The emission of light from a radioisotope occurs during ? decay.

vampirchik [111]

Answer:

a cold air mass and a warm air mass merge together

6 0

3 years ago

A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment. (a) calculate the average force

iragen [17]

(a) $-1.5\cdot 10^6 N$

First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:

$v^2 -u ^2 = 2ad$

where

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d = 1.00 cm = 0.01 m is the displacement of the person

Solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.01)}=-20000 m/s^2$

And the average force on the person is given by

$F=ma$

with m = 75.0 kg being the mass of the person. Substituting,

$F=(75)(-20000)=-1.5\cdot 10^6 N$

where the negative sign means the force is opposite to the direction of motion of the person.

b) $-1.0\cdot 10^5 N$

In this case,

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d = 15.00 cm = 0.15 m is the displacement of the person with the air bag

So the acceleration is

$a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.15)}=-1333 m/s^2$

So the average force on the person is

$F=ma=(75)(-1333)=-1.0\cdot 10^5 N$

7 0

4 years ago

You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If

wlad13 [49]

Answer:

between locations that are 14 cm outboard of the chair edges
the weightless board is centered and end sitters are 25 cm from the ends

Explanation:

We can assume the .5 m-wide chair means that it is comfortable for each student to sit 0.25 m from the end of the board. If the board is centered on the chair, then each student is 1 m from the edge of the chair.

When Dan and Tahreen are seated on the board, their center of mass is ...

(50 kg×2.5 m)/(50 kg +67 kt) = 1.068 m

to the right of the position where Dan is seated. Since this location is over the chair, the board is stable.

Komila can sit as much as x distance from the chair toward Dan, where ...

67(1) +54(x) = 50(1.5)

x = 8/54 ≈ 0.148 . . . . meters

Or, Komila can sit as much as x distance from the chair toward Tahreen, where ...

67(1.5) = 54(x) +50(1)

x = 50.5/54 ≈ 0.935 . . . . meters

Scenario 1

Assuming the (weightless) board is centered on the chair, Komila can sit anywhere between 14.8 cm left of the chair and 93.5 cm right of the chair and the board will remain stable. Sitting on the board centered on the chair is a suitable location. The two students sitting on the ends must become (and stay) seated at the same time. They both must be seated 0.25 m from the end of the board for the other dimensions to remain valid.

Scenario 2

Assuming the (weightless) board is located so its left end is 1.068 m from the chair, and Dan and Tahreen are seated 0.25 m from the ends of the board, Komila can sit anywhere within (117/54×.25 m) = 0.54 m of the chair and the board will remain stable. Again, sitting centered on the chair is a suitable location.

__

There does not appear to be any location where Komila can sit and have the board remain stable with only Dan or Tahreen seated on one end (assuming a width of 0.5 m for each sitter).

_____

Comment on the question

For the board to remain stable, the sum of moments about either edge of the chair must tend to rotate the board toward the chair. This sum will depend on the locations of the sitters relative to each edge of the chair, so there is significant freedom in choosing locations. To make the problem tractable, we have made some specific assumptions about where the board is and what the locations of the sitters might be. YMMV

3 0

4 years ago

If two charged objects in a laboratory are brought to a distance of 0.22 meters away from each other. What is

zysi [14]

Answer:

$q_2=2.47\times 10^{-4}\ C$

Explanation:

The charge on one object, $q_1=9.9\times 10^{-5}\ C$

The distance between the charges, r = 0.22 m

The force between the charges, F = 4,550 N

Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :

$F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{4550\times (0.22) ^2}{9\times 10^9\times 9.9\times 10^{-5}}\\\\q_2=2.47\times 10^{-4}\ C$

So, the charge on the other sphere is $2.47\times 10^{-4}\ C$ .

7 0

3 years ago

A muscle inserts 1.5 cm from the joint axis and exerts 300 N of force at an angle of pull of 60 degrees. How much torque is prod

Amanda [17]

Answer:

torque = 3.897 N-m

Explanation:

given data

force = 300 N

angle = 60 degree

distance = 15 cm

to find out

torque

solution

we will apply here torque formula that is given below

torque = force × sinθ × distance ...................1

put here all these value in equation 1

we get torque

torque = force × sinθ × distance

torque = 300 × sin60 × 1.5 × $10^{-2}$

torque = 300 × 0.8660 × 1.5 × $10^{-2}$

torque = 259.80 × 1.5 × $10^{-2}$

torque = 389.711 × $10^{-2}$

torque = 3.897 N-m

8 0

3 years ago