = 15 mile/hour

So, we want to know the distance traveled in 1.7hours

Then,

Speed = distane / time

Distance = speed × time

Distance =15 × 1.7

Distance = 25.5 miles

So, the distance traveled in 1.7hours is 25.5 miles

5 0

2 years ago

Read 2 more answers

4. Two forces act on a 2 kg object as shown. What is the magnitude of the acceleration of the object?

aleksandrvk [35]

The resultant force on the object is

∑ F = 〈0, 8〉 N + 〈6, 0〉 N = 〈6, 8〉 N

which has a magnitude of

F = √((6 N)² + (8 N)²) = √(100 N²) = 10 N

By Newton's second law, the acceleration has magnitude a such that

F = m a

10 N = (2 kg) a

a = (10 N) / (2 kg)

a = 5 m/s²

so the answer is B.

4 0

2 years ago

A steel bridge is 1000 m long at -20°C in winter. What is the change in length when the temperature rises to 40°C in summer? The

xenn [34]

Answer:

ΔL = 0.66 m

Explanation:

The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:

ΔL = αLΔT

where,

ΔL = Change in Length of the bridge = ?

α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹

L = Original Length of the Bridge = 1000 m

ΔT = Change in Temperature = Final Temperature - Initial Temperature

ΔT = 40°C - (-20°C) = 60°C

Therefore,

ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)

ΔL = 0.66 m

6 0

3 years ago

Alex is standing still and throws a football with a speed of 10 m/s to his friend, who is also standing still. The two friends a

Phantasy [73]

The question is incomplete. It comes with a set of answer choices.

These are the answer choices:

Alex observes it as 10 m/s, and his friend observes it as less than 10 m/s.

Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.

Both Alex and his friend observe it as 10 m/s.

Both Alex and his friend observe it as less than 10 m/s.

Answer: Both Alex and his friend observe it as 10 m/s.

Justification:

1) The speed is relative to the frame of reference.

2) It is said that the both Alex and his friend are standing still.

3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).

Of course, Alex is watching the ball moving away and his friend is seing it approaching, but it is not relevant for the question, as it deals with the speed which is only about magnitude, not direction.

7 0

2 years ago

Read 2 more answers

A boy can swim 3.0 meter a second in still water while trying to swim directly across a river from west to east, he is pulled by

lana66690 [7]

Answer:

Angle: $48.19^o$

Explanation:

Two-Dimension Motion

When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.

Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.

To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

$v_b\ cos\alpha$

where $v_b$ is the speed of the boy in still water and $\alpha$ is the angle respect to the shoreline. If the river flows at speed $v_s$ , we now set

$v_b\ cos\alpha=v_s$

$\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}$

$\alpha=48.19^o$

8 0

2 years ago