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4 years ago

What is the kinetic energy of a 8kg cat running 5m/s?

Physics

1 answer:

ivanzaharov [21]4 years ago

6 0

<span>K.E = 0.5 * m * v^2 ( m = mass(Kg), V = Velocity(m/s)
= 0.5 * 8 * 5^2
= 4 * 25
= 100 J </span>

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Considerable scientific work is currently under way to determine whether weak oscillating magnetic fields such as those found ne

slega [8]

Answer:

$\epsilon = 2.96 \times 10^{-11} \ V$

Explanation:

given,

magnetic field strength = 1.40 ✕ 10⁻³ T

frequency of oscillation = 60 Hz

diameter of RBC = 7.5 μm

EMF = ?

$\epsilon = NBA\omega$

$\epsilon = NB(\pi\ r^2)\ (2\pi f)$

$\epsilon = NB(\pi\ (\dfrac{d}{2})^2)\ (2\pi f)$

$\epsilon = (1)\ 1.4 \times 10^{-3}(\pi\ (\dfrac{7.5 \times 10^{-6}}{2})^2)\ (2\pi\times 60)$

$\epsilon = 2.96 \times 10^{-11} \ V$

maximum emf that can generate around the perimeter of the cell $\epsilon = 2.96 \times 10^{-11} \ V$

5 0

3 years ago

A force of 16.88 N is applied tangentially to a wheel of radius 0.340 m and gives rise to an angular acceleration of 1.20rad / (

Basile [38]

Given.

force = 16.88 N is

radius = 0.340m

an angular acceleration = 1.20rad/s^2

the formula for torque is

F*r = I*a

where I is moment of inertia

16.88*.34 = I*1.2

I = 4.78Kg-m^2

so rotational inertia I = 4.78Kg-m^2

7 0

3 years ago

Four factors that affect the rate heat transfers through solids

MA_775_DIABLO [31]

Answer:

The amount of heat flow is . How does heat flow through solids? Conduction is the transfer of heat through solids. Factors that affect rate of heat flow include the conductivity of the material, temperature difference across the material, thickness of the material, and area of the material.

Explanation:

6 0

3 years ago

A quantity of gas has a volume of 0.20 cubic meter and an absolute temperature of 333 degrees kelvin. When the temperature of th

Nesterboy [21]

Answer:

Option C is the correct answer.

Explanation:

By Charles's law we have

V ∝ T

That is

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Here given that

V₁ = 0.20 cubic meter

T₁ = 333 K

T₂ = 533 K

Substituting

$\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\\\frac{0.20}{333}=\frac{V_2}{533}\\\\V_2=\frac{0.20}{333}\times 533=0.3198m^3$

New volume of the gas = 0.3198 m³

Option C is the correct answer.

7 0

4 years ago

What is the maximum force that could be applied to anterior cruciate ligament (ACL) if it has a diameter of 4.8 mm and a tensile

vovangra [49]

Answer:

Maximum force, F = 1809.55 N

Explanation:

Given that,

Diameter of the anterior cruciate ligament, d = 4.8 mm

Radius, r = 2.4 mm

The tensile strength of the anterior cruciate ligament, $P=100\times 10^6\ N/m^2=10^8\ Pa$

We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,

$F=P\times A\\\\F=10^8\times \pi (2.4\times 10^{-3})^2\\\\F=1809.55\ N$

So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N

4 0

4 years ago