The slope of the graph

If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve o

Scilla [17]

Answer:

Maximum acceleration will be equal to $3.43m/sec^2$

Explanation:

We have given coefficient of kinetic friction $\mu _k=0.7$

And coefficient of static friction $\mu _s=1$

Acceleration due to gravity $g=9.8m/sec^2$

When truck moves maximum force will be equal to $F=\mu _kmg$

It is given that half of the weight is supported by its drive wheels

So force required $=\frac{\mu _kmg}{2}$

From newtons law maximum acceleration will be equal to $a=\frac{\frac{\mu _kmg}{2}}{m}=\frac{\mu _kg}{2}=\frac{0.7\times 9.8}{2}=3.43m/sec^2$

8 0

3 years ago

A stationary boat in the ocean is experiencing waves from a storm. The waves move at 59 km/h and have a wavelength of 145 m . Th

krek1111 [17]

Answer:

The time elapses until the boat is first at the trough of a wave is 4.46 seconds.

Explanation:

Speed of the wave, v = 59 km/h = 16.38 m/s

Wavelength of the wave, $\lambda=145\ m$

If f is the frequency of the wave. The frequency of a wave is given by :

$v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{16.38\ m/s}{145\ m}\\\\f=0.112\ Hz$

The time period of the wave is given by :

$T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.112\ Hz}\\\\T=8.92\ s$

We need to find the time elapses until the boat is first at the trough of a wave. So, the time will be half of the time period of the wave.

$T=\dfrac{8.92}{2}\\\\T=4.46\ s$

Hence, this is the required solution.

5 0

3 years ago

A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times

kodGreya [7K]

Answer:

The sound intensity of train is 1000 times greater than that of the library.

Explanation:

We have expression for sound intensity level,

$L=10log_{10}\left ( \frac{I}{I_0}\right )$

A train whistle has a sound intensity level of 70 dB

We have

$70=10log_{10}\left ( \frac{I_1}{I_0}\right )$

A library has a sound intensity level of about 40 dB

We also have

$40=10log_{10}\left ( \frac{I_2}{I_0}\right )$

Dividing both equations

$\frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000$