What is disaster????

Physics

2 answers:

elixir [45]3 years ago

8 0

Answer:

DISASTER IS A SERIOUS PROBLEM OF OCCURRING OVER A SHORT OR LONG PERIOD. 

telo118 [61]3 years ago

4 0

Disaster:- a calamitous event, especially one occurring suddenly and causing great loss of life, damage, or hardship, as a flood / airplane crash.

:))

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Need help with both questions!

xenn [34]

#14 isn't really a Physics problem. It's more of just reading a graph.

A). When speed changes, acceleration is

(change in speed) / (time for the change) .

To be correct about it, acceleration can be positive ... when speed
is increasing ... or it can be negative ... when speed is decreasing.
So, on this graph, there are two periods of acceleration:

From zero to 2 seconds, acceleration = (8 m/s) / (4 sec) = 2 m/s² .

From 10 to 12 seconds, acceleration = (-4 m/s) / (2 sec) = -2 m/s² .

B). From 12 to16 seconds, you can read the speed right from
the graph. It's 4 m/s .

C). From 2 to 10 seconds, the objects speed is a steady 8 m/s.
Covering 8 m/s every second for 8 seconds, it covers 64 meters.
Do you remember that distance is the area under the speed/time
graph? You can see that plainly on this graph. From 2 to 10 sec,
there are 16 blocks. Each block is (2 m/s) high and (2 sec) wide,
so its area is (2 m/s) x (2 sec) = 4 meters. The area of 16 blocks
is (16) x (4 meters) = 64 meters.
====================================

#15.

a). constant velocity on a distance graph is a line that slopes up;
constant velocity on a velocity graph is a horizontal line;

b). positive constant acceleration on a distance graph is a
line that curves up;
positive constant acceleration on a velocity graph is a
straight line that slopes up;

c). "uniformly slowing down to a stop" on a distance graph
is a line that's less and less curved as time goes on, and
eventually reaches the x-axis.
"uniformly slowing down to a stop" on a velocity graph is
a straight line that slopes down, and stops when it reaches
the x-axis.

7 0

3 years ago

The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 m/s (about 45 mi/h). Initially, the car

Alex73 [517]

Answer:

15.8640053791 s

392.780107582 m

29.5184032275 m/s

Explanation:

0 denotes initial

x denotes displacement

c denotes car

t denotes truck

r denotes rear

$x_0_{cr}=-49.5\ m$

$a_c=0.6\ m/s^2$

$x_0_{t}=0$

$v_0_{c}=v_0_{t}$

For the car

$x_c=x_0+v_0_{c}t+\dfrac{1}{2}at^2$

The displacement of the truck will be

$x_t=v_tt$

From the above two equations we get

$x_c-x_t=x_0+v_0_{c}+\dfrac{1}{2}at-v_{t}t=26\\\Rightarrow 26=x_0+\dfrac{1}{2}at^2\\\Rightarrow 26+49.5=\dfrac{1}{2}0.6t^2\\\Rightarrow t=\sqrt{\dfrac{2(26+49.5)}{0.6}}\\\Rightarrow t=15.8640053791\ s$

The time taken is 15.8640053791 s

$x-x_0=v_{0}_{c}t+\dfrac{1}{2}at^2\\\Rightarrow x-x_0=20\times 15.8640053791+\dfrac{1}{2}0.6\times 15.8640053791^2\\\Rightarrow x-x_0=392.780107582\ m$