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2 years ago

Find the lengths of each of the following vectors

Physics

1 answer:

Irina18 [472]2 years ago

3 0

Answer:

Explanation:

Generally, length of vector means the magnitude of the vector.

So, given a vector

R = a•i + b•j + c•k

Then, it magnitude can be caused using

|R|= √(a²+b²+c²)

So, applying this to each of the vector given.

(a) 2i + 4j + 3k

The length is

L = √(2²+4²+3²)

L = √(4+16+9)

L = √29

L = 5.385 unit

(b) 5i − 2j + k

Note that k means 1k

The length is

L = √(5²+(-2)²+1²)

Note that, -×- = +

L = √(25+4+1)

L = √30

L = 5.477 unit

Note that, since there is no component j implies that j component is 0

L = 2i + 0j - 1k

The length is

L = √(2²+0²+(-1)²)

L = √(4+0+1)

L = √5

L = 2.236 unit

(d) 5i

Same as above no is j-component and k-component

L = 5i + 0j + 0k

The length is

L = √(5²+0²+0²)

L = √(25+0+0)

L = √25

L = 5 unit

(e) 3i − 2j − k

The length is

L = √(3²+(-2)²+(-1)²)

L = √(9+4+1)

L = √14

L = 3.742 unit

(f) i + j + k

The length is

L = √(1²+1²+1²)

L = √(1+1+1)

L = √3

L = 1.7321 unit

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Answer:

The angular velocity of Ball A will be greater than the angular velocity of Ball B when they reach the top of the hill.

Explanation:

Angular velocity can be defined as how fast an object rotates relative to a given point or frame of reference.

The question said the hill encountered by Ball A is frictionless, so Ball A will continue to rotate at the same rate it started with even when it reached the top of the hill.

Ball B on the other hand rolls without slipping over its hill, i.e there's friction to slow down its rotational motion which thus reduces how fast Ball B will rotate at the top of the hill

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2 years ago

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente

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Answer:

$I=68.31\times 10^{-6}\ A$

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

$\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr$

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

By putting the values

$I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A$

$I=68.31\times 10^{-6}\ A$

7 0

2 years ago

Read 2 more answers

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

2 years ago

3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-di

irinina [24]

Answer:

The initial velocity of the water from the tank is 5.42 m/s

Explanation:

By applying Bernoulli equation between point 1 and 2

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L$

At the point 1

P₁=0 ( Gauge pressure)

V₁= 0 m/s

Z₁=3 m

At point 2

P₂=0 ( Gauge pressure)

Z₂= 0 m/s

$h_L=1.5\ m$

Now by putting the values

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L$

$Z_1-h_L=\dfrac{V_2^2}{2g}$

$3-1.5=\dfrac{V_2^2}{2\times 9.81}$

$V_2=\sqrt{2\times 1.5\times 9.81}\ m/s$

V₂= 5.42 m/s

The initial velocity of the water from the tank is 5.42 m/s

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3 years ago

The late news reports the story of a shooting in the city. Investigators think that they have recovered the weapon and they run

Gemiola [76]

Answer:

50000 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of bullet = 0.050 kg

velocity (v) = 400 m/s

Distance (s) = 0.080 m

Force (F) =?

Next, we shall determine the acceleration of the bullet. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 400 m/s

Distance (s) = 0.080 m

Acceleration (a) =?

v² = u² + 2as

400² = 0 + (2 × a × 0.08)

160000 = 0 + 0.16a

160000 = 0.16a

Divide both side by 0.16

a = 160000 / 0.16

a = 1×10⁶ m/s²

Finally, we shall determine the force exerted by the bullet on the target. This can be obtained as follow:

Mass (m) of bullet = 0.050 kg

Acceleration (a) of bullet = 1×10⁶ m/s²

Force (F) =?

F = ma

F = 0.050 × 1×10⁶

F = 50000 N

Thus, the bullet exerted a force of 50000 N on the target.

7 0

2 years ago