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scoundrel [369]
3 years ago
8

On June 9, 1983, the lower part of the Variegated Glacier in Alaska was observed to be moving at a rate of 64 m per day. What is

this speed in kilometers per hour?
Physics
1 answer:
myrzilka [38]3 years ago
3 0

(64 meter/day) x (1 kilometer / 1,000 meters) x (1 day / 24 hours) =

       (64 x 1 x 1) / (1,000 x 24)         (kilometer/hour)  = 

                     1/375  km/hour

                (0.002667 km/hour, rounded)

                 2-2/3 meters/hour

                 (8-ft 9-inches) per hour
 
 
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Please help need this asap
NikAS [45]

Answer:

A)  350 N

B)  58.33 N

C)  35 kg

D)  35 kg

Explanation:

If we use that g = 10 m/s^2, then the acceleration of gravity on the Moon will be 10/6 m/s^2 = 5/3 m/s*2

The weight of the object on Earth is given by:

Weight = mass * g = 35 * 10 = 350 N

The weight of the object on the Moon:

Weight = mass * gmoon = 35 * 5/3 = 58.33 N

The mass of the object on Earth is 35 kg

The mass of the object on the Moon is exactly the same as on the Earth (35 kg) since the mass is a quantity inherent to the object and not to its location.

3 0
2 years ago
Which component of weight is cause for S.H.M of bob simple pendulum
jeka57 [31]

Explanation:

The Simple Pendulum. A simple pendulum is defined to have a point mass, also known as the pendulum bob, which is suspended from a string of length L with negligible mass ((Figure)). Here, the only forces acting on the bob are the force of gravity (i.e., the weight of the bob) and tension from the string.

hope it helps you

6 0
2 years ago
If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?
SVETLANKA909090 [29]

B

Assume north and east as two sides of a right angled triangle. magnitude of the distance is then given by the length of the hypotenuse which is \sqrt{a^2 + b^2}

where a = 1.2 km north

and b = 1.6 km east

magnitude = 2 km

Direction is given by the angle between them, that is atan(a/b) = 36.86 deg north of east = 53.1 deg east of north.

8 0
3 years ago
A car owner forgets to turn off the headlights of his car while it is parked in his garage. If the 12.0-V battery in his car is
avanturin [10]

Answer: 22.6 hours

Explanation:

The power is the measure of the rate of energy.

In this problem, the 12.0 V battery is rated at 51.0 Ah, which means it delivers 51.0 A of current in a time of t = 1 h = 3600 s. The power delivered by the battery can be written as

P=IV

where

I is the current

V = 12.0 V is the voltage of the battery

So the energy delivered by the battery can be written as

E=Pt=VIt

Where

It=51.0 A\cdot h = 51.0 A \cdot 3600 s/h=183,600 A\cdot s

So the energy delivered is

E=(12.0)(183,600)=2.2\cdot 10^6 J

At the same time, the headlight consumes 27.0 W of power, so 27 Joules of energy per second; Therefore, it will remain on for a time of:

t=\frac{2.2\cdot 10^6 J}{27.0 W}=81481 s = 22.6 h

3 0
3 years ago
In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point
Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

=  (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0
3 years ago
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