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zalisa [80]
3 years ago
14

A car is being pulled by a tow truck. What is the car's mass if the net force on the car is 5400 N and it has an acceleration of

5.0 m/s/s?
Physics
1 answer:
Mazyrski [523]3 years ago
8 0

Answer:

1080 m

Explanation:

Divide 5400 over 5.0

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Suppose that a freely falling object were somehow equipped with an odometer. Would the readings of distance fallen each second i
maks197457 [2]

Answer:

1  greater distances fallen in successive seconds

Explanation:

When a body falls freely it is subjected to the action of the force of gravity, which gives an acceleration of 9.8 m / s2, consequently, we are in an accelerated movement

If we use the kinematic formula we can find the position of the body

       Y = Vo t + ½ to t2

Where the initial velocity is zero or constant and the acceleration is the acceleration of gravity

Y = - ½ g t2 = - ½ 9.8 t2 = -4.9 t2

Let's look for the position for successive times

t (s)      Y (m)

  1          -4.9

  2         -19.6

   3        -43.2

The sign indicates that the positive sense is up

It can be clearly seen that the distance is greatly increased every second that passes

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3 years ago
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WINSTONCH [101]
Check the attached file for the answer.

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3 years ago
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Which of the following statements about psychodynamic perspective is not true?
Free_Kalibri [48]
There are NO true statements on that list of choices.
3 0
3 years ago
3 In a television tube, an electron starting from rest experiences a force of 4.0 × 10−15 N over a distance of 50 cm. The final
MAXImum [283]

Answer:

The final speed of the electron = 2.095×10⁸ m/s

Explanation:

From newton's fundamental equation of dynamics,

F = ma ........................Equation 1

Where F = force, m = mass of the electron, a = acceleration of the electron.

making a the subject of the equation,

a = F/m.................... Equation 2

Given: F = 4.0×10⁻¹⁵ N,

Constant: m =  9.109×10⁻³¹ kg.

Substituting into equation 2

a = 4.0×10⁻¹⁵/9.109×10⁻³¹

a = 4.39×10¹⁶ m/s².

Using newton's equation of motion,

v² = u²+2as .......................... Equation 3

Where v = final velocity of the electron, u = initial velocity of the electron, a = acceleration of the electron, s = distance covered by the electron.

Given: u = 0 m/s(at rest), s = 50 cm = 0.5 m, a = 4.39×10¹⁶ m/s²

Substituting into equation 3

v² = 0² + 2(0.5)(4.39×10¹⁶)

v = √(4.39×10¹⁶)

v = 2.095×10⁸ m/s

Thus the final speed of the electron = 2.095×10⁸ m/s

7 0
3 years ago
An airplane is flying overhead at a constant elevation of 3000ft. A man is viewing the plane from a position 4000ft from the bas
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Answer: The distance between the man and the plane increasing at a rate of 400ft/s

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