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nadezda [96]
3 years ago
9

A: If a net force greater than 0 N is applied to an electron and to a proton, which one will accelerate more? The mass of an ele

ctron is roughly 1 2000 th that of a proton. A) The electron will accelerate more than the proton. B) The proton will accelerate more than the electron. C) Both the proton and the electron will move at a constant speed. D) Both the proton and electron will accelerate by the same amount.
Physics
2 answers:
aleksandrvk [35]3 years ago
8 0

according to newton's second law , net force on an object is the product of mass of the object and the acceleration of the object. the formula is given as

F = ma               where F = net force , m = mass and a = acceleration

so acceleration can be given as

a = F/m

for same net force , the acceleration depends on the mass of the object .

greater the mass , smaller will be the acceleration and vice versa.

Since the mass of electron is smaller as compared to the mass of proton, hence the electron will accelerate more as compared to proton.

A) The electron will accelerate more than the proton

navik [9.2K]3 years ago
4 0

A) <u>the electron will accelerate more than the proton.</u>

If the net force is greater than one, SOOO Newton’s second law of motion, an object will accelerate. The mass is inversely proportional to the acceleration. The lower the mass, the faster the acceleration,

hope this helps! :)

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An object covers a distance of 8 meters in the first second of travel, another 8 meters
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A man whose mass is 69 kg and a woman whose mass is 52 kg sit at opposite ends of a canoe 5 m long, whose mass is 20 kg. Suppose
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the canoe moved 1.2234 m in the water

Explanation:

Given that;

A man whose mass = 69 kg

A woman whose mass = 52 kg

at opposite ends of a canoe 5 m long, whose mass is 20 kg

now let;

x1 = position of the man

x2 = position of canoe

x3 = position of the woman

Now,

Centre of mass = [m1x1 + m2x2 + m3x3] / m1 + m2 + m3

= ( 69×0 ) + ( 52×5) + ( 20× 5/2) / 69 + 52 + 20

= (0 + 260 + 50 ) / ( 141 )

= 310 / 141

= 2.19858 m

Centre of mass is 2.19858 m

Now, New center of mass will be;

52 × 2.5 / ( 69 + 52 + 20 )

= 130 / 141

= 0.9219858 m  { away from the man }

To get how far, the canoe moved;

⇒ 2.5 + 0.9219858 - 2.19858

= 1.2234 m

Therefore, the canoe moved 1.2234 m in the water

5 0
3 years ago
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

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