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Montano1993 [528]
2 years ago
13

A vector A~ has a negative x component 2.89 units in length and a positive y component 3.17 units in length. The vector A~ can b

e written in unit vector notation: a1ˆı + a2ˆ. Find the value of (a) a1 (units) (b) a2 (units) (c) What is the magnitude of A~ (units)? (d) What is the direction of A~ (degrees)? 1
Physics
1 answer:
Fantom [35]2 years ago
5 0

Answer:

a1 = -2.89 units

a2 = 3.17 units

magnitude = 4.2896 units

direction = 132.35 degrees

Explanation:

We can write the vector in component form as: <-2.89, 3.17>

That means that its i and j components are:

a1 = -2.89 units

a2 = 3.17 units

The magnitude of the vector is given by:

\sqrt{(-2.89)^2 +3.17^2} \approx 4.2896

The vectors direction can be found from the tangent function, ad noticing that the vector must reside on the Second Quadrant:

arctan(-3.17/2.89) + 180^o\approx 132.35^o

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What types of changes in motion cause acceleration?
xxTIMURxx [149]

Answer:

Change in velocity and direction over a specific period of time.

Explanation:

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate the acceleration of an object.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

a = \frac{v  -  u}{t}

Where,

a is acceleration measured in ms^{-2}

v and u is final and initial velocity respectively, measured in ms^{-1}

t is time measured in seconds.

Hence, the types of changes in motion that cause acceleration is a change in velocity and direction over a specific period of time.

6 0
3 years ago
Which of the following has the largest momentum?
san4es73 [151]

Momentum = mass x velocity

Thus Option A is the correct answer

Momentum (dog)  = 10 kg x (0.447 x 30) m/s

= 134.1 Kg m/s

Momentum ( bullet)  = 0.02 kg x (0.447 x 800) m/s

= 7.152 Kg m/s

Momentum ( truck) = 0, as v = 0

tightrope has both low mass  and low speed, thus its momentum will be low


4 0
3 years ago
Find the volume of a rectangular prism that is 8m long, 4m wide, and 300cm high
konstantin123 [22]
  • L=8m
  • B=4m
  • H=300cm=3m

\\ \bull\tt\longmapsto Volume=LBH

\\ \bull\tt\longmapsto Volume=8(4)(3)

\\ \bull\tt\longmapsto Volume=96m^3

6 0
2 years ago
Read 2 more answers
The drawing shows a large cube (mass = 28.6 kg) being accelerated across a horizontal frictionless surface by a horizontal force
MrRissso [65]

Answer:

P= 454.11 N

Explanation:

Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).

P= (M+m)*a\\a = \frac{P}{28.6 +4.3}\\a = \frac{P}{32.9}

The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:

F = m * a*\mu \\F= 4.3*0.710*\frac{P}{32.9}\\F=3.053*\frac{P}{32.9}

In order for the small cube to not slide down, the friction force must equal the weight of the small cube:

3.053*\frac{P}{32.9} = 4.3 * g\\\\P = \frac{4.3*9.8*32.9}{3.053} \\P= 454.11 N

The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N

8 0
3 years ago
You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation
Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, \omega = 25.13 rad/s

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = \frac{1}{T}

f = \frac{1}{0.25}

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

\omega = 2\pi \times f

\omega = 2\pi \times 4

\omega = 25.13 rad/s

5 0
3 years ago
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