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nika2105 [10]
3 years ago
6

Ms. Kasper is in a panic. Her cat, Penny, is stuck in a tree and about to jump out. In order to save her cat, Ms. Kasper needs t

o run to the tree, 12 meters away. If it takes her cat, 3 seconds to fall, how fast would Ms. Kasper have to run to save her cat?
Physics
1 answer:
Misha Larkins [42]3 years ago
3 0

Answer:

Speed greater than 4 m/s

Explanation:

Given that Ms. Kasper is in a panic. Her cat, Penny, is stuck in a tree and about to jump out. In order to save her cat, Ms. Kasper needs to run to the tree, 12 meters away. If it takes her cat, 3 seconds to fall, how fast would Ms. Kasper have to run to save her cat?

The distance = 12 m

Time = 3s

Speed = distance/time

Speed = 12/3

Speed = 4 m/s

Ms Kasper must run at speed more than 4m/s for her to save the cat.

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Newtown third law applies to blank of objects​
fomenos

Answer:

All

Explanation:

I'm not sure what you meant but Newton's third law which basically states that every action has an equal and opposite reaction applies to <em>all</em> objects. So I think the answer is all.

8 0
3 years ago
What happens when an atom gives up loosely held valence electrons to another atom​
vazorg [7]

Answer:

The number of valence electrons increases to 8 or, the atom gives up loosely held valence electrons.

Explanation:

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3 0
3 years ago
A simple pendulum has length of 820mm. Calculate the frequency (g = 9.8 ms -2)<br>​
Vadim26 [7]

Answer:

\huge\boxed{\sf f=0.55 \ Hz}

Explanation:

<u>Given Data:</u>

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

<u>Required:</u>

Frequency = f = ?

<u>Formula:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }

<u>Solution:</u>

\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}

7 0
2 years ago
1)Determine, in terms of unit vectors, the resultant of the five forces illustrated in the figure, Consider F1=20 N, F2= 12 N, F
LiRa [457]

Explanation:

1) F₁ lies in a plane perpendicular to the xy plane, 60° from the x axis.  The angle between F₁ and the +z axis is 30°.  Therefore, the vector is:

<F₁> = 20 (sin 30° cos 60° i + sin 30° sin 60° j + cos 30° k)

<F₁> = 20 (¼ i + ¼√3 j + ½√3 k)

<F₁> = 5 i + 5√3 j + 10√3 k

F₂ is in the xy plane.  Its slope is -24/7.  The vector is:

<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)

<F₂> = -3.36 i + 11.52 j

F₃ is parallel to the +x axis.  The vector is:

<F₃> = 17 (i + 0 j + 0 k)

<F₃> = 17 i

F₄ is parallel to the -z axis.  The vector is:

<F₄> = 15 (0 i + 0 j − k)

<F₄> = -15 k

F₅ is in the xy plane.  It forms a 15° angle with the -y axis.  The vector is:

<F₅> = 9 (-sin 15° i − cos 15° j + 0 k)

<F₅> = -9 sin 15° i − 9 cos 15° j

The resultant vector is therefore:

<F> = (5 − 3.36 + 17 − 9 sin 15°) i + (5√3 + 11.52 − 9 cos 15°) j + (10√3 − 15) k

<F> = 16.31 i + 11.49 j + 2.32 k

2) Sum of forces at point B in the x direction:

∑F = ma

Tbc cos 40° − ¹⁵/₁₇ Tab = 0

Tbc cos 40° = ¹⁵/₁₇ Tab

Tbc = 1.15 Tab

Sum of forces at point B in the y direction:

∑F = ma

Tbc sin 40° + ⁸/₁₇ Tab − mAg = 0

Tbc sin 40° + ⁸/₁₇ Tab = (2 kg) (10 m/s²)

(1.15 Tab) sin 40° + ⁸/₁₇ Tab = 20 N

1.21 Tab = 20 N

Tab = 16.52 N

Tbc = 19.02 N

Sum of forces at point C in the x direction:

∑F = ma

Tcd sin 25° − Tbc cos 40° = 0

Tcd sin 25° = Tbc cos 40°

Tcd = 1.81 Tbc

Tcd = 34.48 N

3(a) When the crane is on the verge of tipping, the center of gravity is directly over point F.  Relative to point A:

3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm

1336 kgm = 1530 kgm cos θ

θ = 29.17°

3(b) 3.7 m = [ (390 kg) (0.9 m) + (90 kg) (x + 1.7 m) + (80 kg) (x + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + (90 kg) x + 153 kgm + (80 kg) x + 232 kgm

1336 kgm = (170 kg) x

x = 7.86 m

4) Find the lengths of the cables.

Lab = √((2 m)² + (3 m)² + (5 m)²)

Lab = √38 m

Lac = √((2 m)² + (3 m)² + (5 m)²)

Lac = √38 m

Lde = √((2 m)² + (3 m)²)

Lde = √13 m

Sum of forces in the x direction:

∑F = ma

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Sum of forces in the y direction:

∑F = ma

2/√38 Fab − 2/√38 Fac = 0

Fab = Fac

Sum of forces in the z direction:

∑F = ma

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

Sum of moments about the y-axis:

∑τ = Iα

(3/√38 Fab) (5 m) + (3/√38 Fac) (5 m) + (3/√13 Fde) (2 m) − (mg) (2 m) = 0

Substitute Fab = Fac and simplify:

6/√38 Fab + 3/√13 Fde − mg = 0

30/√38 Fab + 6/√13 Fde − 2mg = 0

Double first equation:

12/√38 Fab + 6/√13 Fde − 2mg = 0

Subtract from the second equation:

28/√38 Fab = 0

Fab = 0

Fac = 0

Solve for Fde:

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

3/√13 Fde = mg

3/√13 Fde = (1.7 kg) (10 m/s²)

Fde = 20.43 N

Solve for Rx:

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Rx = 2/√13 Fde

Rx = 11.33 N

8 0
3 years ago
Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a
Viktor [21]

Answer:

Power, P = 924.15 watts

Explanation:

Given that,

Length of the ramp, l = 12 m

Mass of the person, m = 55.8 kg

Angle between the inclined plane and the horizontal, \theta=25^{\circ}

Time, t = 3 s

Let h is the height of the hill from the horizontal,

h=l\ sin\theta

h=12\times \ sin(25)

h = 5.07 m

Let P is the power output necessary for a person to run up long hill side as :

P=\dfrac{E}{t}

P=\dfrac{mgh}{t}

P=\dfrac{55.8\times 9.8\times 5.07}{3}

P = 924.15 watts

So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.

3 0
3 years ago
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