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3 years ago

The value of g is greater at the poles than at the equator why

Physics

1 answer:

babunello [35]3 years ago

6 0

Answer:

because the gravitational pull is maximum at the poles and decreases as it comes down toward the equator.

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Mr. Mehta cut several 275-centimeter lengths of wallpaper to put in his

mihalych1998 [28]

Answer:

2.75 meters.

Explanation:

1 meter = 100 centimeters.

There are 275 centimeters.

275/100=2.75

So, each piece of wallpaper was 2.75 meters long.

6 0

3 years ago

the total positive charge is QQQ = 1.62×10−6 CC , what is the magnitude of the electric field caused by this charge at point P,

balu736 [363]

Answer:

6.1 × 10^9 Nm-1

Explanation:

The electric field is given by

E= Kq/d^2

Where;

K= Coulombs constant = 9.0 × 10^9 C

q = magnitude of charge = 1.62×10−6 C

d = distance of separation = 1.53 mm = 1.55 × 10^-3 m

E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2

E= 14.58 × 10^3/2.4 × 10^-6

E= 6.1 × 10^9 Nm-1

8 0

3 years ago

The cross section of a copper strip is 1.2 mmthick and 20 mm wide. There is a 25-A current through this cross section, with the

Naily [24]

To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

$V_d = \frac{I}{nAq}$

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

$I = 25 A$

$A= 1.2*20 *10^{-6} m^2$

$q= 1.6*10^{-19}C$

$N = 8.47*10^{19} mm^{-3}$

$V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}$

$V_d = 7.68*10^{-5}m/s$

The hall voltage is given by

$V=\frac{IB}{ned}$

Where

B= Magnetic field

n = number of free electrons

d = distance

e = charge of electron

Then using the formula and replacing,

$V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}$

$V = 3.84*10^{-6}V$

5 0

3 years ago

A 81 kg person sits on a 3.8 kg chair. Each leg of the chair makes contact with the floor in a circle that is 1.2 cm in diameter

tankabanditka [31]

Answer:

1.9 MPa

Explanation:

Mass of person = 81 kg

Mass of chair = 3.8 kg

Diameter of contact point = 1.2 cm = D

Radius of contact point = 1.2/2 = 0.6 cm

Total mass of chair and person = 81 + 3.8 = 84.8 kg = M

Acceleration due to gravity = 9.81 m/s²

Force acting on the floor

F = Mg

⇒F = 84.8×9.81

⇒F = 831.888 N

Area of the contact point

A = πR²

⇒A = π0.006²

⇒A = π0.000036 m²

Area of the four points is

4A = 0.000144π m²

Pressure

$p=\frac{F}{A}\\\Rightarrow p=\frac{831.888}{0.000144\pi}\\\Rightarrow p=1838876.21\ Pa=1.83887621\times 10^6\ Pa=1.9 MPa$

Pressure exerted on the floor by each leg of the chair is 1.9 MPa

5 0

3 years ago

A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ

siniylev [52]

Answer:

$m=1.53kg$

Explanation:

To solve this problem we use the Hooke's Law:

$F=k*\Delta x$ (1)

F is the Force needed to expand or compress the spring by a distance Δx.

The spring stretches 0.2cm per Newton, in other words:

1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm

The force applied is due to the weight

$F=mg$

We replace in (1):

$mg=k*\Delta x$

We solve the equation for m:

$m=k*\Delta x/g=5*3/9.81=1.53kg$

7 0

3 years ago

The value of g is greater at the poles than at the equator why ​

The value of g is greater at the poles than at the equator why