Ask question

Ask question

All categories

3 years ago

13

What is another way to describe the vector below?

2 answers:

nignag [31]3 years ago

8 0

Answer:

-40 feet to the left

Explanation:

opposite of 40 feet from right is -40 to left

IrinaVladis [17]3 years ago

8 0

Answer:

B. -40 feet to the left

You might be interested in

HELP ASAP!!!! Which of the following is an example of a hypothesis? A) What will happen if a sugar cube in boiling water? B) I t

Maurinko [17]

Answer:

C) If an ice cube is placed into a boiling water, then it will melt in less than 2 minutes.

Explanation:

7 0

3 years ago

An electron in a cathode-ray beam passes between 2.5cm long parallel-plate electrodes that are 6.0mm apart. A 2.1mT, 2.5-cm-wide

Dmitry_Shevchenko [17]

Answer:

(a). The speed of electron is $1.56\times10^{7}\ m/s$ .

(b). The radius of electron is $4.2\ cm$

Explanation:

Given that,

Length = 2.5 cm

Distance = 6.0 mm

Magnetic field = 2.1 T

Potential difference = 700 V

(a). We need to calculate the electron's speed

Using formula of speed

$v=\sqrt{\dfrac{2eV}{m}}$

Put the value into the formula

$v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times700}{9.1\times10^{-31}}}$

$v=15689290.81\ m/s$

$v=1.56\times10^{7}\ m/s$

(b). We need to calculate the radius of electron

Using formula of centripetal force

$\dfrac{mv^2}{r}=qvB$

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

Where,

m = mass of electron

v = speed of electron

r = radius

q = charge of electron

B = magnetic field

Put the value into the formula

$r=\dfrac{9.1\times10^{-31}\times1.56\times10^{7}}{1.6\times10^{-19}\times2.1\times10^{-3}}$

$r=0.042\ m$

$r=4.2\ cm$

Hence, (a). The speed of electron is $1.56\times10^{7}\ m/s$ .

(b). The radius of electron is 4.2 cm

8 0

3 years ago

An atom of carbon has a radius of 67.0 pm and the average orbital speed of the electrons in it is about 1.3 x 10⁶ m/s.

adoni [48]

Answer :

The least possible uncertainty in an electron's velocity is, $4.32\times 10^{5}m/s$

The percentage of the average speed is, 33 %

Explanation :

According to the Heisenberg's uncertainty principle,

$\Delta x\times \Delta p=\frac{h}{4\pi}$ ...........(1)

where,

$\Delta x$ = uncertainty in position

$\Delta p$ = uncertainty in momentum

h = Planck's constant

And as we know that the momentum is the product of mass and velocity of an object.

$p=m\times v$

or,

$\Delta p=m\times \Delta v$ .......(2)

Equating 1 and 2, we get:

$\Delta x\times m\times \Delta v=\frac{h}{4\pi}$

$\Delta v=\frac{h}{4\pi \Delta x\times m}$

Given:

m = mass of electron = $9.11\times 10^{-31}kg$

h = Planck's constant = $6.626\times 10^{-34}Js$

radius of atom = $67.0pm=67.0\times 10^{-12}m$ $(1pm=10^{-12}m)$

$\Delta x$ = diameter of atom = $2\times 67.0\times 10^{-12}m=134.0\times 10^{-12}m$

Now put all the given values in the above formula, we get:

$\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (134.0\times 10^{-12}m)\times (9.11\times 10^{-31}kg)}$

$\Delta v=4.32\times 10^{5}m/s$

The minimum uncertainty in an electron's velocity is, $4.32\times 10^{5}m/s$

Now we have to calculate the percentage of the average speed.

Percentage of average speed = $\frac{\text{Uncertainty in speed}}{\text{Average speed}}\times 100$

Uncertainty in speed = $4.32\times 10^{5}m/s$

Average speed = $1.3\times 10^{6}m/s$

Percentage of average speed = $\frac{4.32\times 10^{5}m/s}{1.3\times 10^{6}m/s}\times 100$

Percentage of average speed = 33.2 % ≈ 33 %

Thus, the percentage of the average speed is, 33 %

3 0

3 years ago

A block with a mass of 8.7 kg is dropped from rest from a height of 8.7 m, and remains at rest after hitting the ground. 1)If we

Harlamova29_29 [7]

To solve this problem we will apply the concepts related to gravitational potential energy.

This can be defined as the product between mass, gravity and body height.

Mathematically it can be expressed as

$\Delta P = mgh$

$\Delta P = (8.7)(9.8)(3)$

$\Delta P = 255.78J$

Therefore the change in the internal energy of the system is 255.78

7 0

3 years ago

An object is moving at a velocity of 23 m/s. It accelerates to a velocity of 85 m/s over a time of 8.3 s. What acceleration did

Bess [88]

$\frac{v \: final \: - v \: initial }{time} = \: a$
$\\ \frac{85 - 23}{8.3} = 7.467 \: \frac{m}{ {s}^{2} }$

7 0

3 years ago

Read 2 more answers