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2 years ago

At what condition does a body becomes weightless at the equator?

Physics

1 answer:

ArbitrLikvidat [17]2 years ago

4 0

Answer:

The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface

Explanation:

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You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

3 years ago

You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the

podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

$F_x = Fcos30$

$F_y = Fsin30$

Now the normal force on the block is given as

$N = Fsin30 + mg$

$N = 0.5F + (18\times 9.8)$

$N = 0.5F + 176.4$

now the friction force on the cart is given as

$F_f = \mu N$

$F_f = 0.625(0.5F + 176.4)$

$F_f = 110.25 + 0.3125F$

now if cart moves with constant speed then net force on cart must be zero

so now we have

$F_f + F_x = 0$

$Fcos30 - (110.25 + 0.3125F) = 0$

$0.866F - 0.3125F = 110.25$

$F = 199.2 N$

so the force must be 199.2 N

8 0

3 years ago

12. AABC is a right triangle. If AB = 3 and AC = 7, find BC. Leave your answer in simplest radical form.

Dennis_Churaev [7]

Answer: A 2 square root 3

Explanation:

4 0

3 years ago

Consider electrons of kinetic energy 6.0 eV and 600 keV. For each electron, find the de Broglie wavelength, particle speed, phas

irinina [24]

Answer:

For 6.0 eV

0.5 nm, 1.45*10^6 m/s, 6.17*10^10 m/s, 1.45*10^6 m/s

For 600 eV

1.26*10^-3 nm, 2.66*10^8 m/s, 3.37*10^8 m/s, 2.66*10^8 m/s

Explanation:

See attachment for calculation

5 0

3 years ago

A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en

Brut [27]

Answer:

75 N

Explanation:

In this problem, the position of the crate at time t is given by

$y(t)=2.80t+0.61t^3$

The velocity of the crate vs time is given by the derivative of the position, so it is:

$v(t)=y'(t)=\frac{d}{dt}(2.80t+0.61t^3)=2.80+1.83t^2$

Similarly, the acceleration of the crate vs time is given by the derivative of the velocity, so it is:

$a(t)=v'(t)=\frac{d}{dt}(2.80+1.83t^2)=3.66t$ [m/s^2]

According to Newton's second law of motion, the force acting on the crate is equal to the product between mass and acceleration, so:

$F(t)=ma(t)$

where

m = 5.00 kg is the mass of the crate

At t = 4.10 s, the acceleration of the crate is

$a(4.10)=3.66\cdot 4.10 =15.0 m/s^2$

And therefore, the force on the crate is:

$F=ma=(5.00)(15.0)=75 N$

7 0

3 years ago