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sveticcg [70]
2 years ago
5

Six baseball throws are shown below. In each case the baseball is thrown at the same initial speed and from the same height h ab

ove the ground. Assume that the effects of air resistance are negligible. Rank these throws according to the speed of the baseball the instant before it hits the ground.
Physics
2 answers:
AveGali [126]2 years ago
8 0

Answer:

The final velocities of all the six balls will be same.

Explanation:

According to law of conservation of energy:

Gain in K.E = Loss in potential energy

   ½ mv^2 = mgh  

Where “m” and “g” are constant. The interchange in energies will occur only with the change in velocity and height. Since, balls are thrown from the same hight with the same initial velocity so, their final velocities will also be same just before striking the ground.

Assoli18 [71]2 years ago
6 0

Answer:

EXplained

Explanation:

from conservation of energy

change in potential energy = gain in kinetic energy

so as all he balls are throws from the same height thus the change in potential energy is the same for all the balls thus the gain in kinetic energy is the same for all the balls and as they have the same initial velocity thus the final velocity is the same for all the balls.

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A boat is headed with a velocity of 18 meters/second toward the west with respect to the water in a river. If the river is flowi
IrinaK [193]
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.
7 0
1 year ago
A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Volgvan

Answer:

A=0.199

Explanation:

We are given that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=\nu=1.2Hz

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Hence, the amplitude of oscillation=A=0.199

4 0
1 year ago
The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n
qaws [65]
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) 
3 0
2 years ago
A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit
makvit [3.9K]

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.

7 0
2 years ago
A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the
irina [24]

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=\frac{47.8g*1mol}{16g}

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:

2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M

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

molar mass=\frac{52.92 g}{5.975 mol}

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.

3 0
2 years ago
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