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notsponge [240]
2 years ago
15

A body is accelerated continuously. What is the form of the graph?

Physics
1 answer:
Inga [223]2 years ago
6 0
That depends on what quantity is graphed.

It also depends on what kind of acceleration is taking place ...
continuous change of speed or continuous change of direction.

-- If the graph shows speed vs time, and the acceleration is a change
in speed, then the graph is a connected series of straight-line pieces. 
Each straight piece slopes up if speed is increasing, or down if speed
is decreasing.

-- If the graph shows speed vs time, and the acceleration is a change in
direction only, then the graph is a straight horizontal line, since speed is
constant.

-- If the graph shows direction vs time, and the acceleration is a change
in speed only, then the graph is a straight horizontal line, since direction
is constant.

-- If the graph shows direction vs time, and the acceleration is a change
in direction, then the graph is a connected series of pieces of line. 
Each piece may be straight if the direction is changing at a constant rate,
or curved if the direction is changing at a rate which grows or shrinks. 
Each piece may slope up if the angle that defines the direction is growing,
or may slope down if the angle that defines the direction is decreasing.

-- If the graph shows distance vs time, and the acceleration is a
change in speed, then the graph is a connected series of pieces
of curves.  Each piece curves up if speed is increasing, or down if
speed is decreasing.

-- If the graph shows distance vs time, and the acceleration is a change
in direction only, then the graph is a straight line sloping up, since speed
is constant.

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irinina [24]
Pretty sure it’s C. all the others are speeding up. acceleration means gradually (over time) getting faster. So it’s C.
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2 years ago
Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi
PIT_PIT [208]

Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

A.

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

B.

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

5 0
2 years ago
Calculate the constant acceleration a in g’s which the catapult of an aircraft carrier must provide to produce a launch velocity
goblinko [34]
The solution to the problem is as follows:

<span>First, I'd convert 188 mi/hr to ft/s. You should end up with about ~275.7 ft/s.
 
So now write down all the values you know:

Vfinal = 275.7 ft/s

Vinitial = 0 ft/s

distance = 299ft
</span>
<span>Now just plug in Vf, Vi and d to solve
</span>
<span>Vf^2 = Vi^2 + 2 a d 

</span><span>BTW: That will give you the acceleration in ft/s^2. You can convert that to "g"s by dividing it by 32 since 1 g is 32 ft/s^2.</span>
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2 years ago
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Answer:

25

Explanation:

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3 years ago
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<h2>Answer:364 LICKS!</h2><h2 />

Explanation:

7 0
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