Why does it takes the outer planets so long to orbit the Sun? Try to come up with two reasons.

A rifle is aimed horizontally at a target 50.0 m away. The bullet hits the target 2.90 cm below the aim point. . . Whats the bul

Marina CMI [18]

The problem ask to calculate the bullet's flight time and the bullet's speed as it left the barrel. So base on the problem, the answer would be that the flight time is 0.076 seconds and the speed of the bullet is 657.9 m/s. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications.

5 0

3 years ago

COULD SOMEONE PLS DO THESE FOR ME <br> I’LL LOVE U FOREVER AND EVER

sladkih [1.3K]

I can help you do them for yourself :)

7 0

3 years ago

The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of

kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

$\rho_{0\°C}= 730 kg/m^{3}$ is the density of gasoline at $0\°C$

$\beta=9.60(10)^{-4} \°C^{-1}$ is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to $m^{3}$ :

$V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3}$ (1)

Knowing density is given by: $\rho=\frac{m}{V}$ , we can find the mass $m_{1}$ of 8.50 gallons:

$m_{1}=\rho_{0\°C}V$

$m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg$ (2)

Now, we have to calculate the factor $f$ by which the volume of gasoline is increased with the temperature, which is given by:

$f=(1+\beta(T_{f}-T_{o}))$ (3)

Where $T_{o}=0\°C$ is the initial temperature and $T_{f}=21.7\°C$ is the final temperature.

$f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C))$ (4)

$f=1.020832$ (5)

With this, we can calculate the density of gasoline at $21.7\°C$ :

$\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)$

$\rho_{21.7\°C}=745.207 kg/m^{3}$ (6)

Now we can calculate the mass of gasoline at this temperature:

$m_{2}=\rho_{21.7\°C}V$ (7)

$m_{2}=(745.207 kg/m^{3})(0.0323 m^{3})$ (8)

$m_{2}=24.070 kg$ (9)

And finally calculate the mass difference $\Delta m$ :

$\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg$ (10)

$\Delta m=0.4911 kg$ (11) This is the extra mass of gasoline

6 0

3 years ago

An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Exp

Artemon [7]

Explanation:

It is given that,

Mass of the football player, m = 92 kg

Velocity of player, v = 5 m/s

Time taken, t = 10 s

(1) We need to find the original kinetic energy of the player. It is given by :

$k=\dfrac{1}{2}mv^2$

$k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2$

k = 1150 J

In two significant figure, $k=1.2\times 10^3\ J$

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

i.e. $P=\dfrac{W}{t}=\dfrac{\Delta K}{t}$

$P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}$

P = 115 watts

In two significant figures, $P=1.2\times 10^2\ Watts$

Hence, this is the required solution.

6 0

3 years ago

A group of students is measuring the change in momentum of an object. Which set of lab equipment would be most in measuring the

a_sh-v [17]

Answer:

The momentum of an object is defined as the mass of the object times the velocity of the object, as P = m*v.

So the equipment needed would be:

Something to measure the mass of the object, like a balance.

Something to measure the speed of the object, like a doppler radar, or a simpler thing may be a cronometer, with that you can measure the amount of time that the object needs to travel a given distance, and with that you can obtain the speed of the object.

Now you can notice that speed is different than velocity, this is true, velocity is a vector, so this has a direction, then you need something to fix the direction in which the object moves, in this way you can determine the velocity.

4 0

3 years ago