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3 years ago

14

A student is taking an animal science and an astronomy class. In animal sciences, she learns that pumas are called mountain lion

s and cougars in different parts of the world, but they are all the same cat. She makes an analogy, or comparison, to something she learned in her astronomy class. She says that a certain space object is like a puma because it is the same object but has different names depending on where it is. What space object is she thinking of?
1. comet
2. meteoroid
3. asteroid
4. meteoride
I know this isn't Physics but they didn't have a Space option

2 answers:

GarryVolchara [31]3 years ago

8 0

Answer:

2. meteoroid

I saw the question.

user100 [1]3 years ago

8 0

Answer:

2.meteoroid

Explanation:

took test

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In a pig caller can produce a sound intensity level of 107 dB. How many pig callers would be needed to generate an intensity lev

myrzilka [38]

Answer:

20 pig callers

Explanation:

Given that:

A pig caller produced intensity level of a sound = 107 dB

To find how many pig callers required to generate an intensity level of 120 dB;

we have:

120 dB - 107 dB = 13 dB

Taking the logarithm function;

$10 \ log \bigg(\dfrac{I}{I_o} \bigg) = 13 \ dB$

where;

$I_o$ = initial intensity

$log \bigg(\dfrac{I}{I_o} \bigg) = 1.3$

$\dfrac{I}{I_o}= 10^{1.3 }$

I = 19.95 $I_o$

I ≅ 20 pig callers

6 0

3 years ago

Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express

AlladinOne [14]

Answer:

Vb = k Q / r r <R

Vb = k q / R³ (R² - r²) r >R

Explanation:

The electic potential is defined by

ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

Va = - ∫ (k Q / r²) dr

-Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

Vb = k Q / r r <R

We perform the calculation of the power with the expression of the electric field that they give us

Vb = - int (kQ / R3 r) dr

We integrate and evaluate from the starting point r = R to the final point r <R

Vb = ∫kq / R³ r dr

Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

8 0

4 years ago

Paola can flex her legs from a bent position through a distance of 20.1 cm. Paola leaves the ground when her legs are straight,

makkiz [27]

The third equation of free fall can be applied to determine the acceleration. So that Paola's acceleration during the flight is 39.80 m/ $s^{2}$ .

Acceleration is a quantity that has a direct relationship with velocity and also inversely proportional to the time taken. It is a vector quantity.

To determine Paola's acceleration, the third equation of free fall is appropriate.

i.e $V^{2}$ = $U^{2}$ ± 2as

where: V is the final velocity, U is the initial velocity, a is the acceleration, and s is the distance covered.

From the given question, s = 20.1 cm (0.201 m), U = 4.0 m/s, V = 0.

So that since Poala flies against gravity, then we have:

$V^{2}$ = $U^{2}$ - 2as

0 = $(4)^{2}$ - 2(a x 0.201)

= 16 - 0.402a

0.402a = 16

a = $\frac{16}{0.402}$

= 39.801

a = 39.80 m/ $s^{2}$

Therefore Paola's acceleration is 39.80 m/ $s^{2}$ .

Visit: brainly.com/question/17493533

7 0

3 years ago

Read 2 more answers

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

3 years ago

Read 2 more answers

The x vector component of a displacement vector has a magnitude of 86.2 m and points along the negative x axis. The y vector com

Flauer [41]

Answer:

(a) Magnitude of Vector = 207.73 m

(b) Direction = 65.48°

Explanation:

(a)

The formula to find out the magnitude of a resultant vector with the help of its x and y components is given as follows:

$Magnitude\ of\ Vector = \sqrt{d_{x}^{2} + d_{y}^{2}} = \sqrt{(86.2\ m)^{2} + (189\ m)^{2}}\\\\$

<u>Magnitude of Vector = 207.73 m</u>

(b)

For the direction of the vector we have the formula:

$Direction = tan^{-1}(\frac{y}{x})\\\\Direction = tan^{-1}(\frac{189\ m}{86.2\ m})\\\\$

<u>Direction = 65.48°</u>

4 0

3 years ago