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A student of mass 63.4 kg, starting at rest, slides down a slide 21.2 m long, tilted at an angle of 26.1° with respect to thehor

Vera_Pavlovna [14]

Answer:

fr = 65.46 N , a = 8.74 m / s² and vf = 19.25 m / s

Explanation:

We write a reference system with an axis parallel to the slide and gold perpendicular axis, in this system we decompose the weight

sin 21.2 = Wx / W

cos21.2 = Wy / W

Wx = W sin21.2

Wy = W cos 21.2

We form Newton's equations

X axis

Wx -fr = m a

Y Axis

N- Wy = 0

N = Wy

fr = μ N

fr = μ (W cos 21.2)

fr = 0.113 63.4 9.8 cos 21.2

fr = 65.46 N

We replace and calculate the acceleration

W sin 21.2 - μ W cos 21.2 = m a

a = g (sin21.2 - μ cos 21.2)

a = 9.8 (without 21.2 - 0.113 cos 21.2)

a = 8.74 m / s²

This acceleration is along the slope of the slide, so we can calculate the distance

d = 21.2 m

vf² = v₀² + 2 a d

vf² = 0 + 2 a d

vf = √(2 8.74 21.2)

vf = √ (370,576)

vf = 19.25 m / s

3 0

3 years ago

Put these greenhouse effect events in order, starting with light's origin.

Fudgin [204]

Incident infrared radiation is blocked. Visible and ultraviolet radiation heat Earth. Earth radiates infrared radiation. Infrared radiation is blocked and heats Earth. Visible and shortwave radiation heat Earth.Earth radiates longwave radiationLongwave radiation is reflected downward Longwave radiation <span>heats Earth</span>

3 0

3 years ago

Read 2 more answers

The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of

motikmotik

Explanation:

It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,

$F\propto \dfrac{mgv^2}{r}$

$F=\dfrac{kmgv^2}{r}$

mg is the weight of the car

r is the radius of the curve

v is the speed of the car

Case 1.

F = 640 pounds

Weight of the car, W = mg = 2600 pound

Radius of the curve, r = 650 ft

Speed of the car, v = 40 mph

$640=\dfrac{k(2600)(40)^2}{650}$

k = 0.1

Case 2.

Radius of the curve, r = 750 ft

Speed of the car, v = 30 mph

$F=\dfrac{0.1\times 2600\times (30)^2}{750}$

F = 312 N

Hence, this is the required solution.

6 0

3 years ago

(Inelastic) Ballistic Pendulum. A bullet that has a mass of 120g is fired horizontally at a block of jelly of mass 15kg that is

sineoko [7]

Answer:

$\Delta E_{loss} = 9100.262\,J$

Explanation:

The bullet-Ballistic Pendulum system can be modelled after the Principles of Momentum Conservation and Energy Conservation:

Final speed of the ballistic pendulum:

$\frac{1}{2}\cdot (0.12\,kg)\cdot v^{2} = (0.12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.45\,m)$

$v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.45\,m)}$

$v \approx 2.971\,\frac{m}{s}$

Initial speed of the bullet:

$(15\,kg)\cdot \left(0\,\frac{m}{s}\right) + (0.12\,kg)\cdot v = (0.12\,kg)\cdot \left(20\,\frac{m}{s} \right) + (15\,kg)\cdot \left(2.971\,\frac{m}{s} \right)$

$v = 391.375\,\frac{m}{s}$

Energy loss:

$\Delta E_{loss} = \frac{1}{2}\cdot (15\,kg)\cdot \left[\left(0\,\frac{m}{s} \right)^{2}-\left(2.971\,\frac{m}{s} \right)^{2}\right] + \frac{1}{2}\cdot (0.12\,kg)\cdot \left[\left(391.375\,\frac{m}{s} \right)^{2}-\left(20\,\frac{m}{s} \right)^{2}\right]$ $\Delta E_{loss} = 9100.262\,J$

3 0

4 years ago

When does the chunk of ice have the most potential energy? A. right before it breaks off B. a short time after it starts to fall

Anarel [89]

Right before it breaks of, because all the other options are examples of Kinetic energy while the A. answer is the only one describing potential energy.

3 0

3 years ago