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Assoli18 [71]
3 years ago
13

When measuring accelerations with three different graphs (x-t, v-t, a-t) which gives the most reliable results and why?

Physics
1 answer:
kipiarov [429]3 years ago
5 0

Answer:

the best graph to find the acceleration is v-t since calculating the slope averages the different experimental errors.

Explanation:

The different graphics depending on time give various information, let's examine what we can get from some

Graph of x -t. from this graph we can obtain the speed through the slope, but the acceleration is not directly obtainable

v-t chart. We can get the acceleration not through the slope and the distance traveled by the area under the curve. Obtaining acceleration is very accurate since it is an average that avoids possible errors in measurements. This is the best graph to find the acceleration

Graph of a-t In this graph the acceleration is a point on the Y axis, it gives some errors because it depends strongly on the possible experimental errors.

In conclusion, the best graph to find the acceleration is v-t since calculating the slope averages the different experimental errors.

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A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon
Anettt [7]
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed v_x=30 m/s, and an accelerated motion on the y-axis, with initial speed v_y=20 m/s and acceleration g=9.81 m/s^2:
S_x(t)=v_xt
S_y(t)=v_y t- \frac{1}{2} gt^2
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
S_y(t)=0
Therefore:
v_y t -  \frac{1}{2}gt^2=0
which has two solutions:
t=0 is the time of the beginning of the motion,
t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m
4 0
3 years ago
Find the ratio of the final speed of the electron to the final speed of the hydrogen ion, assuming non-relativistic speeds. Take
KiRa [710]

Answer:

\frac{V_{e}}{V_{h}}=0.428*10^{2}

Explanation:

From conservation of energy states that

K_{i}+v_{i}=v_{f}+K_{f}\\ as\\K_{i}=0\\K_{f}=1/2mv^{2}\\ v_{i}=qv\\v_{f}=0\\So\\qv=1/2mv^{2}\\ v=\sqrt{\frac{2qv}{m} }\\ Velocity_{electron}=\sqrt{\frac{2qv}{m_{e}} }\\Velocity_{hydrogen}=\sqrt{\frac{2qv}{m_{h}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{\frac{2qv}{m_{e}}}{\frac{2qv}{m_{h}}}}\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{m_{h}}{m_{e}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{1.67*10^{-27} }{9.11*10^{-31} } }\\\frac{V_{e}}{V_{h}}=0.428*10^{2}

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3 years ago
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andrey2020 [161]
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3 years ago
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Answer:

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2 years ago
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