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2 years ago

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a 60g tennis ball travelling at 30m/s hi a wall and bounce back at 20m/s.calculate (a)momentum of the ball before impact (b)mome

ntum of the ball after impact

1 answer:

frez [133]2 years ago

6 0

Explanation:

the formula for momentum is denoted by p=mv where p is momentum, m is mass and v is velocity. thus, the velocity before impact would be 0.060 x 30 = 1.8 kg/ms

the second one would just be 0.060 x 20 0.72kg/ms

I'm not 100 percent sure this is correct but yeah

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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

If you ran 15 km/h for 20 min, how much distance would you cover?

nikitadnepr [17]

Since we have 15 kilometers per hour, and we're looking for 20 minutes, let's set up proportions.
20/60 minutes = x/15
20/60 = 1/3, so let's leave that simplified.
1/3 = x/15
Look at the denominators, 3 to 15 is a factor of 5, so multiply the numerator by 5.
1 • 5 = 5, so you will cover 5 kilometers in 20 minutes.

I hope this helps!

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2 years ago

Name 5 ways you can deal with a bully without engaging in a physical altercation with him/her.

Delicious77 [7]

Answer:

ignore the bully ,tell the bully to stop,make a joke or laugh with the bully,stick with friend,know how to get out of the bullying situation

Explanation:

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2 years ago

What is an advantage of forced air heating?

Lelu [443]

i do believe the answer is c not sure but correct me if im wrong

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3 years ago

Read 2 more answers

A convex mirror, like the passenger-side rearview mirror on a car, has a focal length of -3.0 m . An object is 6.0 m from the mi

Artist 52 [7]

A) -2.0 m

Look at the ray diagram attached in the picture, where:

p identifies the location of the object

q identifies the location of the image

F identifies the focus of the mirror

Each tick represents 1 m

We have

p = 6.0 m is the distance of the object from the mirror

f = -3.0 m is the focal length

From the ray diagram, we see that q has a distance of 2.0 m from the mirror, and it's on the other side of the mirror compared to the object, so

q = -2.0 m

This can also be verified by using the mirror equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-3.0 m}-\frac{1}{6.0 m}=-\frac{3}{6.0 cm}\\q = \frac{-6.0 cm}{3}=-2.0 cm$

B) Upright and virtual

As we see from the picture, the image is upright, since it has same orientation as the object.

Also, we notice that the image is on the other side of the mirror, compared to the object. For a mirror,

- An image is said to be real if it is on the same side of the object

- An image is said to be virtual if it is on the opposite side of the mirror

Therefore, this means that the image is virtual.

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2 years ago