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olga2289 [7]
3 years ago
6

Name the physical quantity measured by the velocity-time graph?

Physics
1 answer:
VARVARA [1.3K]3 years ago
7 0
The so-called "velocity-time" graph is actually a "speed-time" graph.  At any point
on it, the 'x'-coordinate is a time, and the 'y'-coordinate is the speed at that time.

'Velocity' is a speed AND a direction.  Without a direction, you do not have a velocity,
and these graphs never show the direction of the motion.  It seems to me that it would be
pretty tough to draw a graph that shows the direction of motion at every instant of time,
so my take is that you'll never see a true "velocity-time" graph. 

At best, it would need a second line on it, whose 'y'-coordinate referred to a second
axis, calibrated in angle and representing the 'bearing' or 'heading' of the motion at
each instant. The graph of uniform circular motion, for example, would have a straight
horizontal line for speed, and a 'sawtooth' wave for direction.
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<span>F= ma force is directed on the working surface so 40 = m x 2 hence m must be 20 units needed are probably kg</span>
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3 years ago
(Please help, will give thanks + brainliest answer)
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Answer:

A) 1.9m/s

Explanation:

Use the equation v=d/t

The time is 10 seconds. The distance is more complicated because it’s a circle so u use the equation for circumference C=2•pi•r. So you just do the circumference over time and get the answer. Hope this helps

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4 years ago
The Atomic Mass of an element is
dedylja [7]

Answer:

C) A ratio based on the mass of a carbon-12 atom

Explanation:

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On the other hand we have the number of neutrons, and for one element which difers only on the number of neutrons, we have different isotopes from one element.

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6 0
4 years ago
Read 2 more answers
An iron robot falls from rest at a great height. neglecting air resistance, what is its speed after it has fallen for 3.5 second
algol13
In an uniformly accelerated motion, the velocity of the object follows the law:
v(t) = v_0 + at
where v_0 is the initial velocity, a the acceleration and t the time.

In our problem, the robot starts from rest, so the initial speed is zero: v_i =0. The robot is in free fall, so the acceleraion is the gravitational acceleration g=9.81 m/s^2. therefore, after a time t=3.5 s, the velocity is 
v(3.5 s)= 0 + (9.81 m/s^2)(3.5 s)=34.3 m/s
6 0
4 years ago
Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a
allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2

Acceleration of the frog is 2874.33 m/s²

6 0
4 years ago
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