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crimeas [40]
3 years ago
5

You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position

of the rock is _____ and the velocity of the rock is _____. You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position of the rock is _____ and the velocity of the rock is _____. negative, negative positive, negative negative, positive positive, positive
Physics
1 answer:
Alecsey [184]3 years ago
3 0

Answer:

positive, positive

You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position of the rock is positive and the velocity of the rock is positive

Explanation:

Given that the ground is defined as the origin.

The position of the rock is positive since the rock is thrown upward, the position also increases with time until it reaches the maximum height. Also, since the rock is thrown upward with the ground as the origin, the velocity of the rock is positive but the velocity reduces with time (change in height per unit time as the rock moves up is positive)

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Which of the following would NOT produce a current? A. Moving a bar magnet inside a coil of wire O B. Moving a bar magnet toward
sineoko [7]

Answer:

c

Explanation:

7 0
3 years ago
The human ear canal is about 2.8 cm long. If it is regarded as a tube that is open at one end and closed at the eardrum, what is
Diano4ka-milaya [45]

Answer:

f = 3.1 kHz

Explanation:

given,

length of human canal =2.8 cm = 0.028 m

speed of sound = 343 m/s

fundamental frequency  = ?

The fundamental frequency of a tube with one open end and one closed end is,

f = \dfrac{v}{4L}

f = \dfrac{343}{4\times 0.028}

f = \dfrac{343}{0.112}

       f = 3062.5 Hz

       f = 3.1 kHz

hence, the fundamental frequency is equal to f = 3.1 kHz

8 0
3 years ago
What is the specific heat for the aluminum wire?
Alisiya [41]
0.902

hope this helps :)
4 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
To heat 1g of water by 1 C requires<br> A) 1 calorie<br> b)1 Carlorie<br> c) 1 Joule<br> d) 1 watt
blsea [12.9K]
I think the answer would be 1 watt but i'm not sure
8 0
3 years ago
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