_____ are group of tissue working together to perform a certain job.

stira [4]

Combustion reaction

5 0

1 year ago

A magnetic B field of strength 0.9 T is perpendicular to a loop with an area of 3 m2. If the area of the loop is reduced to zero

Alexeev081 [22]

Answer:

The magnitude of induced emf is 5.4 V

Explanation:

Given:

Magnetic field $B = 0.9$ T

Area of loop $\Delta A = 3$ $m^{2}$

Time take to reduce loop to zero $\Delta t = 0.5$ sec

To find induced emf we use faraday's law,

Induced emf is given by,

$\epsilon = -\frac{\Delta \phi}{\Delta t}$

Here minus sign shows lenz law, for finding magnitude of emf we ignore it.

Where $\phi = B\Delta A$

Put the value of flux and find induced emf,

$\epsilon = \frac{B\Delta A}{\Delta t}$

$\epsilon = \frac{0.9 \times 3}{0.5}$

$\epsilon = 5.4$ V

Therefore, the magnitude of induced emf is 5.4 V

3 0

3 years ago

Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m

otez555 [7]

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

$T = 2 \pi \sqrt\frac{m}{k}$

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

$T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s$

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

5 0

2 years ago

Through what process does water leave the hydrosphere and enter the atmosphere

evablogger [386]

Water and the atmosphere. Water enters the atmosphere through evaporation, transpiration, excretion and sublimation: Transpiration is the loss of water from plants (via their leaves).

8 0

3 years ago

Consider two massless springs connected in series. Spring 1 has a spring constant k1, and spring 2 has a spring constant k2. A c

Andru [333]

Answer:

a. k = (1/k₁ + 1/k₂)⁻¹ b. k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

Explanation:

Since only one force F acts, the force on spring with spring constant k₁ is F = k₁x₁ where x₁ is its extension

the force on spring with spring constant k₂ is F = k₂x₂ where x₁ is its extension

Let F = kx be the force on the equivalent spring with spring constant k and extension x.

The total extension , x = x₁ + x₂

x = F/k = F/k₁ + F/k₂

1/k = 1/k₁ + 1/k₂

k = (1/k₁ + 1/k₂)⁻¹

B

The force on spring with spring constant k₃ is F = k₃x₃ where x₃ is its extension

Let F = kx be the force on the equivalent spring with spring constant k and extension x.

The total extension , x = x₁ + x₂ + x₃

x = F/k = F/k₁ + F/k₂ + F/k₃

1/k = 1/k₁ + 1/k₂ + 1/k₃

k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

8 0

2 years ago

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