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Lina20 [59]
2 years ago
10

An object is travelling at a velocity of 5m/s to the right for 5 seconds. If NO forces are applied to the object, what will the

velocity of the object be after 5 seconds.
Physics
1 answer:
creativ13 [48]2 years ago
6 0

Answer:

Below

Explanation:

We can use this formula to find the final velocity of the object :

     vf = vi + at

Now, because no force is applied to the object, the acceleration of this object will be 0 according to Newton's 1st law!

     vf = 5m/s + (0)(5)

     vf = 5 m/s

Hope this helps!

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vivado [14]
The answer is no I’ve not done that
6 0
2 years ago
It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing
babunello [35]

Answer:

3 km/h

Explanation:

Let's call the rowing speed in still water x, in km/h.

Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

8/(x - 2) + 8/(x + 2) = 48/5      (notice that: time = distance/speed)

Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)

8*(x+2) + 8*(x-2) =  (48/5)*(x² - 4)

Dividing  by 8

(x+2) + (x-2) = (6/5)*(x² - 4)

2*x = (6/5)*x² - 24/5

0 =  (6/5)*x² - 2*x - 24/5

Using quadratic formula

x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}

x = \frac{2 \pm 5.2}{2.4}

x_1 = \frac{2 + 5.2}{2.4}

x_1 = 3

x_2 = \frac{2 - 5.2}{2.4}

x_2 = -1\; 1/3

A negative result has no sense, therefore the rowing speed in still water was 3 km/h

7 0
3 years ago
If a melon has a a mass of 1 kg, how much does the melon weigh?
CaHeK987 [17]

Answer:

A. 10 N

Explanation:

the weight of melon is 1×10=10N

3 0
2 years ago
Read 2 more answers
Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy
8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV

Energy of the second orbit in H atom .

E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV

Energy of the third orbit in H atom .

E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV

Energy of the fifth orbit in H atom .

E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV

Energy of the sixth orbit in H atom .

E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV

Energy of the seventh orbit in H atom .

E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV

During an absorption of energy electron jumps from lower state to higher state.So,  absorption will take place in :

1) n = 2 → n = 3

2) n=  5 → n = 6

Energy absorbed when: n = 2 → n = 3

E=E_3-E_2

E=(-1.51 eV) -(-3.40 eV)=1.89 eV

Energy absorbed when: n = 5 → n = 6

E'=E_6-E_5

E'=(-0.378 eV)-(-0.544 eV) =0.166 eV

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0
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During which segments is kinetic energy decreasing?<br> A)1,2,3<br> B) 4,5<br> C)1,3,5<br> D)2,4
Nutka1998 [239]

Answer:

c

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