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d1i1m1o1n [39]
2 years ago
11

A mass of 10kg is a point a on a table is moved to a point B.if the line joining A and is horizantal what is the done on the obj

ect by the work done on the object by the gravitational force!?
!!!...explain your answer...!!!​
Physics
2 answers:
IrinaK [193]2 years ago
6 0

Answer:

work done on the object by gravitational force = 0 joules

Explanation:

work = force × displacement

force = mass × acceleration

so,

work = mass × acceleration × displacement

we know that mass= 10 kg , gravitational acceleration= 9.8 m/s^2

and displacement= 0 m since the object is not moving vertically.

so,

work = 10 × 9.8 × 0 = 0 joules

arsen [322]2 years ago
3 0

Answer:

<em>I </em><em>hope</em><em> it's</em><em> helpful</em><em> for</em><em> you</em><em>.</em>

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What’s is the movement of one object around another
allsm [11]

Answer:

revolution

Explanation:

8 0
3 years ago
A 20.0 μf capacitor is charged to a potential difference of 850 v. the terminals of the charged capacitor are then connected to
liberstina [14]
  • (a) Q = 1.70\times 10^{-2}\;\text{C};
  • (b) V_\text{final} = 5.31\times 10^{2}\;\text{V};
  • (c) E_\text{final} = 4.52\;\text{J};
  • (d) \Delta E = 2.82\;\text{J}.

All four values are in 3 sig. fig.

<h3>Explanation</h3>

(a)

Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}.

(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial};

(C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial};

\displaystyle V_\text{final} = \frac{C_1}{C_1+C_2}\cdot V_\text{initial}\\\phantom{V_\text{final}} = \frac{20.0\;\mu\text{F}}{20.0\;\mu\text{F} + 12.0\;\mu\text{F}} \times 850\;\text{V}\\\phantom{V_\text{final}} =531\;\text{V}.

(c)

\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}.

\displaystyle E_\text{final} = \frac{1}{2} (C_1 + C_2) \cdot {V_\text{final}}^{2} \\\phantom{E_\text{final}} = \frac{1}{2} \times (20.0\times 10^{-6} + 12.0\times 10^{-6}) \times 531.25\\\phantom{E_\text{final}} = 4.52\;\text{J}.

(d)

Initial energy of the system, which is the same as the initial energy in the 20.0\;\mu\text{F} capacitor:

\displaystyle E_\text{initial} = \frac{1}{2} \times 20.0\times 10^{-6} \times 850^{2} = 7.225\;\text{J}.

Change in energy:

\Delta E = 7.225\;\text{J} - 4.516\;\text{J} = 2.70\;\text{J}.

4 0
3 years ago
an electron and a 0.0320-kg bullet each have a velocity of magnitude 510 m/s, accurate to within 0.0100%. within what lower limi
MArishka [77]

for this we apply, Heisenberg's uncertainty principle.

it states that physical variables like position and momentum, can never simultaneously know both variables at the same moment.

the formula is,

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

by rearranging,

Δx = h / 4π * m(e).Δv

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 5.10*10^-2

Δx = 6.63*10^-34 / 583.9 X 10 ⁻³¹

Δx = 0.011 X 10⁻³

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.032*10^-31 * 5.10*10^-2

Δx = 6.63*10^-34 /2.05

Δx =3.23 X 10⁻³² m

therefore, we can say that the lower limits are   0.011 X 10⁻³  m for the electron and  3.23 X 10⁻³² m   for the bullet

To know more about bullet problem,

brainly.com/question/21150302

#SPJ4

6 0
1 year ago
Which of the following are ohmic materials? a. Batteries Wires Resistors Light bulb filaments b. Batteries, wires, resistors, an
Bingel [31]

Answer:

A., B., and C.

Explanation:

An Ohmic material is a material that obeys Ohm's Law, V = IR.

In contrast, a non-Ohmic material is one that does not obey Ohm's law.

Ohm's law states that the voltage across an electrical object is proportional to the current flowing through it, with the constant of proportionality being Resistance, R (in Ohm's).

The only Non-Ohmic material is the semiconductor, as semiconductors do not obey Ohm's law.

6 0
3 years ago
A 78.5-kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above w
Dima020 [189]

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

F_b be the buoyant force due to the air in the lung

V_a be the volume of air inside man's lungs

w_p be the weight that the buoyant force opposes as a result of the air.

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

F_b = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

4 0
3 years ago
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