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ludmilkaskok [199]
4 years ago
12

An object moving north with an initial velocity of 14 m/s accelerates 5 m/s^2 for 20 seconds.what is the final velocity of the o

bject?
39 m/s
90 m/s
114 m/s
414m/s
Physics
2 answers:
bekas [8.4K]4 years ago
5 0

Answer:

Option C = 114 m/s

Explanation:

a=v-u/t

By substituting,

5=v-14/20

100=v-14

Thus, v=100+14

v= 114 m/s

Hope it helps :)

ELEN [110]4 years ago
5 0
The answer is 114 m/s good luck!!!
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Identify the palindromic base sequence in the DNA donor molecule shown.
Airida [17]

Explanation:

Since a double helix is formed by two paired antiparallel strands of nucleotides that run in opposite directions, and the nucleotides always pair in the same way (adenine (A) with thymine (T) in DNA or uracil (U) in RNA; cytosine (C) with guanine (G)), a (single-stranded) nucleotide sequence is said to be a palindrome

6 0
3 years ago
Suppose two point charges, Q1 = -6.25 x 10-9 C and Q2 = -6.25 x 10-9 C, are separated by a distance d = 0.617 m.
n200080 [17]

Answer:

The answer to your question is the letter A) F =  9.23 x 10⁻⁷ N

Explanation:

Data

q₁ = -6.25 x 10⁻⁹ C

q₂ = -6.25 x 10⁻⁹ C

d = 0.617 m

k = 9 x 10⁹ Nm²/C²

F = ?

Formula

              F = k q₁q₂ /r²

-Substitution

              F = (9 x 10⁹)(-6.25 x 10⁻⁹)(-6.25 x 10⁻⁹) / (0.617)²

-Simplification

              F = 3.512 x 10⁻⁷ / 0.381

-Result

              F = 9.227 x 10⁻⁷ N  ≈ 9.23 x 10⁻⁷ N

8 0
3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
4 years ago
Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite
nalin [4]

Explanation:

Let us assume that the mass of a pitched ball is 0.145 kg.

Initial velocity of the pitched ball, u = 47.5 m/s

Final speed of the ball, v = -51.5 m/s (in opposite direction)

We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :

\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s

So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.

Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :

J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s

Hence, this is the required solution.

7 0
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Vinvika [58]
If the compressor removes 500 j if the 700 the 200 j left would have Ben conducted through the refrigerator/200 j released into the room
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