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3 years ago

Compare the KE of an 80 kg person walking at 4 m/s with that of the same person moving at 6 m/s.

1 answer:

7 0

The correct answer is:
KE at 4 m/s less than KE at 6 m/s

In fact, the formula for the kinetic energy is:
 $K= \frac{1}{2} mv^2$
where m is the mass of the person and v his speed.

When the person is moving at 4 m/s, his kinetic energy is:
 $K= \frac{1}{2} (80 kg)(4 m/s)^2=640 J$
instead, when he's moving at 6 m/s, his kinetic energy is:
$K= \frac{1}{2}(80 kg)(6 m/s)^2=1440 J$

So, the kinetic energy at 4 m/s is less than the kinetic energy at 6 m/s.

You might be interested in

Can a body having zero velocity move with uniform speed? Give an example.

GaryK [48]

Sure. Body can move with uniform speed, and having zero velocity, when velocity becomes zero due to change in direction over time t.

For Example. - An Object is moving with uniform speed in a circular path, then after one complete revolution, it's velocity is zero, but speed still remains uniform

Hope this helps!

3 0

3 years ago

11. A vector M is 15.0 cm long and makes an angle of 20° CCW from x axis and another vector N is 8.0 cm long and makes an angle

Alja [10]

Answer:

The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

Explanation:

To find the resultant vector, you first calculate x and y components of the two vectors M and N. The components of the vectors are calculated by using cos and sin function.

For M vector you obtain:

$M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}$

For N vector:

$N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}$

The resultant vector is the sum of the components of M and N:

$F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}$

The magnitude of the resultant vector is:

$|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm$

And the direction of the vector is:

$\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°$

hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

4 0

3 years ago

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

3 0

3 years ago

According to the Doppler affect what happens when a light surfaceMoves further away from an observer

Naily [24]

The correct answer is The electromagnetic waves appear more in red color.
Since red is at the low-frequency end of the visible spectrum, we say that light from a receding star is shifted toward red, or redshifted.

8 0

3 years ago

Read 2 more answers

What is linear momemtum

Genrish500 [490]

Answer:

Its momentum thats linear

Explanation:

from my secret analysis i would say this is really linear

7 0

2 years ago