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Vitek1552 [10]
2 years ago
10

A tin can whirled on the end of a string moves in a circle because

Physics
1 answer:
Ilya [14]2 years ago
8 0

Answer:

There is an inward force acting on the can

Explanation:

This inward force is known as Centripetal force and it is responsible for making the can whirl on the end of a string in circle and it is also directed towards the center around which the can is moving.

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In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric
Viktor [21]

To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are

Comet mass M = 1.0*10^{13} kg

Radius r = 1.6km = 1600 m

Rock was dropped from a height 'h' from surface = 1m

The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is

g = \frac{GM}{R^2}

Where G means gravitational universal constant and M the mass of the planet

g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}

g = 2.607*10^{-4} m/s^2

Now calculate the value of the time

h = \frac{1}{2} gt^2

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}

t = 87.58s

The time taken for the rock to reach the surface is t = 87.58s

8 0
2 years ago
An object begins at position x = 0 and moves one-dimensionally along the x-axis witļi a velocity v
Liula [17]

Answer:

The answer is "between 20 s and 30 s".

Explanation:

Calculating the value of positive displacement:

\ (x_{+ve}) = \frac{1}{2} \times 15 \times  20 \\\\

          = \frac{1}{2} \times 300 \\\\=  150 \\\\

Calculating the value of negative displacement upon the time t:

(x_{-ve}) = \frac{1}{2} \times 5 \times 20- 20(t-20) \\\\

          = \frac{1}{2} \times 100- 20t+ 400 \\\\= 50- 20t+ 400 \\\\

\to X= X_{+ve} + X_{-ve} \\\\

\to  150 - 50 -20t+400 =0\\\\\to 100 -20t+400 =0 \\\\\to 500 -20t =0\\\\\to 20t =500 \\\\\to t=\frac{500}{20}\\\\\to t=\frac{50}{2}\\\\\to t= 25

That's why its value lie in "between 20 s and 30 s".

6 0
2 years ago
Most workers in nanotechnology are actively monitored for excess static charge buildup. the human body acts like an insulator as
irga5000 [103]
Sometimes arithmetic problems can be solved much more easily using the dimensional analysis approach. You focus on the units of the given information. Then, you manipulate them applying the laws of algebra where like units cancel, in order to end up with the unit of the unknown.

Given:
-50 nc/step
31 steps
Unknown: charge

Thus,
Charge = -50 nc/step * 31 steps =<em> -1550 nc</em>
7 0
2 years ago
A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
dlinn [17]

Answer:

a = 4.72 m/s²  

Explanation:

given,

mass of the box (m)= 6 Kg

angle of inclination (θ) = 39°

coefficient of kinetic friction (μ) = 0.19

magnitude of acceleration = ?

box is sliding downward so,

F - f = m a                        

f is the friction force

m g sinθ - μ N = ma                        

m g sinθ - μ m g cos θ = ma            

a = g sinθ - μ g cos θ                    

a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°

a = 4.72 m/s²                                      

the magnitude of acceleration of the box down the slope is a = 4.72 m/s²  

3 0
2 years ago
An automobile being tested on a straight road is 400 feet from its starting point when the stopwatch reads 8.0 seconds and is 55
m_a_m_a [10]

Answer:

a)   v = 75 ft / s , b)  v = 55 ft / s , c)   Δx = 1000 ft

Explanation:

We can solve this exercise with the expressions of kinematics

a) average speed is defined as the distance traveled in a given time interval

        v = (x₂-x₁) / (t₂-t₁)

         v = (550 - 400) / (10 -8)

         v = 75 ft / s

b) we repeat the calculations for this interval

   v = (550 - 0) / (10 -0)

   v = 55 ft / s

c)  we clear the distance from the average velocity equation

     Δx = v (t₂ -t₁)

     Δx = 100 (20-10)

     Δx = 1000 ft

4 0
2 years ago
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