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Natali5045456 [20]
4 years ago
5

A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti

on force of 12 N. How far does it slide on the snow before coming to rest?
Physics
1 answer:
uysha [10]4 years ago
3 0

Answer:The sled slides 16.875m before rest.

Explanation:

a=\frac{F}{m} =\frac{12N}{20kg}

a=0.6 m/s²

Vf=0=Vi-a.t

t=\frac{Vi}{a} =t=\frac{4.5m/s}{0.6 m/s2} =t=7.5seg

d= Vi.t - \frac{a.t^{2}}{2}

d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m

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7200 joules of heat and light energy was dissipated into the air. But no work was done ... no force moved through no distance.

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3 years ago
Silver has a mass of 10.5 grams and a volume of 19.3 cm3. What is its density?
Neko [114]

Answer:

<h3>The answer is 0.54 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question we have

density =  \frac{10.5}{19.3}  \\  = 0.54404145...

We have the final answer as

<h3>0.54 g/cm³</h3>

Hope this helps you

7 0
3 years ago
On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth's atmosphere over Chelyabinsk, Russia, and expl
fredd [130]

Answer:

156.67 m/s

0.45676 times the speed of sound

Explanation:

Distance from the ground = 23.5 km = 23500 m

Time taken by the blast waves to reach the ground = 2\ minutes\ 30\ seconds=2\times 60+30=150\ s

Spedd of the wave would be

Speed=\dfrac{Distance}{Time}\\\Rightarrow v_b=\dfrac{23500}{150}\\\Rightarrow v-b=156.67\ m/s

The velocity of the blast wave is 156.67 m/s

v = Velocity of sound = 343 m/s

\dfrac{v_b}{v}=\dfrac{156.67}{343}\\\Rightarrow v_b=v\dfrac{156.67}{343}\\\Rightarrow v_b=0.45676v

The blast wave is 0.45676 times the speed of sound

7 0
4 years ago
A skillet is removed from an oven whose temperature is 450°F and placed in a room whose temperature is 70°F. After 5 minutes, th
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Answer:

attached below

Explanation:

3 0
3 years ago
A cube has a drag coefficient of 0.8. what would be the terminal velocity of a sugar cube 1 cm on a side in air (ρ = 1.2 kg/m3)?
kramer
To answer this question, we should know the formula for the terminal velocity. The formula is written below:

v = √(2mg/ρAC)
where
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient

Let's determine the mass, m, to be density*volume.
Volume = s³ = (1 cm*1 m/100 cm)³ = 10⁻⁶ m³
m = (1.6×10³ kg/m³)(10⁻⁶ m³) = 1.6×10⁻³ kg
A = (1 cm * 1 m/100 cm)² = 10⁻⁴ m²

v = √(2*1.6×10⁻³ kg*9.81 m/s²/1.6×10³ kg/m³*10⁻⁴ m²*0.8)
<em>v = 0.495 m/s</em>
7 0
3 years ago
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