1)
The acceleration of the car is the rate of change of velocity of the car; it can be calculated as:
[tex]a=\frac{v-u}{t}[/tex]
where
u is the initial velocity
v is the final velocity
t is the time taken for the velocity of the car to change from u to v
In this problem, for this car we have:
u = 37.1 m/s
v = 29.8 m/s
t = 3 s
So, the acceleration is:
[tex]a=\frac{29.8-37.1}{3}=-2.43 m/s^2[/tex]
2)
The work done in lifting the box is equal to the potential energy transferred to the box during the process; it is given by:
[tex]W=Fd[/tex]
where
F is the force applied
d is the displacement of the box
Here we have:
F = 87.3 N is the force applied
d = 2.04 m is the displacement of the box
So, the work done to lift the box is:
[tex]W=(87.3)(2.04)=178.1 J[/tex]
3)
The power is the rate of work done per unit time. It is calculated as:
[tex]P=\frac{W}{t}[/tex]
where
W is the work done
t is the time taken to do the work
For the child in this problem, we have:
W = 1250 J is the work done by the child running up the stairs
P = 267 W is the power used
Therefore, re-arranging the equation, we find the time taken:
[tex]t=\frac{W}{P}=\frac{1250}{267}=4.68 s[/tex]
4)
The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by
[tex]KE=\frac{1}{2}mv^2[/tex]
where
m is the mass of the object
v is its speed
For the rabbit in this problem, we have:
m = 8.642 kg is the mass of the rabbit
KE = 125.6 is its kinetic energy
Solving the formula for v, we find the speed of the rabbit:
[tex]v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(125.6)}{8.642}}=5.4 m/s[/tex]
5)
The efficiency of a machine is the ratio between the energy produced in output by the machine and the work done in input. Mathematically, it is given by
[tex]\eta = \frac{E_{out}}{W_{in}}\cdot 100[/tex]
where
[tex]E_{out}[/tex] is the energy in output
[tex]W_{in}[/tex] is the work in input
For the machine in this problem,
[tex]W_{in}=120 J[/tex] is the work in input
[tex]E_{out}=93 J[/tex] is the energy in output
Therefore, the efficiency of this machine is:
[tex]\eta=\frac{93}{120}\cdot 100=77.5\%[/tex]
6)
During a collision, the total momentum of the system is always conserved before and after the collision. So we can write:
[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 =(m_1+m_2)v[/tex]
where
[tex]m_1=212 kg[/tex] is the mass of the first car
[tex]u_1=8.00 m/s[/tex] is the initial velocity of the first car
[tex]m_2=196 kg[/tex] is the mass of the 2nd car
[tex]u_2=6.75 m/s[/tex] is the initial velocity of the 2nd car
[tex]v[/tex] is the final velocity of the two cars stuck together (after the collision, they move together)
Solving the equation for v, we find:
[tex]v=\frac{m_1 u_1 +m_2 u_2}{m_1 +m_2}=\frac{(212)(8.00)+(196)(6.75)}{212+196}=7.40 m/s[/tex]
7)
The relationship between speed, frequency and wavelength of a wave is given by the wave equation:
[tex]v=f\lambda[/tex]
where
v is the speed of the wave
f is the frequency of the wave
[tex]\lambda[/tex] is the wavelength
For the wave in the string in this problem we have:
[tex]\lambda=0.23 m[/tex] (wavelength)
f = 12 Hz (frequency)
So, the speed of the wave is:
[tex]v=(12)(0.23)=2.76 m/s[/tex]
8)
The relationship between frequency and wavelength for an electromagnetic wave is given by
[tex]c=f\lambda[/tex]
where:
c is the speed of light in a vacuum
f is the frequency of the wave
[tex]\lambda[/tex] is the wavelength of the wave
For the blue light in this problem, we have
[tex]f=6.2\cdot 10^{14}Hz[/tex] (frequency)
while the speed of light is
[tex]c=3.0\cdot 10^8 m/s[/tex]
So, the wavelength of blue light is:
[tex]\lambda=\frac{c}{f}=\frac{3.0\cdot 10^8}{6.2\cdot 10^{14}}=4.8\cdot 10^{-7} m[/tex]
9)
The sound wave in this problem travels with uniform motion (=constant velocity), therefore we can use the following equation:
[tex]d=vt[/tex]
where
d is the distance covered by the wave
v is the speed of the wave
t is the time elapsed
In this problem:
v = 343 m/s is the speed of the sound wave
t = 0.287 s is the time elapsed
So, the distance covered by the wave is
[tex]d=(343)(0.287)=98.4 m[/tex]
