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dmitriy555 [2]
2 years ago
13

The skateboard stops after covering some distance bwhich law of newton is it ?

Physics
1 answer:
MrRa [10]2 years ago
6 0

Friction... Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force

Means it will keep going until acted on by an outside force, in this case friction

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A hummingbird can a flutter its wings 4800 times per minute what is the frequency of wing flutters per second
Scilla [17]

the answer i got is 80 i hope this helpss!!!!

7 0
2 years ago
Read 2 more answers
A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20.0ºC to 80.0ºC. (a) How much heatis required? Wh
Jet001 [13]

Answer:

heat used to rise temperature pan =  30.1%

heat used to rise temperature water =  69.9%

Explanation:

Given data

mass of water = 0.250 liter = 0.250 kg

aluminum pan mass = 0.500 kg

initial temperature = 20.0ºC

final temperature =  80.0ºC

to find out

heat used to rise temperature of  pan and water

solution

we find here heat transferred to the water that is

heat transferred to the water = mass of water × specific heat of water × change in temperature    ...........1

specific heat of water is 4186 J/kgºC

so

heat transferred to the water = 0.250 × 4186 × (80-20) kJ

heat transferred to the water = 62.8 kJ

and

heat transferred to the aluminum that is

heat transferred to the aluminum = mass of aluminum × specific heat of aluminum × change in temperature    ...........2

here specific heat of aluminum is 900 J/kgºC

heat transferred to the aluminum = 0.500 × 900 × (80-20) kJ

heat transferred to the aluminum = 27 kJ

so

total heat = 62.8 + 27 = 89.8 kJ

so

heat used to rise temperature pan = 27/89.8 ×100% = 30.1%

heat used to rise temperature water = 62.8 / 89.8 ×100% = 69.9%

8 0
2 years ago
To what potential should you charge a 2.0 μF capacitor to store 1.0 J of energy?
Bess [88]
E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V

You should charge it to 1000 volts to store 1.0 J of energy.
6 0
2 years ago
In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053nm around a stationary proton. par
Alenkasestr [34]
<span>Here the force that is applied between the electron and proton is centripetal, so equate the two forces to determine the velocity.
 We know charge of the electron which for both Q1 and Q2, e = 1.60 x 10^-19 C
 The Coulombs Constant k = 9.0 x 10^9
 Radius r = 0.053 x 10^-9m = 5.3 x 10^-11 m
  Mass of the Electron = 9.11 x 10^-31
  F = k x Q1 x Q2 / r^2 = m x v^2 / r(centripetal force)
  ke^2 / r^2 = m x v^2 / r => v^2 = ke^2 / m x r
 v^2 = ((1.60 x 10^-19)^2 x 9.0 x 10^9) / (9.11 x 10^-31 x 5.3 x 10^-11 )
 v^2 = 4.77 x 10^12 = 2.18 x 10^6 m/s
 Since one orbit is the distance,
  one orbit = circumference = 2 x pi x r; distance s = v x t.
 v x t = 2 x pi x r => t = (2 x 3.14 x 5.3 x 10^-11) / (2.18 x 10^6)
  t = 33.3 x 10^-11 / 2.18 x 10^6 = 15.27 x 10^-17 s
 Revolutions per sec = 1 / t = 1 / 15.27 x 10^-17 = 6.54 x 10^15 Hz</span>
6 0
2 years ago
A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant
OverLord2011 [107]

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

a_t = ar

where;

a is the angular acceleration and

r is the radius of the circular path

a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2

Determine time of the rotation;

\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2

The magnitude of total linear acceleration is given by;

a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27  \ m/s^2

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

5 0
2 years ago
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