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AleksandrR [38]
3 years ago
12

A tornado lifts a truck 252 m above the ground ñ. As the storm continues, the tornado throws the truck horizontally. It lands 56

0 m away from where it was picked up. How fast was the truck traveling horizontally through its fight? Round your answer to the nearest tenth.
Physics
2 answers:
sashaice [31]3 years ago
8 0

i think its 200 i hope that helps

Lubov Fominskaja [6]3 years ago
7 0

Answer:

78.1 m/s

Just took it.

Explanation:

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What is the magnitude of the velocity of a 25 kg mass that is moving with a momentum of 100 kg*m/s?
Gekata [30.6K]

Answer:

v= 4 m/s

Explanation:

Momenutm is, by definition, the product of mass and velocity.

p = mv

Let's replace what we know and solve for whatever's left

100 kg\cdot m/s = 25kg \cdot v \rightarrow v= 4 m/s

7 0
2 years ago
An egg is thrown horizontally off the top row bleachers at the Brickyard,
Ludmilka [50]

Answer:

4

Explanation:

Divide 30 meters by 7.5 and you´re answer is 4. This is how I would think you solve the problem

6 0
2 years ago
he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?
Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}

With the following condition :

  • \sf{\omega} = angular frequency (rad/s)
  • \sf{\theta} = change of angle value (rad)
  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • \sf{\theta} = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
  • t = interval of the time = 54.9 s.

What was asked :

  • \sf{\omega} = angular frequency = ... rad/s

Step by step :

\sf{\omega = \frac{\theta}{t}}

\sf{\omega = \frac{2,000 \pi}{54.9}}

\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0
2 years ago
How can i calculate distances between objects by using the concepts of echo location
dangina [55]
Send wave from your location to the object and wait until echo is back.
Measure the time taken.

If you know the speed of wave (say sound wave), than just multiply by half time taken wave to return
5 0
3 years ago
1. Mr. Ure has a mass of 65 kg, due to the fact that he is WAY too skinny! What is the force of Earth's gravity on him?
shepuryov [24]

1.

m = mass of Mr. Ure = 65 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Ure is given as

F = mg

F = 65 x 9.8

F = 637 N


2.

F = force of gravity on car = 3050 N

m = mass of the car = ?

g = acceleration due to gravity = 9.8 m/s²

force of gravity on car is given as

F = mg

3050 = m (9.8)

m = 3050/9.8

m = 311.22 kg


3.

m = mass of Mr. Rees = 90 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Rees is given as

F = mg

F = 90 x 9.8

F = 882 N


7 0
3 years ago
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