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3 years ago

13

WHICH SHOULD NOT BE PART OF SCIENTIFIC INQUIRY

1 answer:

umka21 [38]3 years ago

5 0

Answer:

Not collecting the info

Explanation:

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Can someone help me with this

yKpoI14uk [10]

bonded pairs of electrons, lone pairs of electrons.

4 0

3 years ago

If there are 50 people per square kilometer in a city, and the area of the city is 1.5 × 10 square kilometers. What is the total

qaws [65]

Answer:

750 people

Explanation:

From the question,

Number of people in the city = population density×Area of the city

N = D×A.......................... Equagtion 1

Where N = Number of people in the city, D = population density, A = Area of the city.

Given: D = 50 people per square kilometer, A = 1.5×10 square kilometer.

Substitute into equation 1

N = 50(1.5×10)

N = 750 people.

Hence the total number of people in the city is 750 people.

6 0

3 years ago

The number of electrons in an element with atomic number 20 is

Alex787 [66]

Answer:

the answer is calcium....

5 0

3 years ago

Read 2 more answers

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

The average distance from the sun to Pluto is approximately 6.10 × 109 km. How long does it take light from

Scorpion4ik [409]

$V= \frac{S}{t}$
$t= \frac{S}{V}$ <u />
$t= \frac{S}{c}$
$t= \frac{6.1*10^{12}}{299792458}$
$t=20347.4098071s$

It takes 20347.4098071s for light from the sun to reach Pluto.
The 6.1*10^9 is replaced by 6.1*10^12 on line 4 because we convert the distance from km to m.
c = speed of light. If a different value was given in the previous question then use that instead of the value I used to do the final calculation.

3 0

3 years ago