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Pavlova-9 [17]
2 years ago
6

Suppose you ran 53 km in 67 min. With what speed did you run?

Physics
2 answers:
jekas [21]2 years ago
8 0

0.8 km/h

speed = distance / time

melomori [17]2 years ago
4 0

Answer:

47.46 kilometers an hour

Explanation:

hope this helped

You might be interested in
How much heat will be needed to warm 187 grams of water from 10 0C to 90 0C?
kvv77 [185]
<h3>Hello there!</h3>

Here, you are looking for the amount of heat put in for water, at a mass of 187 grams, to change by 80 degrees.

The equation commonly accepted to find the answer to questions like these is the specific heat formula.

The equation is Q = mc∆T, where Q is the amount of energy put in to raise the temperature by a certain amount, m is the mass, c is the specific heat capacity, and ΔT is the amount of temperature change.

The information given:

m = 187 grams

c = specific heat capacity of water, or in this case 1 calorie, or 4.184 joules (which is what we will be using)

ΔT = 80 degrees

Now just plug everything in to solve.

Q = 187 * 4.184 * 80

Q = 62592.64

So you have your answer: 62592.64 joules.

Hope this helped!

5 0
3 years ago
A block of amber is placed in water and a laser beam travels from the water through the amber. The angle of incidence is 35 degr
7nadin3 [17]
Answer: 1.88

Explanation

Applying Snell’s Law, sin(1)/sin(2) = n(2)/n(1), where n is the index of refraction and sin 1 and 2 being of incidence and refracted respectively.

1) sin35/sin24 = n(2)/1.33
2) 1.41 = n(2)/1.33
3) n(2) = 1.41 x 1.33
4) n(2) = 1.88

Hope this helps :)
7 0
2 years ago
If an object has a mass of 38 kg, what is its approximate weight on earth?
klasskru [66]
38*10=380 N
To be more exact, 38 should be multiplied by 9.8 instead of 10.
3 0
3 years ago
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 12.8 m/s2 . At t1 the rocket e
Simora [160]

Answer:4.39 s

Explanation:

Given

initial velocity u=0

acceleration a=12.8 m/s^2

velocity acquired by sled in t_1 time

v=0+at

v=12.8t_1

distance traveled by sled in t_1 s

v^2-u^2=2as

(12.8t_1)^2-0=2\times 12.8\times s_1

s_1=6.4\cdot t_1^2

distance traveled in t_2 time with velocity v=12.8t_1

s_2=v\times t_2

s_2=12.8\times t_1\times t_2

s_2=12.8\cdot t_1\cdot t_2

s_1+s_2=5.37\times 10^3

6.4t_1^2+12.8t_1t_2=5370----1

t_1+t_2=97.7 s

t_2=97.7-t_1

substitute the value of t_2 in 1

we get

6.4t_1^2-1250.56t_1+5370=0

thus t_1=\frac{1250.56-1194.33}{12.8}=4.39 s

t_1=4.39 s

5 0
3 years ago
What's a difference between mercury and oxygen?<br> (They both are elements)
borishaifa [10]

--  There are 80 protons in the nucleus of every atom of Mercury,
but only 8 of them in the nucleus of an atom of Oxygen.

-- Mercury must be warmer than 357°C in order to boil, but Oxygen
must only be warmer than -183°C.

--  Mercury must be colder than -39°C in order to freeze, but Oxygen
must be colder than -219°C.

--  Oxygen is required for human life.  Mercury is a deadly poison.

8 0
3 years ago
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