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3 years ago

Express each of the following as indicated:

1 answer:

4 0

1 decimeter is equal to 100 millimeters. In this case, 2 decimeters is equal to 200 millimeters. 1 hour is equal to 3600 seconds while 1 minute is equal to 60 seconds. In this case, 2 h 10 minutes is equal to 7800 seconds. 1 gram is equal to 1000 micrograms. Thus, 16 grams is equal to 16000 mg and 0.675 mg is equal to 0.000675 grams. 1 centimeter is equal to 10000 micrometers. Hence 462 micrometers is equal to 0.0462 centimetes. 35 meter per hour is equal to 0.009722 m/s

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Leta stetter hollingworth conducted pioneering work on __________. a. identity development in ethnic minorities b. cognitive pro

Elena L [17]

D. Leta stetter hollingworth conducted pioneering work on <u>adolescent development and gifted children</u>.

<h3>Who is Leta stetter hollingworth?</h3>

Leta stetter hollingworth is an early feminist and active member of the Women's Suffrage Party.

Leta Stetter Hollingworth is best known for her landmark contributions to the psychology of women and to education of the gifted. That is adolescent development and gifted children.

Thus, we can conclude that Leta stetter hollingworth conducted pioneering work on <u>adolescent development and gifted children</u>.

Learn more about Leta stetter hollingworth here: brainly.com/question/2680369

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4 0

1 year ago

a 63 kg object needs to be lifted 7 meters in a matter of 5 seconds. approximately how much horsepower is required to achieve th

Natali [406]

Power is defined as the rate of doing work or the work per unit of time. The first step to solve this problem is by calculating the work which can be determined by the equation:

W = Fd

where:

F = force exerted = ma
d = distance traveled
m = mass of object
a = acceleration

Acceleration is equivalent to the gravitational constant (9.81 m/s^2) if the force exerted has a vertical direction such as lifting.

W = Fd = mad = 63(9.81)(7) = 4326.21 Joules

Now that we have work, we can calculate power.

P = W/t = 4325.21 J / 5 seconds = 865.242 J/s or watts

Convert watts to horsepower (1 hp = 745.7 watts)

P = 865.242 watts (1hp/745.7 watts) = 1.16 hp

3 0

3 years ago

Read 2 more answers

A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta

nevsk [136]

Answer:

1.25 kgm²/sec

Explanation:

Disk inertia, Jd =

Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

Disk angular speed =

ωd = 0.1047 * 30 = 3.1416 rad/sec

Hollow cylinder inertia =

Jc = 3.7 * 0.40² = 0.592 kgm²

Initial Kinetic Energy of the disk

Ekd = 1/2 * Jd * ωd²

Ekd = 0.148 * 9.87

Ekd = 1.4607 joule

Ekd = (Jc + 1/2*Jd) * ω²

Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

ω² = 1.4607/0.74

ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

4 0

4 years ago

I would like to know why this is the correct answer

Helen [10]

The acceleration of the object if the net force is decreased = 0.13 m/s²

<h3>Further explanation</h3>

Given

A net force of 0.8 N acting on a 1.5-kg mass.

The net force is decreased to 0.2 N

Required

The acceleration of the object if the net force is decreased

Solution

Newton's 2nd law :

$\tt \sum F=m.a$

The mass used in state 1 and 2 remains the same, at 1.5 kg

state 1

ΣF=0.8 N

m=1.5 kg

The acceleration, a:

$\tt a=\dfrac{\sum F}{m}\\\\a=\dfrac{0.8}{1.5}\\\\a=0.53`m/s^2$

state 2

ΣF=0.2 N

m=1.5 kg

The acceleration, a:

$\tt a=\dfrac{\sum F}{m}\\\\a=\dfrac{0.2}{1.5}\\\\a=0.13~m/s^2$

8 0

3 years ago

A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 5 nC is located at a distance r = 30 c

valina [46]

Answer:

$q_1 =7.08*10^{-9}C.$

Explanation:

Gauss's Law says that the electric flux $\Phi_E$ through a closed surface is directly proportional to the charge $Q_{enc}$ inside it. More precisely,

$\Phi_E=\oint_S E\cdot dA = \dfrac{Q_{enc}}{\epsilon_0}.$

This means what is outside this closed surface $S$ does not contribute to the flux through it because field lines that go in must come out, <em>resulting a zero flux from an external charge. </em>

In our context, this means the charge $q_2$ which is outside the sphere will have zero flux through the surface; therefore, Gauss's law will only be concerned with charge $q_1$ which is inside the sphere; Hence,

$\Phi_E=\oint_S E\cdot dA = \dfrac{q_1}{\epsilon_0} = 800 N\cdot m^2/C.$

Solving for $q_1$ gives

$q_1= (800 N\cdot m^2/C)\epsilon_0,$

$q_1= (800 N\cdot m^2/C)*(8.85*10^{-12}C^2/N\cdot m^2)$

$\boxed{q_1 =7.08*10^{-9}C. }$

which is the charge inside the sphere.

5 0

3 years ago