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3 years ago

10

What is the value of sin square 60

1 answer:

maksim [4K]3 years ago

5 0

Answer:

$sin {}^{2} 60 = \frac{3}{4}$

Explanation:

sin^2 60° = ( \|3 / 2 ) ^2 = 3 / 4.

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Why does the sun appear to rise in the east and set in the west every day?

anygoal [31]

Answer: even heating

Explanation:

the earth rotates

4 0

3 years ago

Read 2 more answers

Determine the magnitude and direction of the resultant force of the following free body diagram.

Papessa [141]

Answer:

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

Explanation:

First, we must calculate the resultant force ( $\vec F$ ), in newtons, by vectorial sum:

$\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j}$ (1)

$\vec F = 182.843\,\hat{i} + 356.048\,\hat{j}$

Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:

$\|\vec F\| = \sqrt{(482.843\,N)^{2}+(356.048\,N)^{2}}$

$\|\vec F\| \approx 599.923\,N$

Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:

$\theta = \tan^{-1}\left(\frac{356.048\,N}{482.843\,N} \right)$

Where $\theta$ is the direction of the resultant force, in sexagesimal degrees.

$\theta \approx 36.405^{\circ}$

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

4 0

3 years ago

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

3 years ago

What type of material does not transfer heat well?

scZoUnD [109]

Insulators such as wood

6 0

3 years ago

Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the

joja [24]

As it is given that the air bag deploy in time

$t = 10 ms = 0.010 s$

total distance moved by the front face of the bag

$d = 20 cm = 0.20 m$

Now we will use kinematics to find the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$0.20 = 0 + \frac{1}{2}a*0.010^2$

$0.20 = 5 * 10^{-5}* a$

$a = 4000 m/s^2$

now as we know that

$g = 10 m/s^2$

so we have

$a = 400g$

so the acceleration is 400g for the front surface of balloon

3 0

3 years ago