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2 years ago

Describe a structural adaption that a snake has. What challenge does it overcome? PLZ HELP ASAP

1 answer:

6 0

It has a flexible vertibrae which allows it to move faster and fit into places . Plus it also allows it to get its prey easier

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In the equation for the gravitational force between two objects, what does G represent? the gravitational constant the gravitati

kumpel [21]

G is the gravitational constant, which is approximately 6.6x10^-11 Nm/s^2. It has the same value regardless of the masses of both objects or the distance between them.

3 0

2 years ago

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How much voltage is in the primary coil if there are 3200 windings in the

Lesechka [4]

Answer:

Voltage in primary coil is 3.91 V

Explanation:

For transformer we know that the working principle is given as

$\frac{V_1}{V_2} = \frac{N_1}{N_2}$

here we know that

$V_1 [tex] = voltage in primary coil[tex]V_2 = 25 V$

$N_1 = 500$

$N_2 = 3200$

Now we have

$\frac{V_1}{25} = \frac{500}{3200}$

$V_1 = 3.91 V$

8 0

3 years ago

Which of the following cannot be determined by analyzing the periodic table?

dybincka [34]

Your answer is C) The year the element was discovered :)

4 0

2 years ago

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You walk into a darkened room and turn on a flashlight. You see an image of the flashlight reflecting off a plane mirror in fron

user100 [1]

Answer:

4.2 is the answer

Explanation

The image formed in a plane mirror is an equal distance behind the mirror as the object in front of it.

Step 1: the equation to this problem would be: 8.4/2

Step 2: 8.4 ÷ 2 = 4.2

6 0

1 year ago

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0

2 years ago