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aliina [53]
2 years ago
11

2. What is the water pressure at a depth of 24 m in a lake? [Density of water, p = 1 000 kg m- and gravitational acceleration, g

= 9.81 ms-²]​
Physics
1 answer:
Andrei [34K]2 years ago
4 0

Answer:

Pressure = ρgh

pressure (p) is measured in pascals (Pa)

density (ρ) is measured in kilograms per metre cubed (kg/m3)

The fore of gravitational field strength (g) is measured in N/kg or m/s 2

height of column (h) is measured in metres (m)

Answer = 235,200 Pa

Explanation:

Pressure = ρgh

Pressure = 1,000 x 9.8 x 24

Pressure = 235,200 Pa

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You have two lightweight metal spheres, each hanging from an insulating nylon thread. One of the spheres has a net negative char
kkurt [141]

Answer:attract each other

Explanation:

When two-sphere, one with a negative charge and another neutral is brought close together but do not touch then they try to attract each other.

This because of the polarization of the neutral sphere as it is placed in the vicinity of a negatively charged sphere. The negatively charged sphere will induce the positive charge in the neutral sphere and they will attract each other according to Columb law.

8 0
3 years ago
An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get
Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

mV_i = (m-m_O)V_f - m_OV_O

where,

m=Total mass

m_O = Mass of Object

V_i = Velocity before throwing

V_f = Final Velocity

V_O = Velocity of Object

Our values are:

m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg

Solving to find the final speed, after throwing the object we have

V_f=\frac{mV_0+m_TV_O}{m-m_O}

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) 5.3Kg\rightarrow 15m/s

V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}

V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}

V_{f1}=0.8511m/s

B) 7.9Kg\rightarrow 11.2m/s

V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}

V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}

V_{f2} = 2.0173m/s

C) 10.5Kg\rightarrow 7m/s

V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}

V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}

V_{f3} = 3.63478m/s

Therefore the final velocity of astronaut is 3.63m/s

7 0
3 years ago
A cello string 0.75 m long has a 220 hz fundamental frequency. find the wave speed along the vibrating string. answer in units o
maxonik [38]
For fundamental frequency of a string to occur, the length of the string has to be half the wavelength. That is,

1/2y = L, where L = length of the string, y = wavelength.

Therefore,
y = 2L = 2*0.75 =1.5 m

Additionally,
y = v/f Where v = wave speed, and f = ferquncy

Then,
v = y*f = 1.5*220 = 330 m/s
4 0
3 years ago
A 2.0 kg block is pulled across a horizontal surface by a 15 N force at a constant velocity. What is the force of friction actin
lianna [129]

Answer:

<em>The force of friction acting on the block has a magnitude of 15 N and acts opposite to the applied force.</em>

Explanation:

<u>Net Force </u>

The Second Newton's law states that an object acquires acceleration when an unbalanced net force is applied to it.

The acceleration is proportional to the net force and inversely proportional to the mass of the object.

If the object has zero net force, it won't get accelerated and its velocity will remain constant.

The m=2 kg block is being pulled across a horizontal surface by a force of F=15 N and we are told the block moves at a constant velocity. This means the acceleration is zero and therefore the net force is also zero.

Since there is an external force applied to the box, it must have been balanced by the force of friction, thus the force of friction has the same magnitude acting opposite to the applied force.

The force of friction acting on the block has a magnitude of 15 N opposite to the applied force.

6 0
2 years ago
Fill in the blank
aliina [53]

Answer:

hypothesis

Explanation:

8 0
3 years ago
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