When the direction of an object changes, what happens to its velocity?

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

Are all forms of radiation safe for humans?

Evgen [1.6K]

Not all forms of radiation is safe for humans

7 0

3 years ago

Read 2 more answers

A 62.5 kg carpenter at a construction site plans to swing in a circular arc from one roof top to an adjacent roof at the end of

denis23 [38]

Answer:

v = 9.45 m/s

Explanation:

given,

mass of the carpenter = 62.5 Kg

length of rope = 13.4 m

Capable of exerting force = 1034 N

centripetal force acting on the body

$F = \dfrac{mv^2}{r}$

$F = \dfrac{62.5\times v^2}{13.4}$

F =4.664 v² N

Gravitational force on her =

F = m g

F = 62.5 x 9.81

F = 613.125 N

now,

4.664 v² + 613.125 = 1034

4.664 v² = 420.875

v² = 90.24

v = 9.45 m/s

Maximum speed which she can tolerate = v = 9.45 m/s

7 0

3 years ago

You break a piece of Styro foam packing material, and it releases lots of little spheres whose electric charge makes them stick

stich3 [128]

Answer:

The magnitude of charge on each sphere is $q=2.50\times 10^{-8}\ C$ .

Explanation:

Given that,

Force of repulsion between the charges, F = 22 mN

The distance between spheres, r = 16 mm = 0.016 m

It is mentioned that both the spheres carry equal charges. The force between charges is given by :

$F=\dfrac{kq^2}{r^2}$

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{22\times 10^{-3}\times (0.016)^2}{9\times 10^9}}$

$q=2.50\times 10^{-8}\ C$

So, the magnitude of charge on each sphere is $q=2.50\times 10^{-8}\ C$ . Hence, this is the required solution.

3 0

3 years ago

Phosphorus 32 has a half life of 14.3 days how much of a 4mg sample of phosphorus - 32 will remain after 71.5 days

andrew-mc [135]

Let us first find out the radioactive constant of Phosphorus 32.

Radioactive constant, λ = $\frac{ln2}{t_{1/2}}$

Here, ${t_{1/2}$ is half life of phosphorus 32 = 14.3 days

λ = $\frac{ln2}{14.3}$

The amount of 4 mg of phosphorus remain after 71.5 days can be found using the formula,

m=m₀e⁻(λt)

= $4*e^{-\frac{ln2}{14.3}71.5}$

= $4*e^{-ln2*5}$

= $4*e^{-ln(2)^{5}}$

= $4*e^{-ln32}$

= $\frac{4}{32}$

= 0.125 mg

The mass of 4 mg of phosphorus 32 remains after 71.5 days will be 0.125 mg.

6 0

3 years ago