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Sunny_sXe [5.5K]
3 years ago
14

If a 4.5 kg object is dropped from a height of 6.0 m, what will be its velocity when it is halfway toward the ground? (Use g = 9

.80 m/s2, and ignore air resistance.) 7.7 m/s 11 m/s 16 m/s 29 m/s
Physics
2 answers:
dangina [55]3 years ago
7 0
When an object falls or is dropped from rest it's initial velocity is zero.
Using the equations for a motion in straight line. I can find the time it takes to reach 3.0 m down (half way).
x = vt - 4.9t²
-3 = 0 - 4.9t²
-3/-4.9 = t²
0.6122 = t²
0.7825 sec = t

v = v - gt
v = 0 - 9.8(0.7825)
v = -7.67 m/s
the negative denotes downward direction.

You  could also solve the problem using potential and kinetic energy.

Since it starts with maximum PE and gets converted to KE when it hits the ground. mgh = mv²/2
mass cancels, use 3 meters for the halfway distance
-9.8(-3) = v²/2
29.4 * 2 = v²
√(58.8) = 7.67 m/s downwards
notsponge [240]3 years ago
6 0

Answer:

7.7 rounded

Explanation:

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Electric field.
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How much is 1.75 inches in cm?
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4.44 cm

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A very good insulator has an R-value of 29. The material's heat transfer coefficient is
Fittoniya [83]

Answer:

b. 0.034

Explanation:

The heat transfer coefficient of a material (U-value) is equal to the reciprocal of its R-value, therefore:

U = \frac{1}{R}

where

R is the R-value of the material

For the insulator in this problem,

R = 29

Substituting into the equation, we find the heat transfer coefficient:

U=\frac{1}{29}=0.034

4 0
2 years ago
A child is riding in a child-restraint chair, securely fastened to the seat of a car. Assume the car has speed 47 km/h when it h
Elina [12.6K]

Answer: F = 1235 N

Explanation: Newton's Second Law of Motion describes the effect of mass and net force upon acceleration: F_{net}=m.a

Acceleration is the change of velocity in a period of time: a=\frac{\Delta v}{\Delta t}

Velocity of the car is in km/h. Transforming it in m/s:

v=\frac{47.10^{3}}{36.10^{2}}

v = 13 m/s

At the moment the car decelerates, acceleration is

a=\frac{13}{0.2}

a = 65 m/s²

Then, force will be

F_{net}=19(65)

F_{net} = 1235 N

The horizontal net force the straps of the restraint chair exerted on the child to hold her is 1235 newtons.

5 0
2 years ago
Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each
AURORKA [14]

Answer: A (\sqrt{61},309.8°)

              B (2\sqrt{2}, 315°)

             C (3\sqrt{5}, 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

r=\sqrt{x^{2}+y^{2}} and \theta=tan^{-1}(\frac{y}{x})

For point A:

r=\sqrt{(-5)^{2}+6^{2}}

r=\sqrt{61}

\theta=tan^{-1}(\frac{6}{-5})

\theta=tan^{-1}(-1.2)

\theta=-50.2°

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

\theta=360-50.2

\theta= 309.8°

Polar coordinates for point A is (\sqrt{61}, 309.8°)

For point B:

r=\sqrt{2^{2}+(-2)^{2}}

r=\sqrt{8}

r=2\sqrt{2}

\theta=tan^{-1}(\frac{-2}{2} )

\theta=tan^{-1}(1)

\theta=-45°

Point B is in IV quadrant, so:

\theta=360-45

\theta= 315°

Polar coordinates for point B is (2\sqrt{2}, 315°)

For point C:

r=\sqrt{(-6)^{2}+(-3)^{2}}

r=\sqrt{45}

r=3\sqrt{5}

\theta=tan^{-1}(\frac{-3}{-6} )

\theta=tan^{-1}(0.5)

\theta= 26.56°

Polar coordinates for point C is (3\sqrt{5}, 26.56°)

3 0
3 years ago
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