3 years ago

What is the relationship between the kinetic energy of molecules in an object and the object temperature

Physics

1 answer:

andrew11 [14]3 years ago

7 0

I think it might be friction

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How would you write the number 6,500,000,000 in scientific notation?

photoshop1234 [79]

Scientific form = 6.5 x 109.

8 0

3 years ago

A dragster race car can accelerate from rest to incredible speeds. In one case a dragster is able to finish the 305 m run in 3.6

Dafna1 [17]

Answer:

45.89m/s²

Explanation:

Given

Distance S = 305m

Time t = 3.64s

To get the acceleration during this run, we will apply the equation of motion:

S = ut+1/2at²

Substitute the given parameters into the formula and calculate the value of a

305 = 0+1/2 a(3.64)²

304 = 1/2(13.2496)a

304 = 6.6248a

a = 304/6.6248

a = 45.89m/s²

Hence the average acceleration during this run is 45.89m/s²

4 0

2 years ago

The school bag of four students A,B,C,D measures 9kg, 2800gm, 2kg and 8000gm respectively. Whose bag is the lightest

inna [77]

Answer:

Student C

Explanation:

order from heaviest to lightest is...

9 kg (A) , 8000g (8 kg) (D) , 2800g (2.8kg) (B), 2 kg (C)

5 0

2 years ago

In 1970, a rocket powered car called Blue Flame achieved a maximum speed of 1.00(10 km/h (278m/s).Suppose the magnitude of the c

sammy [17]

Answer:

Distance traveled during this acceleration will be 6950 m

Explanation:

Wear have given maximum speed tat will be equal to final speed of the car v = 278 m/sec

Constant acceleration $a=5.56m/sec^2$

As the car starts initially starts from rest so initial velocity of the car u = 0 m/sec

From third equation of motion $v^2=u^2+2as$

Putting all values in equation

$278^2=0^2+2\times 5.56\times s$

s = 6950 m

So distance traveled during this acceleration will be 6950 m

3 0

3 years ago

a current of 180 mini amphere passes through a conductor for 5minute calculate the quantity of electricity transported

oksano4ka [1.4K]

Answer:

Explanation:

You can calculate the total electric charge that passes through the conductor as $q=It=(180\times 10^{-3})(5\times 60)= 54 C$ . It means that the number of electron that passes through the conductor is:

$n=\frac{q}{e}=\frac{54}{1.6\times 10^{-19}}=33.75\times 10^{19}$

8 0

1 year ago