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3 years ago

8

One of the most common diseases a pet gets intestate with, is a worm infection in the stomach. 'Tape worm' is one such worm whic

h lives on the nutrients from the stomach of an animal. Tape worms are, therefore, (a) Saprophytes
(b) Autotrophs

(c) Heterotrophs

(d) Parasites

1 answer:

GREYUIT [131]3 years ago

6 0

D. Parasites

I hope this helped good luck!

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A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop s

skelet666 [1.2K]

Answer:

frictional force = 0.52 N

Explanation:

diameter of turn table (D1) = 30 cm = 0.3 m

mass of turn table (M1) = 1.2 kg

diameter of shaft (D2) = 1.2 cm = 0.012 m

mass of shaft (M2) = 450 g = 0.45 kg

time (t) = 15 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

radius of turn table (R1) = 0.3 / 2 = 0.15 m

radius of shaft (R2) = 0.012 / 2 = 0.006 m

total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft

I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}

I = 0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}

I = 0.0135 + 0.0000081 = 0.0135081

ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s

α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}

torque = I x α

torque = 0.0135081 x (-0.23) = - 0.00311 N.m

torque = frictional force x R2

- 0.00311 = frictional force x 0.006

frictional force = 0.52 N

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3 years ago

I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency. I then stretch it to twice its length an

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Answer:

The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)

Explanation:

Fundamental frequency = wave velocity/2L

where;

L is the length of the stretched rubber

Wave velocity = $\sqrt{\frac{T}{\frac{M}{L}}}$

Frequency (F₁) = $\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}$

To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.

Given:

L₂ =2L₁ = 2L

T₂ = 2T₁ = 2T

(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)

F₂ = $\frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1$

Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).

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However, sometimes levers are designed to increase the input force needed to move an object. Which of the following is most like

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Cups
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3 years ago

What is the specific heat of a material that has a mass of 5 grams and increases 125ºC when it receives 150J of heat?

rjkz [21]

Answer:

0.24 ? I hope that was the answer you were looking for.

Explanation:

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3 years ago

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