The stratospheric chemical that prevents much of the solar ultraviolet radiation from penetrating to earth's surface is:

In Shinto, there is s strong emphasis on purification and recognition of the role environment plays in creating harmony. Explore

Savatey [412]

Answer:

Explanation:

i have no idea

8 0

3 years ago

What must the charge (sign and magnitude) of a particle of mass 1.41 gg be for it to remain stationary when placed in a downward

Yuri [45]

Answer:

$q = 2.067 \times 10^{-5}\ C$

Explanation:

Given,

mass = 1.41 g = 0.00141 Kg

Electric field,E = 670 N/C.

We know,

Force in charge due to Electric field.

F = E q

And also we know

F = m g

Equating both the equation of motion

m g = E q

$q =\dfrac{mg}{E}$

$q =\dfrac{0.00141 \times 9.81}{670}$

$q = 2.067 \times 10^{-5}\ C$

Charge of the particle is equal to $q = 2.067 \times 10^{-5}\ C$

6 0

3 years ago

How many electrons make up a charge of - 30.0 μc?

Ymorist [56]

Each electron has a charge of

$q=-1.6 \cdot 10^{-19}C$

In this problem, the total charge is

$Q=-30.0 \mu C=-30.0 \cdot 10^{-6} C$

Therefore, the number of electrons contained in this total charge will be given by the total charge divided by the charge of a single electron:

$N=\frac{Q}{q}=\frac{-30.0 \cdot 10^{-6} C}{-1.6 \cdot 10^{-19} C}=1.88 \cdot 10^{14}$

3 0

3 years ago

Read 2 more answers

Suppose a baseball pitcher throws the ball to his catcher.

amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

$\Delta v = |v-u|$

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

$\Delta v=|v-0|=|v|$

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

$\Delta v =|0-v|=|-v|$

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

$\Delta p = m \Delta v$

where

m is the mass of the object

$\Delta v$ is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

$|\Delta p|=m|\Delta v|$

- In situation 1 (pitcher throwing the ball), the change in momentum is

$\Delta p = m|\Delta v|=m|v|=mv$

- In situation 2 (catcher receiving the ball), the change in momentum is

$\Delta p = m\Delta v = m|-v|=mv$

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

$I=\Delta p$

where

I is the impulse

$\Delta p$ is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

$I=F\Delta t$

where

I is the impulse

F is the force applied

$\Delta t$ is the time of impact

This can be rewritten as

$F=\frac{I}{\Delta t}$

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- $\Delta t$ when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to $\Delta t$ , this means that the force is larger when the ball is catched.

6 0

3 years ago

Latitude and longitude picture.

TiliK225 [7]

Answer:

where? please comment it

Explanation:

7 0

3 years ago