Dirt can increase friction and cause the worker to slip a)-True b)- false

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

3 years ago

A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next

zalisa [80]

Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

$V=1.4*10^5m/s$

Explanation:

From the question we are told that:

Electric field $B=1.5*10N/C$

Distance $d=2 x 10^{-3}$

At negative plate

Generally the equation for Velocity is mathematically given by

$V^2=2as$

Therefore

$V^2=\frac{2*e_0E*d}{m}$

$V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}$

$V=\sqrt{19.2*10^9}$

$V=1.4*10^5m/s$

5 0

3 years ago

Weightlessness is experienced by an astronaut in space. This means that the astronaut's muscles have to be stronger to move his

just olya [345]

The answer is false. The speed of the astronaut cancels out the force of gravity, causing a 'stationary freefall'. While under these effects, it is not required for an astronaut to 'strengthen' his body.

4 0

3 years ago

Read 2 more answers

The manufacturer of a 6V dry-cell flashlight battery says that the battery will deliver 15 mA for 60 continuous hours. During th

Vaselesa [24]

Answer:

16200 J

Explanation:

t = Time the battery is on = 60 hours

I = Current = $15\times 10^{-3}\ A$

Average voltage

$V=\dfrac{6+4}{2}=5\ V$

Energy is given by

$E=V\times I\times t$

$\\\Rightarrow E=5\times 15\times 10^{-3}\times 60\times 3600$

$\\\Rightarrow E=16200\ J$

The energy delivered in the given time is 16200 J

6 0

3 years ago

Many television sets show 25 images, called 'frames, each second. What is the time interval between one

Firlakuza [10]

Answer:

Given,

Frame rate = 25 frames per second

To find,

Time interval between one frame and the next.

Solution,

We can simply solve this numerical problem by using the following process.

Now,

Number of frames = 25

Total time taken to display the given number of frames (ie. 25 frames) = 1 second

To calculate the time interval between one frame and next, we need to divide the time taken to display total number of frames by total number of frames.

So,

Time interval between one frame and next :

= Time taken to display total number of frames / Total frames

= 1/25

= 0.04 second

Hence, time interval between one frame and next is 0.04 second.

8 0

2 years ago