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4 years ago

If you want to lift a 5-kg box to a height of 2 meters, how much work must be

Physics

2 answers:

kramer4 years ago

7 0

Answer:100joules

Explanation:

Work= force×distance

Work=mass×acceleration×distance

Work=5×10×2

Work=100joules

grandymaker [24]4 years ago

6 0

Answer:

98J is the correct answer

Explanation:

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Which example involves the transformation of chemical energy directly into light energy?

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burning sodium or magnesium

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3 years ago

What are the six different character traits scientists should have?

frozen [14]

careful observation, curiosity, logic, creativity, skepticism, and objectivity

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3 years ago

Read 2 more answers

Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of

lions [1.4K]

Answer:

Explanation:

Magnitude of force per unit length of wire on each of wires

= μ₀ x 2 i₁ x i₂ / 4π r where i₁ and i₂ are current in the two wires , r is distance between the two and μ₀ is permeability .

Putting the values ,

force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )

= .67 i² x 10⁻⁴

force on 3 m length

= 3 x .67 x 10⁻⁴ i²

Given ,

8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²

i² = 3.98 x 10⁻²

i = 1.995 x 10⁻¹

= .1995

= 0.2 A approx .

2 i = .4 A Ans .

6 0

3 years ago

In a hydraulic lift, if the pressure exerted on the liquid by one piston is increased by 100 N/m2 , how much additional weight c

shepuryov [24]

Answer:

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Explanation:

By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:

Hydraulic Lift - Before change

$P = \frac{F}{A}$

Hydraulic Lift - After change

$P + \Delta P = \frac{F + \Delta F}{A}$

Where:

$P$ - Hydrostatic pressure, measured in pascals.

$\Delta P$ - Change in hydrostatic pressure, measured in pascals.

$A$ - Cross sectional area of the hydraulic lift, measured in square meters.

$F$ - Hydrostatic force, measured in newtons.

$\Delta F$ - Change in hydrostatic force, measured in newtons.

The additional weight is obtained after some algebraic handling and the replacing of all inputs:

$\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}$

$\Delta P = \frac{\Delta F}{A}$

$\Delta F = A\cdot \Delta P$

Given that $\Delta P = 100\,Pa$ and $A = 25\,m^{2}$ , the additional weight is:

$\Delta F = (25\,m^{2})\cdot (100\,Pa)$

$\Delta F = 2500\,N$

The additional mass needed for the additional weight is:

$\Delta m = \frac{\Delta F}{g}$

Where:

$\Delta F$ - Additional weight, measured in newtons.

$\Delta m$ - Additional mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

If $\Delta F = 2500\,N$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }$

$\Delta m = 254.92\,kg$

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

3 0

3 years ago

While visiting the planet Mars, Moe leaps straight up off the surface of the Martian ground with an initial velocity of 105 m/s

yawa3891 [41]

Answer:

L = 0 m

Therefore, the cricket was 0m off the ground when it became Moe’s lunch.

Explanation:

Let L represent Moe's height during the leap.

Moe's velocity v at any point in time during the leap is;

v = dL/dt = u - gt .......1

Where;

u = it's initial speed

g = acceleration due to gravity on Mars

t = time

The determine how far the cricket was off the ground when it became Moe’s lunch.

We need to integrate equation 1 with respect to t

L = ∫dL/dt = ∫( u - gt)

L = ut - 0.5gt^2 + L₀

Where;

L₀ = Moe's initial height = 0

u = 105m/s

t = 56 s

g = 3.75 m/s^2

Substituting the values, we have;

L = (105×56) -(0.5×3.75×56^2) + 0

L = 0 m

Therefore, the cricket was 0m off the ground when it became Moe’s lunch.

5 0

3 years ago