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2 years ago

7

type 2 diabetes runs in familes, but its frequency has been rising in recent years. what factors might explain this increase in

type 2 diabetes?

1 answer:

FrozenT [24]2 years ago

8 0

Answer:

poor diet andlack of exersise

Explanation:

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The process of changing the energy of a system by means of force. Force * Diatance

Dmitry_Shevchenko [17]

This is known as work

7 0

3 years ago

Seema knows the mass of a basketball. What other information is needed to find the ball’s potential energy?

adelina 88 [10]

Potential energy can be found using this formula:

PE= m * g * h

where:

PE= potential energy

m=mass

g=gravitational acceleration constant (9.8 m/s^2)

h= height

So your answer is height because you also use the gravitational constant.

7 0

2 years ago

6 Fig. 6.1 is a full-scale diagram that represents a sound wave travelling in air

Oxana [17]

From the measured wavelength from diagram, the frequency of the sound is 6660 Hz.

<h3>What is the frequency of a wave?</h3>

The frequency of a wave is the number of complete oscillation per second completed by a wave.

Frequency is related to wavelength and speed by the following formula:

Frequency = velocity/wavelength

Velocity of sound in air = 330 m/s

The measured wavelength = 5.0 cm = 0.05 m

Frequency = 330/0.05 = 6660 Hz

Therefore, based on the measured wavelength from diagram, the frequency of the sound is 6660 Hz.

Learn more about frequency of sound at: https://brainly.in/question/15373132
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7 0

1 year ago

A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

Drupady [299]

Answer:

$4.5\times 10^{-5} T$

Explanation:

We are given that

Current in wire=40 A

Magnetic field= $B_1=3.5\times 10^{-5}$ T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

$B_{wire}=B_2=\frac{\mu_0I}{2\pi R}$

We have R=29 cm= $\frac{29}{100}=0.29 m$

1 m=100 cm

Substitute the values in the given formula

$B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T$

The resultant magnetic field is given by

$B=\sqrt{B^2_1+B^2_2}$

Substitute the values then we get

$B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}$

$B=4.5\times 10^{-5} T$

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire= $4.5\times 10^{-5} T$

Hence, the resultant magnitude of the magnetic field 29 cm above the wire= $4.5\times 10^{-5} T$

7 0

2 years ago

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

2 years ago