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3 years ago

What is the shallowest part of the ocean ecosystem is called

Physics

1 answer:

bearhunter [10]3 years ago

7 0

Answer:

intertidal zone.

the shallowest part of the ocean ecosystem, where the ocean floor os covered and uncovered as the tide goes in and out. Plankton. the organisms that float on the water in aquatic ecosystems and are unable to swim.

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An electric dipole of dipole moment 'p' is placed in the position of stable equilibrium in a uniform field of intensity 'E'. The

larisa86 [58]

Answer:

Torque by electric dipole = pEcos thita

5 0

3 years ago

Marc and Linh stretch out a long spring on the classroom floor. Marc holds one end of the spring still. Linh creates waves in th

Soloha48 [4]

The speed of the wave created by Linh in the spring by moving the other end right and left with a frequency of 2 Hz is 1m/s.

<h3>How to calculate speed of a wave?</h3>

The speed of a wave can be calculated by using the following formula:

Speed = Wavelength x Frequency

According to this question, Linh creates waves in the spring by moving the other end right and left with a frequency of 2 Hz. If wave crests are 0.5 m apart, the speed can be calculated as follows:

speed = 2Hz × 0.5m

speed = 1m/s

Therefore, the speed of the wave created by Linh in the spring by moving the other end right and left with a frequency of 2 Hz is 1m/s.

Learn more about speed at: brainly.com/question/10715783

#SPJ1

7 0

2 years ago

About once every 30 minutes, a geyser known as Old Faceful projects water 11.0 m straight up into the air. Use g = 9.80 m/s^2, a

Maurinko [17]

Answer:

The speed of the water is 14.68 m/s.

Explanation:

Given that,

Time = 30 minutes

Distance = 11.0 m

Pressure = 101.3 kPa

Density of water = 1000 kg/m³

We need to calculate the speed of the water

Using equation of motion

$v^2=u^2+2gs$

Where, u = speed of water

g = acceleration due to gravity

h = height

Put the value into the formula

$0=u^2-2\times9.8\times11.0$

$u=\sqrt{2\times9.8\times11.0}$

$u=14.68\ m/s$

Hence, The speed of the water is 14.68 m/s.

7 0

3 years ago

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s$

b) the change in the combined kinetic energy of the two-car system during this collision

$\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))$

substitute the value in the equation above

$=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J$

Hence, the change in combine kinetic energy is -2534.78J

8 0

3 years ago

A sample of copper with a mass of 1.80 kg, initially at a temperature of 150.0°C, is in a well-insulated container. Water at a t

user100 [1]

Answer:

the mass of water is 0.3 Kg

Explanation:

since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:

Q water + Q copper = Q surroundings =0 (insulated)

Q water = - Q copper

since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature

and denoting w as water and co as copper :

m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) = m co * c co * (T co - Ti eq)

m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]

We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C

if we assume that both specific heats do not change during the process (or the change is insignificant)

m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]

m w= 1.80 kg * 0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))

m w= 0.3 kg

7 0

4 years ago