What are some examples of aerobic exercise?

1) Little Timmy wants to measure how tall his house is. He doesn’t have a tape measure but does have a stopwatch. He recruits Bi

kifflom [539]

1) 13.7 m

The motion of the rock is a free fall, with constant acceleration g=9.8 m/s^2 towards the ground, so the total distance it covers is given by the SUVAT equation:

$S=\frac{1}{2}gt^2$

where S is the height of the house, and t is the time the rock takes to reach the ground. Substituting t=1.67 s, we find:

$S=\frac{1}{2}(9.8 m/s^2)(1.67 s)^2=13.7 m$

2) 105.5 m

The motion of the stuffed chicken is a projectile motion, with a uniform horizontal motion (with constant velocity of v=36.0 m/s) and a vertical accelerated motion (with constant acceleration of g=9.8 m/s^2).

First of all, we can find the total time of the ball by considering the vertical motion only. We know the vertical distance covered, S=42.2 m, so the time of the fall is

$S=\frac{1}{2}gt^2\\t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(42.2 m)}{9.8 m/s^2}}=2.93 s$

And now we can consider the horizontal motion to find the horizontal distance covered by the stuffed chicken:

$d=vt=(36.0 m/s)(2.93 s)=105.5 m$

3) 49.4 m

Again, the motion of the ball is a projectile motion, with a horizontal motion and a vertical motion.

The range of a projectile launched from the ground can be found by using the formula:

$d=\frac{v^2}{g} sin 2 \theta$

where, in this case:

v = 22.0 m/s is the initial velocity

$\theta=45^{\circ}$

Substituting into the formula, we find

$d=\frac{(22.0 m/s)^2}{9.8 m/s^2}(sin (2\cdot 45^{\circ}))=49.4 m$

4) 9.6 m/s^2

The frictional force acting on the monkey is given by:

$F_f = \mu mg=(0.16)(31.0 kg)(9.8 m/s^2)=48.6 N$

where $\mu$ is the coefficient of friction and m is the mass of the monkey.

We have two forces acting on the monkey: the push of F=345 N and the frictional force acting in the opposite direction. According to Newton's second law, the net force will be equal to the product between the monkey's mass and its acceleration, so we can find the acceleration:

$F-F_f=ma\\a=\frac{F-F_f}{m}=\frac{345 N-48.6 N}{31.0 kg}=9.6 m/s^2$

5) 462.3 N

The horizontal component of the pushing force is:

$F_x = F cos \theta = (648 N)(cos 25^{\circ})=587.3 N$

The frictional force, acting in the opposite direction, is

$F_f = \mu mg=(0.17)(75.0 kg)(9.8 m/s^2)=125.0 N$

where $\mu$ is the coefficient of friction and m is the mass of the box.

The net force on the box is therefore given by the net force on the horizontal direction:

$F_{net}=F_x -F_f=587.3 N -125.0 N=462.3 N$

6) 89.5 N

First of all we need to calculate the total weight of the table and the items above it.

The weight of the table is:

$W=mg=(25.0 kg)(9.8 m/s^2)=245 N$

So the total weight of the table and the items is

$W=245 N+63 N+12 N+44 N+24 N+9N+10N=407 N$

The force needed to get the table moving must be at least equal to the frictional force, which is equal to the product between the coefficient of friction and the weight of the all stuff:

$F=F_f = \mu W=(0.22)(407 N)=89.5 N$

7 0

2 years ago

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 2.4

NeX [460]

When person is observing destructive interference at 0.20 m distance from the equidistant position then we can say that path difference must be equal to half of the wavelength

now we will have

$\frac{\lambda}{2} = \frac{yd}{L}$