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adell [148]
3 years ago
8

A soccer ball is released from rest at the top of a grassy incline. After 8.6 seconds, the ball travels 87 meters and 1.0 s afte

r this, the ball reaches the bottom of the incline. (a) What was the magnitude of the ball's acceleration, assume it to be constant
Physics
1 answer:
Arturiano [62]3 years ago
6 0

Answer:

a) a = 2.35 m/s^2

Explanation:

(a) In order to calculate the magnitude of the acceleration of the ball, you use the following formula, for the position of the ball:

x=v_ot+\frac{1}{2}at^2     (1)

x: position of the ball after t seconds = 87 m

t: time  = 8.6 s

a: acceleration of the ball = ?

vo: initial velocity of the ball = 0 m/s

You solve the equation (1) for a:

x=0+\frac{1}{2}at^2\\\\a=\frac{2x}{t^2}

You replace the values of the parameters in the previous equation:

a=\frac{2(87m)}{(8.6s)^2}=2.35\frac{m}{s^2}

The acceleration of the ball is 2.35 m/s^2

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A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn
KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

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4 years ago
What is the freezing point of water in Kelvin?
monitta

Answer:

the correct answer is 273.2 k

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3 years ago
A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (
Anika [276]

Answer:

a=9.8 rad/s^{2}

Explanation:

Torque, \tau is given by

\tau=Fr where F is force and r is perpendicular distance

R=0.5Lcos\theta where \theta is the angle of inclination

Torque, \tau can also be found by

\tau=Ia where I is moment of inertia and a is angular acceleration

Therefore, Fr=Ia and F=mg where m is mass and g is acceleration due to gravity

Making a the subject, a=\frac {Fr}{I}=\frac {mgr}{I} and already I is given as  

I=\frac {mL^{2}}{3} and r is 0.5Lcos\theta hence  

a=\frac {0.5mgLcos\theta}{1/3 mL^{2}}

a=\frac {3gcos\theta}{2L}

Taking g as 9.81, \theta is given as 37 and L is 1.2

a=\frac {3*9.81cos37}{2*1.2}=9.7932679419

a=9.8 rad/s^{2}

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Answer:

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