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Alika [10]
3 years ago
7

A box experiencing a gravitational force of 600 N is being pulled to the right with force of 250 N. A 25 N frictional force acts

on the box as it moves to the right.
Physics
1 answer:
bearhunter [10]3 years ago
6 0

Answer: 0 NEWTONS

Explanation:

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Step2247 [10]
It accelerates in the y component (bc of gravity) AND the x-component (b/c of the friction force).
5 0
3 years ago
How does the period affects the centripetal force?
Anna71 [15]

Answer:According to the Equation (2), centripetal force is proportional to the square of the speed for an object of given mass M rotating in a given radius R.

Explanation:The Period T. The time T required for one complete revolution is called the period. For. constant speed. v = 2π r T holds.

8 0
3 years ago
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For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

4 0
3 years ago
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An electric Kettle is rated at 25 W. Calculate the quantity of heat generated in 2s
MatroZZZ [7]

Answer:

Energy consumed by the electric kettle in 9.5 min =Pt=(2.5×10

3

)×(9.5×60)=14.25×10

5

J

Energy usefully consumed =msΔT=3×(4.2×10

3

)×(100−15)=10.71×10

5

where s=4.2J/g

o

C= specific heat of water and boiling point temp=100

o

C

Heat lost =14.25×10

5

−10.71×10

5

=3.54×10

5

4 0
3 years ago
Radioactive isotopes can be used to find the age of rocks, fossils, or other artifacts. Carbon-14 has a half-life of 5,730 years
bazaltina [42]

Answer:

1/8 = (1/2)^3

This implies the sample has decayed for 3 half lives

3 * 5730 yrs = 17,200 years

8 0
2 years ago
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