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kvv77 [185]
3 years ago
6

What do you think a force diagram might look like for a hit that causes a concussion? Draw a diagram. Consider only the person’s

head in your diagram. Be sure to draw arrows to the proper scale. In a text box, state whether your diagram shows balanced or unbalanced forces.
if someone is answering this don't put a link.
Physics
1 answer:
Alexandra [31]3 years ago
5 0

Answer:

\huge\color{green}\boxed{\colorbox{red}{Don't know}}

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Why does the chord of an electric heater not glow while the heating element does​
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A can of beans that has mass M is launched by a spring-powered device from level ground. The can is launched at an angle of α0 a
scZoUnD [109]

Hi there!

A.

Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.

Thus, the time to its highest point:
T_h = \frac{T}{2}

Now, we can determine the velocity at which the can was launched at using the following equation:
v_f = v_i + at

In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.

Therefore:
0 = v_y + at\\\\0 = vsin\theta - g\frac{T}{2}

***vsinθ is the vertical component of the velocity.

Solve for 'v':
vsin(\alpha_0) = g\frac{T}{2}\\\\v = \frac{gT}{2sin(\alpha_0)}

Now, recall that:
W = \Delta KE = \frac{1}{2}m(\Delta v)^2

Plug in the expression for velocity:
W = \frac{1}{2}M (\frac{gT}{2sin(\alpha_0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{8sin^2(\alpha _0)}}

B.

We can use the same process as above, where T' = 2T and Th = T.

v = \frac{gT}{sin(\alpha _0)} }\\\\W = \frac{1}{2}M(\frac{gT}{sin(\alpha _0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{2sin^2(\alpha _0)}}

C.

The work done in part B is 4 times greater than the work done in part A.

\boxed{\frac{W_B}{W_A} = \frac{4}{1} = 4.0}

4 0
2 years ago
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