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telo118 [61]
3 years ago
12

Imagine tying a string to a ball and twirling it around you. How is this similar to the moon orbiting Earth? In this example, wh

at is providing the constantly changing, inward force?
Physics
1 answer:
leonid [27]3 years ago
5 0
The string kind of acts like gravity 
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An astronaut weighs 104 newtons on the moon, where the strength of gravity is 1.6N/kg. What is her mass?
Leona [35]
Her mass is 65 kilograms
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3 years ago
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A car mass of 1.2 x 10 kilograms starts from rest and attains a speed of 20 meters/seconds in 5 seconds. What net force acted on
lorasvet [3.4K]

Net force on the car=F=4.8 x 10³ N

Explanation:

mass of car= 1.2 x 10³ Kg

initial velocity= Vi=0

Final velocity= Vf= 20 m/s

time = t= 5 s

Using kinematic equation,

Vf= Vi + at

20= 0 + a (5)

5 a=20

a= 20/5

a= 4 m/s²

Now force is given by F = ma

F= 1.2 x 10³ (4)

F=4.8 x 10³ N

7 0
3 years ago
Which statement is true regarding a distance vs. time graph? (1 point)
Nadusha1986 [10]

The statement that is true regarding a distance vs. time graph is option A: The graph should show distance on the vertical axis.

<h3>Where is the plot of distance?</h3>

How far an object has come in a certain amount of time is displayed on a distance-time graph. Time is represented on the X-axis and Distance is plotted on the Y-axis (left) (bottom).

On a distance-time graph, an object's motion is indicated by a sloping line. The slope or gradient of the line in a distance-time graph is equal to the object's speed. The object is travelling more quickly the steeper the line is (and the bigger the gradient).

Note that the distance-time graph shows the relationship between distance and time by plotting distance on the y-axis and time on the x-axis.

Learn more about distance vs. time graph from

brainly.com/question/16825120
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6 0
1 year ago
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g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p
kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r = \frac{mv}{qB}

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0
3 years ago
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The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on
alekssr [168]

Answer:

The change in temperature is \Delta T  = 1795 K

Explanation:

From the question we  are told that

   The temperature coefficient is  \alpha  =  4 * 10^{-3 }\  k^{-1 }

The resistance of the filament is mathematically represented as

           R  =  R_o [1 + \alpha  \Delta T]

Where R_o is the initial resistance

Making the change in temperature the subject of the formula

     \Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]

Now from ohm law

           I = \frac{V}{R}

This implies that current varies inversely with current so

           \frac{R}{R_o} = \frac{I_o}{I}

Substituting this we have

       \Delta T  = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]

From the question we are told that

    I  = \frac{I_o}{8}

Substituting this we have

   \Delta T  = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]

=>     \Delta T  = \frac{1}{3.9 * 10^{-3}} (8 -1 )

        \Delta T  = 1795 K

6 0
3 years ago
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