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3 years ago

50 points for 5 questions p.s if u get brainliest u get 50 moe and then u have 100 SO YAAAAAAY

Physics

2 answers:

cestrela7 [59]3 years ago

4 0

Answer:

#1 = B, C, & D

#2 = A

#3 = C

#4 = C & A

#5 = C

Hope this helps! Have a good day :D!!

Explanation:

bogdanovich [222]3 years ago

3 0

Answer:

The correct answers are listed below:

#1 = B, C, & D

#2 = A

#3 = C

#4 = C & A

#5 = C

Explanation:

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A bug completes one lap along the edge of a circular planter of radius 18.3 cm in 13.81 s. answer to the nearest thousandth in u

algol13

Answer:

Speed of the bug is 0.0832 m/s.

Explanation:

We have,

Radius of the circular planet is 18.3 cm or 0.183 m and time taken by the bug to complete one lap is 13.81 s.

It is required to find the speed of the bug in the circular path. It is given by the displacement in one lap divided by time taken. So,

$v=\dfrac{d}{t}\\\\v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 0.183 }{13.81}\\\\v=0.0832\ m/s$

So, the speed of the bug is 0.0832 m/s.

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3 years ago

44.2 cm + 0.123 cm = cm

trapecia [35]

The answer is 44.323 cm

3 0

3 years ago

Read 2 more answers

A roadrunner at rest suddenly spots a rattlesnake slithering directly away at a constant speed of 0.75 m/s. At the moment the sn

Shalnov [3]

Answer:

The roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

Explanation:

From the statement we notice that:

1) Rattlesnake moves a constant speed ( $v_{S} = 0.75\,\frac{m}{s}$ ), whereas the roadrunner accelerates uniformly from rest. ( $v_{o, R} = 0\,\frac{m}{s}$ , $a = 1\,\frac{m}{s^{2}}$ )

2) Initial distance between the roadrunner and rattlesnake is 10 meters. ( $x_{o, R} = 0\,m$ , $x_{o,S} = 10\,m$ )

3) The roadrunner catches up to the snake at the end. ( $x_{S} = x_{R}$ )

Now we construct kinematic expression for each animal:

Rattlesnake

$x_{S} = x_{o,S}+v_{S}\cdot t$

Where:

$x_{o, S}$ - Initial position of the rattlesnake, measured in meters.

$x_{S}$ - Final position of the rattlesnake, measured in meters.

$v_{S}$ - Speed of the rattlesnake, measured in meters per second.

$t$ - Time, measured in seconds.

Roadrunner

$x_{R} = x_{o,R} +v_{o,R}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$

Where:

$x_{o, R}$ - Initial position of the roadrunner, measured in meters.

$x_{R}$ - Final position of the roadrunner, measured in meters.

$v_{o,R}$ - Initial speed of the roadrunner, measured in meters per second.

$a$ - Acceleration of the roadrunner, measured in meters per square second.

$t$ - Time, measured in seconds.

By eliminating the final positions of both creatures, we get the resulting quadratic function:

$x_{o,S}+v_{S}\cdot t = x_{o,R}+v_{o,R}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$

$\frac{1}{2}\cdot a \cdot t^{2} + (v_{o,R}-v_{S})\cdot t + (x_{o,R}-x_{o,S}) = 0$

If we know that $a = 1\,\frac{m}{s^{2}}$ , $v_{o, R} = 0\,\frac{m}{s}$ , $v_{S} = 0.75\,\frac{m}{s}$ , $x_{o, R} = 0\,m$ and $x_{o,S} = 10\,m$ , the resulting expression is:

$0.5\cdot t^{2}-0.75\cdot t -10=0$

We can find its root via Quadratic Formula:

$t_{1,2} = \frac{-(-0.75)\pm \sqrt{(-0.75)^{2}-4\cdot (0.5)\cdot (-10)}}{2\cdot (0.5)}$

$t_{1,2} = \frac{3}{4}\pm \frac{\sqrt{329}}{4}$

Roots are $t_{1} \approx 5.285\,s$ and $t_{2}\approx -3.785\,s$ , respectively. Both are valid mathematically, but only the first one is valid physically. Hence, the roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

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3 years ago

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beks73 [17]

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3 years ago

Match each force abbreviation to the correct description. Fg Fp Ff Fn force exerted by a push or pull. Support force at a right

qwelly [4]

Explanation:

Force Description

1. $F_g$ It is also known as the weight of an object. It is the force that is exerted on an object due to its mass