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3 years ago

Pleaseee someone help me TEST TOMORROW

Physics

2 answers:

Liono4ka [1.6K]3 years ago

4 0

3.A

4.c

5.A
hope this helps

evablogger [386]3 years ago

4 0

3. D
4. A
5. I think its B?
I hope you do good on your test tomorrow!

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Science: Could a dysfunction or disease of the nervous system affect other body systems, such as the muscular or digestive syste

SashulF [63]

Answer:

You may experience the sudden onset of one or more symptoms, such as: Numbness, tingling, weakness, or inability to move a part or all of one side of the body (paralysis). Dimness, blurring, double vision, or loss of vision in one or both eyes. Loss of speech, trouble talking, or trouble understanding speech.

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2 years ago

What are the signs of the velocity, force, and acceleration (components) afterthe cart is released and is moving qway from the m

SVETLANKA909090 [29]

Answer:

car is moving away its direction is negative.

the speed must also be negative.

speed this distance decreases the acceleration is negative and if the speed is increasing the acceleration is positive.

Explanation:

In the exercise they indicate that the direction to the motion sensor is positive, as they indicate that the car is moving away its direction is negative.

The speed of the car is

v = (x₂-x₁) / t

As the positions are negative, and the car moves away the speed must also be negative.

The analysis for acceleration must be very careful if the speed this distance decreases the acceleration is negative and if the speed is increasing the acceleration is positive.

3 0

2 years ago

Thermal conductivity of a material is given as 129Btuft–¹ h–¹°F–¹.Calculate this thermal conductivity in Jm–¹s–¹°C–¹(Given: 1Btu

ss7ja [257]

Answer:

223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

Explanation:

The thermal conductivity of an object is defined as the measure or the ability of the object to transfer heat or conduct heat through its body.

In the context, the thermal conductivity of the material is given as

$=129 \text{ Btu ft}^{-1}\text{h}^{-1}^\circ\text{F}^{-1}$

And it is given that :

1 Btu = 1055 J

1 ft = 0.3048 m

$1^\circ F = \frac{5}{9}^\circ C$

We know that 1 h = 3600 s

So the thermal conductivity of the material in $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$ is :

Thermal conductivity :

$=\frac{129 \text{ Btu}}{1 \text{ ft }\times \text{1 h}\times 1^\circ\text{F}}$

$=\frac{129 \times 1055 \text{ J}}{0.3048 \text{ m} \ \times 3600 \text{ s}\ \times \frac{5}{9}^\circ \text{C}}$

= 223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

3 0

3 years ago

(PLEASE HELP) What area of the Earth's mantle is the densest?

Nady [450]

The area nearest to the earth's core is the densest..
So your answer would be letter choice ( D ) . . .

Hope it Helped :)

7 0

3 years ago

Read 2 more answers

What fraction of an iceberg is submerged? (ρice = 917 kg/m3, ,ρsea = 1030 kg/m3.)

aksik [14]

Answer:

Choice d. Approximately $89\%$ of the volume of this iceberg would be submerged.

Explanation:

Let $V_\text{ice}$ denote the total volume of this iceberg. Let $V_\text{submerged}$ denote the volume of the portion that is under the liquid.

The mass of that iceberg would be $\rho_\text{ice} \cdot V_\text{ice}$ . Let $g$ denote the gravitational field strength ( $g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth.) The weight of that iceberg would be: $\rho_\text{ice} \cdot V_\text{ice} \cdot g$ .

If the iceberg is going to be lifted out of the sea, it would take water with volume $V_\text{submerged}$ to fill the space that the iceberg has previously taken. The mass of that much sea water would be $\rho_\text{sea} \cdot V_\text{submerged}$ .

Archimedes' Principle suggests that the weight of that much water will be exactly equal to the buoyancy on the iceberg. By Archimedes' Principle:

$\text{buoyancy} = \rho_\text{sea} \cdot V_\text{submerged} \cdot g$ .

The buoyancy on the iceberg should balance the weight of this iceberg. In other words:

$\underbrace{\rho_\text{ice} \cdot V_\text{ice} \cdot g}_\text{weight of iceberg} = \underbrace{\rho_\text{sea} \cdot V_\text{submerged} \cdot g}_\text{buoyancy on iceberg}$ .

Rearrange this equation to find the ratio between $V_\text{submerged}$ and $V_\text{ice}$ :

$\begin{aligned} &\frac{V_\text{submerged}}{V_\text{ice}} \\&= \frac{\rho_\text{ice} \cdot g}{\rho_\text{sea} \cdot g}\\ &= \frac{\rho_\text{ice}}{\rho_\text{sea}}\ = \frac{917\; \rm kg \cdot m^{-3}}{1030\; \rm kg \cdot m^{-3}} \approx 0.89 \end{aligned}$ .

In other words, $89\%$ of the volume of this iceberg would have been submerged for buoyancy to balance the weight of this iceberg.

8 0

2 years ago