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2 years ago

What were the three factors that affected density

Physics

1 answer:

Stolb23 [73]2 years ago

7 0

Answer:

Pressure, temperature and humidity all affect air density. And you can think of air density as the mass of air molecules in a given volume.

Explanation:

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10-kg box is sliding across an ice rink at 10 m/s . A skater exerts a constant force of 10 N against it. How long will it take f

Taya2010 [7]

Answer:

Time, t = 10 seconds

Explanation:

Given the following data;

Mass = 10kg

Force = 10N

Final velocity = 10m/s

Initial velocity = 0m/s

To find the time;

First of all, we would find the acceleration of the box.

Force = mass * acceleration

10 = 10 * acceleration

Acceleration = 10/10 = 1m/s²

Now, we can find the time by using the first equation of motion;

V = U + at

10 = 0 + 1t

10 = t

Time, t = 10 seconds

Therefore, it will take 10 seconds for the box to come to a complete stop.

7 0

3 years ago

1. The potential difference between the terminals a cell in open circuit is 2.2volt. With resistance of 5ohm across the terminal

nadya68 [22]

Explanation:

10 /9 Ω

potential difference across the cell in open circuit is the emf of the cell.

Hence, emf E=2.2V

when, circuit is closed, potential difference across cell is given by V=E−Ir

And,

I= E/ R+r

Hence, V= E− Er/ R+r

⟹ V= ER/ R+r

⟹ 1.8= 2.2×5 /5+r

⟹9+1.8r=11

⟹ r= 2/ 1.8 Ω

⟹ r= 10/9 Ω

7 0

2 years ago

if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav

allochka39001 [22]

Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, $R_{max}$ = 3 m

The acceleration due to gravity on the Moon = 1 ÷ 16 of that on Earth

The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, $R_{max}$ , of a projectile motion is found by using the following formula

$R_{max} = \dfrac{u^2}{g}$

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

$R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }$

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, $g_{Moon}$ = 1/16 × g

∴ $g_{Moon}$ = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

$R_{max \ Moon} = \dfrac{u^2}{g_{Moon}} = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m$

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is $R_{max \ Moon}$ ≈ 48 m

8 0

2 years ago

A car is traveling at a 20.0 m/s for 7.00 s and then suddenly comes to a stop over a 3 s period.

Delicious77 [7]

Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.

B) The total distance the car travels is 200 meter during the 10 s period.

Explanation:

Given Data

Initial velocity of the car ( $v_{i}$ ) = 20.0 m/s

Final velocity of the car ( $v_{f}$ ) = 0 m/s

Time (in motion) =7.00 s

Time (in rest) =3 s

To find - A) car's deceleration while it came to a stop

B) the total distance the car travels in 10 s

A) The formula to find the deceleration is

Deceleration = (( final velocity - initial velocity ) ÷ Time) (m/s²)

Deceleration = (( $v_{f}$ ) - ( $v_{i}$ )) ÷ time (m/s²)

Deceleration = ( 0.0 - 20 ) ÷ 3 (m/s²)

Deceleration = (- 20) ÷ 3 (m/s²)

Deceleration = - 6.6667 m/s²

(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)

The car's deceleration is - 6.6667 m/s² while it came to a stop

B) The formula to find the distance traveled by the car is

Distance traveled by the car is equals to the product of the speed and time

Distance = Speed × Time (meter)

Distance = 20.0 × 10

Distance = 200 meters

The total distance the car travels during the period of 10 s is 200 meters

7 0

3 years ago

f your risk-aversion coefficient is A = 4 and you believe that the entire 1926–2015 period is representative of future expected

siniylev [52]

Answer:

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

Explanation:

As the complete question is not given here ,the table of data is missing which is as attached herewith.

From the maximized equation of the utility function it is evident that

$Weight=\frac{E_M-r_f}{A\sigma_M^2}$

For the equity, here as

$Weight$ is percentage of the equity which is to be calculated
${E_M-r_f}$ is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
$A$ is the risk aversion factor which is given as 4.
$\sigma_M$ is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59