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3 years ago

Answers to question 2,3,4

Physics

2 answers:

tino4ka555 [31]3 years ago

8 0

2 is B, and 3 is c and 4 is c I hope this helps because I’ve had this before

Sati [7]3 years ago

5 0

Answer:

OMG

these are sooooooo hard

Explanation:

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Which describes the average velocity of an ant traveling at a constant speed

yanalaym [24]

Answer: A. The total displacement divided by the time and C. The slope of the ant's displacement vs. time graph.

Explanation:

Hi! The question seems incomplete, but I found the options on the internt:

A. The total displacement divided by the time.

B. The slope of the ant's acceleration vs. time graph.

C. The slope of the ant's displacement vs. time graph.

D. The average acceleration divided by the time.

Now, since we know the ant is travelling at a constant speed, its average velocity $V$ will be expressed by the following equation:

$V=\frac{d}{t}$

Where:

$d$ is the ant's total displacement

$t$ is the time it took to the ant to travel to the kitchen

Hence one of the correct options is: A. The total displacement divided by the time

On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is: C. The slope of the ant's displacement vs. time graph.

3 0

3 years ago

A 50.0-kg skater begins a spin with an angular speed of 4.0 rad/s. By changing the position of her arms, the skater decreases he

Pani-rosa [81]

Answer:

The Skater's final angular speed =8rad/s

Explanation:

Let initial momentum of inertia be I1

Let final momentum of inertia be I2

But I2= I1/2

It is decreased by one-half

There is no external torque on the system, therefore

I1w1=I2w2

I1×4=(I1/2)×w2

Dividing through by I1 on both sides leaves you with:

4=w2/2

Cross multiply

W2=4×2=8rad/s

4 0

4 years ago

What is the frequency of an X-ray with wavelength 0.13 nm ? Assume that the wave travels in free space. Express your answer to t

dem82 [27]

Answer:

Frequency, $f=2.30\times 10^{18}\ Hz$

Explanation:

Given that,

The wavelength of the x-rays, $\lambda=0.13\ nm=0.13\times 10^{-9}\ m$

We need to find the frequency of an x-ray. All electromagnetic wave travel with a speed of light. It is given by the formula as :

$c=f\lambda$

f is the frequency

$f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.13\times 10^{-9}}\\\\f=2.30\times 10^{18}\ Hz$

So, the frequency of an x-ray is $2.30\times 10^{18}\ Hz$ . Hence, this is the required solution.

5 0

3 years ago

I will gib brainlyest or whatever.

astraxan [27]

Answer:

Range of the projectile: approximately $1.06 \times 10^{3}\; {\rm m}$ .

Maximum height of the projectile: approximately $80\; {\rm m}$ (approximately $45.0\; {\rm m}$ above the top of the cliff.)

The projectile was in the air for approximately $7.07\; {\rm s}$ .

The speed of the projectile would be approximately $155\; {\rm m \cdot s^{-1}}$ right before landing.

(Assumptions: drag is negligible, and that $g = 9.81\; {\rm m\cdot s^{-1}}$ .)

Explanation:

If drag is negligible, the vertical acceleration of this projectile will be constantly $a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}$ . The SUVAT equations will apply.

Let $\theta$ denote the initial angle of elevation of this projectile.

Initial velocity of the projectile:

vertical component: $u_{y} = u\, \sin(\theta) = 153\, \sin(11.2^{\circ}) \approx 29.71786\; {\rm m\cdot s^{-1}}$
horizontal component: $u_{x} = u\, \cos(\theta) = 153\, \cos(11.2^{\circ}) \approx 150.086\; {\rm m\cdot s^{-1}}$ .

Final vertical displacement of the projectile: $x_{y} = (-35)\; {\rm m}$ (the projectile landed $35\: {\rm m}$ below the top of the cliff.)

Apply the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to find the final vertical velocity $v_{y}$ of this projectile:

${v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}$ .

$\begin{aligned} v_{y} &= -\sqrt{{u_{y}}^{2} + 2\, a_{y} \, x_{y}} \\ &= -\sqrt{(29.71786)^{2} + 2\, (-9.81)\, (-35)} \\ &\approx (-39.621)\; {\rm m\cdot s^{-1}}\end{aligned}$ .

(Negative since the projectile will be travelling downward towards the ground.)

Since drag is negligible, the horizontal velocity of this projectile will be a constant value. Thus, the final horizontal velocity of this projectile will be equal to the initial horizontal velocity: $v_{x} = u_{x}$ .

The overall final velocity of this projectile will be:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \sqrt{(150.086)^{2} + (-39.621)^{2}} \\ &\approx 155\; {\rm m\cdot s^{-1}} \end{aligned}$ .

Change in the vertical component of the velocity of this projectile:

$\begin{aligned} \Delta v_{y} &= v_{y} - u_{y} \\ &\approx (-39.621) - 29.71786 \\ &\approx 69.3386 \end{aligned}$ .

Divide the change in velocity by acceleration (rate of change in velocity) to find the time required to achieve such change:

$\begin{aligned}t &= \frac{\Delta v_{y}}{a_{y}} \\ &\approx \frac{69.3386}{(-9.81)} \\ &\approx 7.0682\; {\rm s}\end{aligned}$ .

Hence, the projectile would be in the air for approximately $7.07\; {\rm s}$ .

Also the horizontal velocity of this projectile is $u_{x} \approx 150.086\; {\rm m\cdot s^{-1}}$ throughout the flight, the range of this projectile will be:

$\begin{aligned}x_{x} &= u_{x}\, t \\ &\approx (150.086)\, (7.0682) \\ &\approx 1.06 \times 10^{3}\; {\rm m} \end{aligned}$ .

When this projectile is at maximum height, its vertical velocity will be $0$ . Apply the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to find the maximum height of the projectile (relative to the top of the $35\; {\rm m}$ cliff.)

$\begin{aligned}x &= \frac{{v_{y}}^{2} - {u_{y}}^{2}}{2\, a} \\ &\approx \frac{0^{2} - 29.71786^{2}}{2\, (-9.81)} \\ &\approx 45.0\; {\rm m}\end{aligned}$ .

Thus, the maximum height of the projectile relative to the ground will be approximately $45.0\; {\rm m} + 35\; {\rm m} = 80\; {\rm m}$ .

5 0

1 year ago

Can u help me. thank you

katrin2010 [14]

I can give you a search engine that could help you with all ir hw its called socratic it uses everything on the internet to search for answers it’s literally a search engine

6 0

2 years ago

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