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sveticcg [70]
3 years ago
7

Does the KE of a car change more when it accelerates from 22 km/h to 32 km/h or when it accelerates from 32 km/h to 42 km/h

Physics
1 answer:
mote1985 [20]3 years ago
6 0

Answer:

The change in kinetic energy (KE) of the car is more in the second case.

Explanation:

Let the mass of the car = m

initial velocity of the first case, u = 22 km/h = 6.11 m/s

final velocity of the first case, v = 32 km/h = 8.89 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(8.89² - 6.11²)

                                                    = 20.85m J

 

initial velocity of the second case, u = 32 km/h = 8.89 m/s

final velocity of the second case, v = 42 km/h = 11.67 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(11.67² - 8.89²)

                                                    = 28.58m J

The change in kinetic energy (KE) of the car is more in the second case.

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When magnesium fluoride reacts with iodine, the products are fluorine and magnesium iodide. What would the equation be?
likoan [24]

Answer:

Explanation:

Every chemical reaction can be represented by simple chemical equations that shows how compounds are combining to give some products.

Such reactions are usually made up of:

     Reactants on the left hand side

     Products on the right hand side

            Reactants  →  Products

In this given problem;

            Reactants are :   Magnesium fluoride  = MgF₂

                                             Iodine in form of Iodide   = I₂

           Product : Magnesium iodide  = MgI₂

             

               MgF₂   +     I₂    →       MgI₂   +   F₂

   

                             

   

7 0
3 years ago
Car a has a mass of 5 kg and 10 m/s .cart B has a mass of 10 kg and 5 m/s .which of the following statements best compares the i
weeeeeb [17]

The two cars have same momentum

Explanation:

The momentum of an object is given by the equation

p=mv

where

m is the mass of the object

v is its velocity

For the car A in this problem,

m = 5 kg

v = 10 m/s

So its momentum is

p_A = (5 kg)(10 m/s) = 50 kg m/s

For car B we have,

m = 10 kg

v = 5 m/s

So its momentum is

p_B = (10 kg)(5 m/s)=50 kg m/s

Therefore, the two cars have same momentum.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

7 0
3 years ago
A 63.3 kg wood board is resting on very smooth ice in the middle of a frozen lake. A 35.7 kg boy stands at one end of the board.
Vlada [557]

The velocity of the board relative to the ice is zero, since both are at rest.

<h3>What is relative velocity?</h3>

Relative velocity is the velocity of an object in relation to another reference object or point.

When two objects are travelling or moving with the same velocity in the same direction, the relative velocity one relative to the other is zero.

Also, when two objects are at rest, the relative velocity one relative to the other is zero.

Therefore, the velocity of the board relative to the ice is zero, since both are at rest.

Learn more about relative velocity at: brainly.com/question/24337516

#SPJ1

8 0
2 years ago
Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca
Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*(\sqrt{\frac{r^3}{Gm}})

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

\frac{Tsaturn}{Tearth} = \sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be \sqrt{10^3}} years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

1year * \frac{360degrees}{31.62years} = 11.38 degrees

or about 12 degrees.

6 0
3 years ago
What do you notice from the picture?​
Schach [20]
Answer:there talking about the house
5 0
4 years ago
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