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Sindrei [870]
3 years ago
7

Consider this line graph that represents speed in meters/second. The X-axis is labelled in seconds; the Y-axis is labelled in me

ters. Imagine that the moving object stops after 2.5 seconds and stays stopped for 0.5 seconds. What would the line for that time interval look like on this graph?
Physics
2 answers:
Mrac [35]3 years ago
8 0
If the x-axis represent time and the y-axis represents distance,
then the SPEED in meters/second is the SLOPE of the line
on the graph.

The object is moving, so the line is either rising or falling, but
it's definitely not staying horizontal, because the distance is
changing.

The object stops moving at 2.5 seconds, and stays stopped
for 0.5 second.  So right at the 2.5 sec point, the line stops
sloping, and becomes horizontal until 3.0 sec.

We don't know what happens then.  If the object stays still
for longer, then the line stays horizontal until the object starts
moving again.  Then the line starts sloping again.
Ivanshal [37]3 years ago
8 0

To represent no movement of distance/time graph, the line would have no slope. In this case, the line would be straight and horizontal at Y equals 14, to {3, 14}.

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Ivahew [28]

Gamma radiation could be used to sterilize plastic petri plates in a plastic wrapper.

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<u>Explanation: </u>

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8 0
4 years ago
In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. the dish is on a shelf above the poi
Georgia [21]

components of the speed of the coin is given as

v_x = v cos60

v_x = 6.4 cos60 = 3.2 m/s

v_y = vsin60

v_y = 6.4 sin60 = 5.54 m/s

now the time taken by the coin to reach the plate is given by

t = \frac{\delta x}{v_x}

t = \frac{2.1}{3.2}

t = 0.656 s

now in order to find the height

h = vy * t + \frac{1}{2} at^2

h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2

h = 1.52 m

so it is placed at 1.52 m height

3 0
3 years ago
A cannon fires a 0.652 kg shell with initial
laila [671]
 Mass have no effect for the projectile motion and  u want to know the  height "h"   
first,
        find the vertical and horizontal components of velocity 
 vertical component of velocity = 12 sin 61                  
horizontal component of velocity = 12 cos 61
now for the vertical motion ;               
             S = ut + (1/2) at^2
where
 s = h 
u = initial vertical component of velocity 
t = 0.473 s 
a = gravitational deceleration (-g) = -9.8 m/s^2    
       
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u can simplify this and u will get the answer

h=.5Gt2 

H=1.09m
6 0
3 years ago
Read 2 more answers
Which statement best describes how work and power are different? a. To find work we need to know force and distance; to find pow
weqwewe [10]
A. To find work we need to know F and S; to find power we need to know F and V
6 0
3 years ago
Explanation A 5000 kg rocket is at rest in deep space. The rocket burns fuel pushing 10kg of exhaust gases rearward at 4000 m/s.
katovenus [111]

Answer:

F = 4000 N

Explanation:

given,

mass of rocket (M)= 5000 Kg

10 Kg gas burns at speed (m)= 4000 m/s

time = 10 s

average force = ?

at the end the rocket is at rest

by conservation of momentum

 M v + m v' = 0

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speed of rocket = 8 m/s

now,

we know

change in momentum = F x Δ t

F = \dfrac{m(v_i-v_f)}{\Delta t}

F = \dfrac{5000(8-0)}{10}

      F = 4000 N

Hence, the average force applied to the rocket is equal to F = 4000 N

4 0
4 years ago
Read 2 more answers
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