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IrinaK [193]
3 years ago
12

PLEASE HELP! I NEED TO ANSWER THIS QUICKLY!

Physics
1 answer:
rodikova [14]3 years ago
4 0

Answer:

I'm not a genius ok?

Explanation:

1. Radar communication, Analysis of the molecular and atomic structure, telephone communication

2. c

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A rocket moves through outer space at 11,000 m/s. At this rate, how much time would be required to travel the distance from Eart
Charra [1.4K]
11,000 m = 11 km

11 km/s over 380,000km

380,000 / 11 = 34545.4 seconds

34545.4 / 60 = 575.7 minutes
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3 years ago
(Figure 1) shows the angular-velocity-versus-time graph for a particle moving in a circle. How many revolutions does the object
Travka [436]

Answer: 10.34

Explanation:

Given

\omega -t graph for a particle is given

angle turned by the particle in radians is given by the area under \omega -t graph

The area is given by

A=20\times (2-0)+10(4-2)+\dfrac{1}{2}\times (20-10)\times (3-2)\\A=40+20+5=65\ rad

Revolutions(N) made by the object is given by

N=\dfrac{65}{2\pi }=10.34

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2 years ago
Extend the life of your __________ by avoiding fast starts, stops and sharp turns.
skelet666 [1.2K]

Answer:

Tires.

Explanation:

There are the few steps which are discussed below should be taken to increase or extend the life of tires.

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(2)  Avoid fast stop: Fast stop of the vehicle will also increase the pressure on the tires due to the friction between the tires and the road will decrease the life of tires.

(3) Avoid sharp turns: The alignment of the wheels and tires are in such a way that they work properly when vehicle is drive in a straight path but sharp turn will increase the uneven pressure on the tires will lead to decrease the life of tires.

Therefore, the life of tires can be extend by avoiding all the above mention actions such as fast stop, start and sharp turns.

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3 years ago
How do you draw an ear
Neporo4naja [7]

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3 years ago
A highway patrol car traveling a constant speed of 105 km/h is passed by a speeding car traveling 140 km/h. Exactly 1.00 s after
vodka [1.7K]

Answer:

The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.

Explanation:

Hi there!

The position of the patrol car at a time "t" can be calculated using this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the patrol car at a time "t"

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:

x = x0 + v · t

Where "v" is the constant velocity.

First, let´s convert the velocity units into m/s:

140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s

105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s

We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.

When the patrol car catches the speeder, the position of both cars is the same:

position of the patrol car = position of the speeder

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:

x = v · t

x = 38.9 m/s · 1 s = 38.9 m

When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t

Let´s agrupate terms and equalize to zero:

-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0

-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0

Solving the quadratic equation for t using the quadratic formula:

t = 8.24 s  (the other solution is discarded because it is negative)

The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.

3 0
3 years ago
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