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jekas [21]
3 years ago
9

You drive your car 3 hrs at an average of 130 km how far did you go

Physics
1 answer:
Bingel [31]3 years ago
7 0
You have traveled 390 km. In order to find your answer all you have to do is multiply your speed which is 130 km/h by 3 hrs. That's my friend is how you get your answer! Have a great day!
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So what would the answer for D be?
Mademuasel [1]
For what, exactly? XD
8 0
3 years ago
Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m
Natali [406]
<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

7 0
2 years ago
A straight line is drawn on the surface of a 14-cm-radius turntable from the center to the perimeter. A bug crawls along this li
sleet_krkn [62]

Answer:

v = \left[\begin{array}{c}0.66&0\end{array}\right]m/s

Explanation:

The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:

\overrightarrow{r}=vtcos(\omega t)\hat{x}+vtsin(\omega t)\hat{y}

The velocity vector v is the first derivative of the position vector r with respect to time:

\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}

The given values are:

t=\frac{x}{v}=\frac{14}{3.8}=3.7 s

\omega=\frac{45\times2\pi}{60s}=4.7\frac{1}{s}

8 0
3 years ago
Which bond(s) shown are double bonds?
Nikitich [7]

Answer:

I'm pretty sure its the 2nd one.

3 0
3 years ago
A jogger runs at a speed of 3 m/s. How far does he run in 120 seconds?<br>​
Licemer1 [7]

Answer: *360 mph*

Explanation:

I am pretty sure that it is 360 mph

3 times 120 = 360

8 0
3 years ago
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