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3 years ago

What is the Deepest point of the ocean and where is it?

Physics

2 answers:

qaws [65]3 years ago

5 0

Answer:

marinas trench in Pacific Ocean

<h2>mark it as brainliest</h2>

Rashid [163]3 years ago

3 0

Explanation:

Located in the western Pacific Ocean, the Marina Trench

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The student in item 1 moves the box up a ramp inclined at 12° with the horizontal. if the box starts from rest at the bottom of

Nata [24]

Answer: Fg,y = mg(cos q1) Fg,x = mg(sin q1) Fapplied,x = Fapplied(cos q2) Fapplied,y = Fapplied(sin q2) Fn = Fg,y â’ Fapplied,y Fk =mkFn Fapplied,x â’ Fk â’ Fg,x ax=fapplied-fk-fg,x/ m

8 0

3 years ago

A 1.30-kg object is held 1.10 m above a relaxed, massless vertical spring with a force constant of 315 N/m. The object is droppe

pshichka [43]

Answer:

0.345m

Explanation:

Let x (m) be the length that the spring is compress. If we take the point where the spring is compressed as a reference point, then the distance from that point to point where the ball is held is x + 1.1 m.

And so the potential energy of the object at the held point is:

$E_p = mgh$

where m = 1.3 kg is the object mass, g = 10m/s2 is the gravitational acceleration and h = x + 1.1 m is the height of the object with respect to the reference point

$E_p = 1.3 * 10 * (x + 1.1) = 13(x + 1.1) = 13x + 14.3 J$

According to the conservation law of energy, this potential energy is converted to spring elastic energy once it's compressed

$E_p = E_k = kx^2/2 = 13x + 14.3$

where k = 315 is the spring constant and x is the compressed length

$315x^2 = 26x + 28.6$

$315x^2 - 26x - 28.6 = 0$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{26 \pm \sqrt{26^2 - 4*(-28.6)*315}}{2*315}$

$x = \frac{26 \pm 191.6}{630}$

x = 0.345 m or x = -0.263 m

Since x can only be positive we will pick the 0.345m

6 0

3 years ago

A 3 inch fire hose has a water flow of 200 gallons per minute. What is the flow in liters per second? Note: Use US gallons not U

stepladder [879]

Answer:

Case I: 12.617 L/s

Case II: 161.406 cubic meters per hour

Case III: 1.062 Pound inches

Explanation:

Given:

Speed of water flow = 200 gallons per minute
Speed of air blow = 95 cubic feet per minute
Measure of Torque = 12 Newton centimeter

Assumptions:

1 US gallon = 3.785 L
1 min = 60 s
1 ft = 0.3048 m
1 h = 60 min
1 inch = 2.54 cm
1 N = 0.2248 lb

Case I:

$Speed\ of\ water\ flow = 200 \dfrac{gallon}{min}\\\Rightarrow V_{water} = 200\times \dfrac{3.785\ L}{60\ s}\\\Rightarrow V_{water} = 12.617\ L/s$

Case II:

$Speed\ of\ air\ blow = 95 \dfrac{ft^3}{min}\\\Rightarrow V_{air} = 95\times \dfrac{(0.3048\ m)^3}{\dfrac{1}{60}\ h}\\\Rightarrow V_{air} = 95\times (0.3048)^3\times 60\ m^3/h\\\Rightarrow V_{air} = 161.406\ m^3/h$

Case III:

$Measure\ of\ torque = 12\ N cm\\\Rightarrow \tau = 12\times (0.2248\ lb)\times \dfrac{1}{2.54}\ in\\ \Rightarrow \tau = 1.062\ lb in$

6 0

3 years ago

Write down the equation that links energy transferred, charge flow, and potential difference.

laiz [17]

Answer:

The description of the given scenario is described in the explanation segment below.

Explanation:

As a charge passes through some kind of potential gap, the electrical task is undertaken as well as energy will be transferred.
The Potential difference seems to be an amount of work performed per unit charge and is denoted by "V" as well as a charge is denoted by "Q".

The energy transferred could be determined by using the following equation or formula:

⇒ Energy transferred = charge × potential difference

6 0

3 years ago

Calculate the change in temperature of 20 kg of water if 30 kg of aluminum is dropped in the water and the aluminum changes temp

vivado [14]

ΔE = 0

(m_alum) (c_alum) Δt - (m_water)(c_water) Δt' = 0

(30) (897) (20) - (20)(4186) Δt' = 0

Δt' = 6.42857 °C

6 0

4 years ago