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LenaWriter [7]
3 years ago
14

The ____ of a position-time graph represents an object’s velocity.

Physics
2 answers:
xenn [34]3 years ago
8 0
<span>The slope of a position-time graph represents an object’s speed.

You can't determine the object's velocity from the graph, because
the graph only tells you the distance the object moves, but not the
direction.
</span>
Marianna [84]3 years ago
3 0

Answer:

The slope of a position-time graph represents an object’s velocity.

Explanation:

In a position-time graph, the values on the x-axis represent the time, while the values on the y-axis represent the position of the object.

Velocity is defined as the ratio between the displacement of an object and the time taken:

v=\frac{\Delta s}{\Delta t}

However, we can see that this definition corresponds to the slope of the curve in a position-time graph. In fact:

\Delta s, the displacement, corresponds to the difference in position, so the difference between the values on the y-axis: \Delta s=y_2 -y_1

\Delta t, the time interval, corresponds to the difference in times, so the difference between the values on the x-axis: \Delta t= t_2 -t_1=x_2 -x_1

So, the velocity is

v=\frac{\Delta s}{\Delta t}=\frac{y_2 -y_1}{x_2 -x_1}

which corresponds to the slope of the curve.

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An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of
Elan Coil [88]

Solution:

Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg

Temperature = 200 °C

And 25% of the volume by liquid water is steam.

State 1

$m=\frac{V}{v}$

$m=m_f+m_g$

$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$

$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$       (taking the value of $v_g$ and $v_g$ at 200°C  )

$V=6.304 \times 10^{-3}$

Now quality of vapor

$x=\frac{m_g}{m}$

  $=3.377 \times 10^{-3}$

Internal energy at state 1 can be found out by

$u_1=u_f+xu_{fg}$

    $=850.65+3.377\times10^{-3}\times 1744.65$

    = 856.54 kJ/kg

After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor $v_2=v_g \text { and }\ x=1$

Tank is rigid, so volume of tank is constant.

$v_g=v_2=\frac{V}{m}$

$v_g=\frac{6.304\times 10^{-3}}{1.4}$

$v_g=4.502 \times 10^{-3} \ m^3 /kg$

Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$

   = 369.11° C

Internal energy at state 2

$u_2=2154.9 \ kJ/kg$

Now power rating of the resistor

$P=\frac{m(u_2-u_1)}{t}$

$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$

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3 years ago
Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the
Masja [62]

Answer:

distance of 2nd team from 1st team will be:  58.2

Direction of 2nd team from 1st team will be:  14.90 deg North of east

Explanation:

ASSUME Vector is R and  makes angle A with +x-axis,

therefore component of vector R is

R_x = Rcos A

R_y = Rsin A

From above relation

Assuming base camp as the origin, location of 1st team is

R_1 = 37 km away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km

R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km

location of 2nd team is at

R_2 = 32 km, at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km

R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km

Now position of 2nd team with respect to 1st team will be given by:

R_3 = R_2 - R_1

R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j

Using above values:

R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j

R_3 = 51.49 i + 13.71 j

distance of 2nd team from 1st team will be:

\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)

\left | R_3 \right | = 53.28 km = 58.2 km

Direction of 2nd team from 1st team will be:

Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]

Direction = 14.90 deg North of east

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Answer:

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A block of wood 3.0 cm on each side has a mass of 27g. what is the density of this block?
schepotkina [342]
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