Ask question

Ask question

All categories

2 years ago

8

An electron moves with a constant horizontal velocity of 3.0 × 106 m/s and no initial vertical velocity as it enters a deflector

inside a TV tube. The electron strikes the screen after traveling 11 cm horizontally and 34 cm vertically upward with no horizontal acceleration. What is the constant vertical acceleration provided by the deflector? (The effects of gravity can be ignored.)

1 answer:

Ghella [55]2 years ago

7 0

Answer:

a = 5.05 x 10¹⁴ m/s²

Explanation:

Consider the motion along the horizontal direction

$v_{x}$ = velocity along the horizontal direction = 3.0 x 10⁶ m/s

t = time of travel

X = horizontal distance traveled = 11 cm = 0.11 m

Time of travel can be given as

$t = \frac{X}{v_{x}}$

inserting the values

t = 0.11/(3.0 x 10⁶)

t = 3.67 x 10⁻⁸ sec

Consider the motion along the vertical direction

Y = vertical distance traveled = 34 cm = 0.34 m

a = acceleration = ?

t = time of travel = 3.67 x 10⁻⁸ sec

$v_{y}$ = initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

Y = $v_{y}$ t + (0.5) a t²

0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²

a = 5.05 x 10¹⁴ m/s²

You might be interested in

A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the

cupoosta [38]

Answer: deceleration of $1.904\ rad/s^2$

Explanation:

Given

Car is traveling at a speed of u=20 m/s

The diameter of the car is d=70 cm

It slows down to rest in 300 m

If the car rolls without slipping, then it must be experiencing pure rolling i.e. $a=\alpha \cdot r$

Using the equation of motion

$v^2-u^2=2as\\$

Insert $v=0,u=20,s=300$

$0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2$

Write acceleration as $a=\alpha \cdot r$

$-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2$

So, the car must be experiencing the deceleration of $1.904\ rad/s^2$ .

4 0

2 years ago

At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti

dalvyx [7]

Answer:

Part(a): The angular acceleration is $5.63~rad~s^{-2}$ .

Part(b): The angular displacement is $2629~rad$ .

Explanation:

Part(a):

If $\omega_{1},~\omega_{2}~and~\alpha$ be the initial angular speed, final angular speed and angular acceleration of the centrifuge respectively, then from rotational kinematic equation, we can write

$\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)$

where ' $t$ ' is the time taken by the centrifuge to increase its angular speed.

Given, $\omega_{i} = 250~rad~s^{-1}$ , $\omega_{f} = 750~rad~s^{-1}$ and $t = 9.5~s$ . From equation ( $I$ ), the angular acceleration is given by

$\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}$

Part(b):

Also the angular displacement ( $\Delta \theta$ ) can be written as

$&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad$

8 0

2 years ago

Two cyclists start on a race between points A and D on two different routes. Cyclist X takes the route passing through the equid

ollegr [7]

The displacement is the shortest distance between two points, which is 546.41. The displacement for both is 546.41 meters

Average velocity of X = (200 + 200 + 200) / 30
Average velocity of X = 20 m/s

Average velocity of Y = 546.41 / 30 = 18.2 m/s

8 0

2 years ago

Read 2 more answers

the distance between the sun and earth is about 1.5X10^11 m. express this distance with an SI prefix and kilometers

Angelina_Jolie [31]

First, we write the SI prefixed. The SI unit for distance is meters.

Kilo = 10³
Mega = 10⁶
Giga = 10⁹
Terra = 10¹²

Because our value has ten to the power of 11, we will use the closest and lowest power prefix, which is giga.

1.5 x 10¹¹ / 10⁹
= 1.5 x 10² Gm or 150 Gm

Writing in kilometers, we simply repeat the procedure except we divide by 10³ this time.

1.5 x 10¹¹ / 10³
= 1.5 x 10⁸ km

5 0

2 years ago

Tara is giving a speech on the how to get a U.S. passport for the first time. During her speech, she tells the audience the requ

Alchen [17]

Answer:

a. 1. she informed the audience when she was concluding

2. She simply outlined the steps tying her conclusion to the topic

3. her conclusion was concise and simple.

b. 13.1 grams

Explanation:

a. signalling the end of a speech prepares the mind of the audience and leaves them satisfied, and Kara correctly signaled the end of her speech when she said "In conclusion".

Also, Kara summarized the ideas in her speech by simply outlining the requirements needed to get a U.S passport.

Her conclusion was concise and simple which is very important especially for lengthy speeches because the audience at this point must be getting tired and distracted.

b. Amount of cholesterol in 100 mL of blood = 248 mg

Total Volume of blood = 5.3 Liters.

the question is asking us to find the total grams of cholesterol in the individual's body, that is the total grams of cholesterol in 5.3 Liters of the individuals blood.

To solve, let us use the following steps:

1. unify the units of measuring volume.

since the first volume is given in Milliliters, I will convert the second volume (5.3 Liters) to milliliters too. To do this, note that:

1 liter = 1000 Milliliter (mL)

∴ 5.3 Liters = 5.3 × 1000 = 5300 Milliliters (mL)

2. Calculate the amount of cholesterol in 5300 mL.

100 mL = 248 mg of cholesterol

∴ 1 mL = $\frac{248}{100\\}$ = 2.48 mg

∴ 5300 mL = 2.48 × 5300 = 13,144 mg of cholesterol.

3. Convert the amount of cholesterol from Milligram to gram

Since the question asked for the gram of cholesterol, not the milligram of cholesterol, converting from milligram to gram, we use the following steps:

1000 mg = 1 g

∴ 1 mg = $\frac{1}{1000}$ g

∴ 13144 mg = 13144 × $\frac{1}{1000}$ = $\frac{13144}{1000}$ = 13.144 ≅ 13.1 g of cholesterol ( 1 decimal place).

I suggest you go and study the conversion relationships between the sub units milli, micro, centi and deci.

7 0

2 years ago