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REY [17]
3 years ago
13

I need 4 answers for Earth's layers and 4 answers for the Descriptions for Earth's layers graphic organizer.

Physics
1 answer:
Arada [10]3 years ago
3 0
Ok well have a nice day
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Coins were developed as a medium of exchange because other items like cows, grain, and land were more difficult to move from pla
miv72 [106K]
True, they used them because its easier to trade coins than products
7 0
3 years ago
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A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface
givi [52]

Answer:

a = 0.009 J

b = 0.19 m/s

c = 0.005 J and 0.004 J

Explanation:

Given that

Mass of the object, m = 0.5 kg

Spring constant of the spring, k = 20 N/m

Amplitude of the motion, A = 3 cm = 0.03 m

Displacement of the system, x = 2 cm = 0.02 m

a

Total energy of the system, E =

E = 1/2 * k * A²

E = 1/2 * 20 * 0.03²

E = 10 * 0.0009

E = 0.009 J

b

E = 1/2 * k * A² = 1/2 * m * v(max)²

1/2 * m * v(max)² = 0.009

1/2 * 0.5 * v(max)² = 0.009

v(max)² = 0.009 * 2/0.5

v(max)² = 0.018 / 0.5

v(max)² = 0.036

v(max) = √0.036

v(max) = 0.19 m/s

c

V = ±√[(k/m) * (A² - x²)]

V = ±√[(20/0.5) * (0.03² - 0.02²)]

V = ±√(40 * 0.0005)

V = ±√0.02

V = ±0.141 m/s

Kinetic Energy, K = 1/2 * m * v²

K = 1/2 * 0.5 * 0.141²

K = 1/4 * 0.02

K = 0.005 J

Potential Energy, P = 1/2 * k * x²

P = 1/2 * 20 * 0.02²

P = 10 * 0.0004

P = 0.004 J

4 0
2 years ago
Can someone please help, ty!
lyudmila [28]

Answer:hypothesis: the plant died of lack  of light,moisture,or water

Explanation:

to test my hypothesis i put a plant in a room at room temp and repeat how i grew it,and observe what went wrong. hope this helps=)

5 0
3 years ago
Your office has a 0.025 m^3 cylindrical container of drinking water. The radius of the container is about 13 cm. Required:a. Whe
uranmaximum [27]

Answer:

<em>a) 105935.7 Pa</em>

<em>b) 103630.35 Pa</em>

Explanation:

The volume of the container = 0.025 m^3

The radius of the container = 13 cm = 0.13 m

We have to find the height of the tank

From the equation for finding the volume of the cylinder,

V = \pi r^2h

where

V is the volume of the cylinder

h is the height of the cylinder

substituting values, we have

0.025 = 3.142 x 0.13^2 x h

0.025 = 0.0531h

h = 0.025/0.0531 = 0.47 m

Pressure at the bottom of the tank P = ρgh

where

ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the depth of water which is equal to the height of the tank

substituting values, we have

P = 1000 x 9.81 x 0.47 = 4610.7 Pa

atmospheric pressure = 101325 Pa

therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = <em>105935.7 Pa</em>

b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m

pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa

This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = <em>103630.35 Pa</em>

7 0
3 years ago
A closed system’s internal energy changes by 178 J as a result of being heated with 658 J of energy.
statuscvo [17]
Internal energy of the system changes by ΔE = 178 J.
Heat given to the system = Q = +658 J.

According to the first law of thermodynamics, 
ΔE = Q + W
178 = 658 + W 
∴ W = 178-658 = -480 J

Minus sign indicates that work is done by the system. 
7 0
3 years ago
Read 2 more answers
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