Ask question

Ask question

All categories

3 years ago

15

On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max

pulls on the sled with 12 n of force, directed at an angle of 15° above the ground. when he comes to a recently plowed section of road, he continues to pull the sled with the same force across the road while the road exerts a frictional force of 4 n on sled. what is the net work done on the sled while max pulls it 5 m across the road?

1 answer:

sesenic [268]3 years ago

7 0

Work done by a given force is given by

$W = F.d$

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = $W_1 = Fdcos\theta$

$W_1 = 12*5cos15$

$W_1 = 57.96 J$

Now similarly work done by frictional force

$W_2 = Fdcos\theta$

$W_2 = 4*5cos180$

$W_2= -20 J$

Now total work done on sled

$W_{net}= W_1 + W2$

$W_{net} = 57.96 - 20 = 37.96 J$

You might be interested in

Infrared waves from the sun are what make our skin feel warm on a sunny day. If an infrared wave has a frequency of 3.0 x 1012 H

djyliett [7]

Answer:

The wavelength of the infrared wave is <u>0.0001 m</u>.

Explanation:

Given:

Frequency of an infrared wave is, $f=3.0\times 10^{12}\ Hz$

We know that, infrared waves are electromagnetic waves. All electromagnetic waves travel with the same speed and their magnitude is equal to the speed of light in air.

So, speed of infrared waves coming from the Sun travels with the speed of light and thus its magnitude is given as:

$v=c=3.0\times 10^8\ m/s$

Where, 'v' is the speed of infrared waves and 'c' is the speed of light.

Now, we have a formula for the speed of any wave and is given as:

$v=f\lambda$

Where, $\lambda \to \textrm{Wavelength of infrared wave}$

Now, rewriting the above formula in terms of wavelength, $\lambda$ , we get:

$\lambda=\dfrac{v}{f}$

Now, plug in $3.0\times 10^8$ for 'v', $3.0\times 10^{12}$ for 'f' and solve for $\lambda$ . This gives,

$\lambda=\frac{3.0\times 10^8}{3.0\times 10^{12}}\\\\\lambda=0.0001\ m$

Therefore, the wavelength of the infrared wave is 0.0001 m.

5 0

3 years ago

Shelly rolls ball A in the positive x direction with a velocity of 7.5 meters/second. It hits stationary ball B and they undergo

blagie [28]

<span>If Shelly rolls ball A in the positive x direction with a velocity of 7.5 meters/second, and It hits stationary ball B and they undergo elastic collision, thus the two balls have different masses, then the following statement which is true is the statement that stated that there was no y-momentum initially.</span>

7 0

3 years ago

A whistling sound that is audible when a hearing aid is held in the hand with the power on and the volume high indicates that th

zzz [600]

The whistling sound from the hearing aids represents that your hearing aids is working perfectly ad is known as the "feedback". So, the given statement is true.

Answer: Option A

<u>Explanation:</u>

It's often sounds irritating when a hearing aids of your grandpa or Grandma whistles. especially, when they put them out of their ears. Actually, this feedback sound from hearing aids occur when the sounds from the outer side bounces back to the microphone of the hearing aids.

The sound bounces back when it doesn't gets inside of your ear canal so that one can hear the sound through the hearing aid. When the sounds bounces back in the hearing aid, it get re-amplified and thus we hear the whistle sound which is known as the feedback of the device.

It's not always the feedback sound though. Sometimes the device whistles when it has some mechanical defect or when one hugs the other one or water gets inside and damaged the whole system.

7 0

3 years ago

Physics B 2020 Unit 3 Test

weqwewe [10]

Answer:

1)

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

$F=qvB sin \theta$

where here:

For the proton in this problem:

$q=1.602\cdot 10^{-19}C$ is the charge of the proton

v = 300 m/s is the speed of the proton

B = 19 T is the magnetic field

$\theta=65^{\circ}$ is the angle between the directions of v and B

So the force is

$F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N$

2)

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

- The density of field lines is higher near the Poles

- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

3)

The right hand rule gives the direction of the force experienced by a charged particle moving in a magnetic field.

It can be applied as follows:

- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)

In this problem:

- Direction of motion = to the right (index finger)

- Direction of field = downward (middle finger)

- Direction of force = into the screen (thumb)

4)

The radius of a particle moving in a magnetic field is given by:

$r=\frac{mv}{qB}$

where here we have:

$m=6.64\cdot 10^{-22} kg$ is the mass of the alpha particle

$v=2155 m/s$ is the speed of the alpha particle

$q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C$ is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

$r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m$

5)

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

$f=\frac{qB}{2\pi m}$

where here:

$q=1.602\cdot 10^{-19}C$ is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

$m=9.31\cdot 10^{-31} kg$ is the mass of the electron

Substituting, we find:

$f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz$

6)

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

1- A uniform motion in a certain direction

2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

Therefore in this case, when the particle gains the opposite charge, the circular motion 2) changes sign, so the path will remains helical, but it reverses direction.

7)

The electromotive force induced in a conducting loop due to electromagnetic induction is given by Faraday-Newmann-Lenz:

$\epsilon=-\frac{N\Delta \Phi}{\Delta t}$

where

N is the number of turns in the loop

$\Delta \Phi$ is the change in magnetic flux through the loop

$\Delta t$ is the time elapsed

From the formula, we see that the emf is induced in the loop (and so, a current is also induced) only if $\Delta \Phi \neq 0$ , which means only if there is a change in magnetic flux through the loop: this occurs if the magnetic field is changing, or if the area of the loop is changing, or if the angle between the loop and the field is changing.

8)

The flux is calculated as

$\Phi = BA sin \theta$

where

B = 5.5 T is the strength of the magnetic field

A is the area of the coil

$\theta=18^{\circ}$ is the angle between the direction of the field and the plane of the loop

Here the loop is rectangular with lenght 15 cm and width 8 cm, so the area is

$A=(0.15 m)(0.08 m)=0.012 m^2$

So the flux is

$\Phi = (5.5)(0.012)(sin 18^{\circ})=0.021 Wb$

See the last 7 answers in the attached document.

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>

5 0

3 years ago

The rate at which an object’s velocity changes is called its

topjm [15]

I believe it is call “Acceleration”

6 0

3 years ago

Read 2 more answers