If the boat's velocity is 18m/sec relative to the water in the river and not the shore, it would need to be added the river speed of 2.5m/sec to get a total of 20.5m/sec. The 20.5m/sec would then be the total velocity of the boat relative to the shore. From personal experience, I know that when one runs with the tide, one is adding the tide flow speed to one's boat speed (what it would be in neutral waters) to get a sometimes much faster speed.
Answer:
A=0.199
Explanation:
We are given that
Mass of spring=m=450 g=
Where 1 kg=1000 g
Frequency of oscillation=
Total energy of the oscillation=0.51 J
We have to find the amplitude of oscillations.
Energy of oscillator=
Where
=Angular frequency
A=Amplitude

Using the formula



Hence, the amplitude of oscillation=A=0.199
Felectric = q*E
<span> Ftranslational = m*a
</span><span> Felectric = Ftranslational
</span> <span>q*E = m*a
</span><span> Solve for a
</span><span> a = q/m*E </span>
<span> Our sign convention is "up is positive"
</span><span> q = 1.6*10^-19 C
</span><span> m = 1.67*10^-27 kg
</span><span> E = -150 N/C (- because it is down and up is positive)
</span> a =<span>
-6,4*10^5</span><span> m/s^2 (downward)
</span> answer
a = -6,4*10^5 m/s^2 (downward)
Answer:
Final velocity of the block = 2.40 m/s east.
Explanation:
Here momentum is conserved.
Initial momentum = Final momentum
Mass of bullet = 0.0140 kg
Consider east as positive.
Initial velocity of bullet = 205 m/s
Mass of Block = 1.8 kg
Initial velocity of block = 0 m/s
Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s
Final velocity of bullet = -103 m/s
We need to find final velocity of the block( u )
Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u
We have
2.87 = -1.442 + 1.8 u
u = 2.40 m/s
Final velocity of the block = 2.40 m/s east.
Answer:
The atomic weight in g/mole of the metal (molar mass) is 8.87.
Explanation:
To begin, it is possible to assume that, as a sample, it has 100 g of the compound. This means that:
- 52.92% metal: 52.92 g M
- 47.80% oxygen: 47.80 g O
Using the molar mass of oxygen, which is 16 g / mol, it is possible to calculate the amount of moles of oxygen present in the sample using the rule of three:

moles of oxygen=2.9875
The chemical formula of metal oxide tells you that:
2 M⁺³ + 3 O²⁻ ⇒ M₂O₃
In the previous equation you can see that you need 3 oxygen anions to react with two metal cations. Then:

You have 52.92 g of metal in the sample, then the molar mass of the metal is:

molar mass≅ 8.87 g/mol
<u><em> The atomic weight in g/mole of the metal (molar mass) is 8.87.</em></u>
The closest match to this value is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.