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3 years ago

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

1 answer:

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.

You might be interested in

Quinn accelerates her skateboard along a straight path from 0 m/s to 4.0 m/s

umka2103 [35]

initial velocity=u=0m/s
Final velocity=v=4m/s
Time=t=2.5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{4-0}{2.5}$

$\\ \sf\longmapsto Acceleration=\dfrac{4}{2.5}$

$\\ \sf\longmapsto Acceleration=1.6m/s^2$

8 0

3 years ago

A marshmallow is dropped from a 5.71 meter high pedestrian bridge, and 0.921 seconds later, it lands right on the head of an uns

Natalka [10]

let the height of the person with marshmallow on her head be "h"

consider the motion of the marshmallow after it is dropped from bridge.

Y₀ = initial position of the marshmallow above the ground = 5.71 m

Y = final position of marshmallow on head of person = h

v₀ = initial velocity of the marshmallow = 0 m/s

a = acceleration due to gravity = - 9.8 m/s²

t = time of travel for marshmallow = 0.921 sec

Using the kinematics equation

Y = Y₀ + v₀ t + (0.5) a t²

inserting the values

h = 5.71 + 0 (0.921) + (0.5) (-9.8) (0.921)²

h = 5.71 - 4.16

h = 1.55 m

5 0

3 years ago

At the end of the adiabatic expansion, the gas fills a new volume V₁, where V₁ > V₀. Find W, the work done by the gas on the

tino4ka555 [31]

Answer:

$W=\frac{p_0V_0-p_1V_1}{\gamma-1}$

Explanation:

An adiabatic process refers to one where there is no exchange of heat.

The equation of state of an adiabatic process is given by,

$pV^{\gamma}=k$

where,

$p$ = pressure

$V$ = volume

$\gamma=\frac{C_p}{C_V}$

$k$ = constant

Therefore, work done by the gas during expansion is,

$W=\int\limits^{V_1}_{V_0} {p} \, dV$

$=k\int\limits^{V_1}_{V_0} {V^{-\gamma}} \, dV$

$=\frac{k}{\gamma -1} (V_0^{1-\gamma}-V_1^{1-\gamma})\\$

(using $pV^{\gamma}=k$ )

$=\frac{p_0V_0-p_1V_1}{\gamma-1}$

4 0

3 years ago

A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$

The locations are +5.9 cm or -5.9 cm

4 0

3 years ago

A tire hanging from a tree is at rest. The force of gravity is pulling down on the tire. Another force, called tension, is pulli

Anton [14]

Answer:

body will stop moving if the gravity will be equal to the tension produced in the string so both the forces will cancel the influence of each other leading to the state when the body is in equilibrium

8 0

3 years ago