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LenaWriter [7]
3 years ago
6

A gry is an old English measure for length, defined as 1/10 of a line, where line is another old English measure for length, def

ined as 1/12 inch. A common measure for length in the publishing business is a point, defined as 1/72 inch. What is an area of 0.16 gry2 in points squared?
Physics
1 answer:
Rashid [163]3 years ago
8 0

Answer:

0.0576 points²

Explanation:

1\ gry=\frac{1}{10}\ line

1\ line=\frac{1}{12}\ inch

\frac{1}{10}\ line=\frac{1}{10}\times \frac{1}{12}\ inch\\\Rightarrow \frac{1}{10}\ line=\frac{1}{120}\ inch

\therefore 1\ gry=\frac{1}{120}\ inch

1\ point=\frac{1}{72}\ inch\\\Rightarrow 72\ points=1\ inch

\\\Rightarrow \frac{1}{120}\ inch=\frac{1}{120}\times 72\ points

1\ gry=\frac{72}{120}\ point\\\Rightarrow 1\ gry=0.6\ point

0.16\ gry^2=0.6^2\times 0.16\\\Rightarrow 0.16\ gry^2=0.0576\ points^2

∴ Area of 0.16 gry² is 0.0576 points²

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Oksana_A [137]

Answer:

i) 0.9504

ii) 0.0452

Explanation:

Given data: reliability of hydraulic brakes= 0.96

reliability of mechanical brakes = 0.99

So the probability of stopping the truck = 0.96×0.99= 0.9504

At low speed

case: A works and B does not

= 0.96×(1-0.99) = 0.0096

case2 : B works and A does not

= 0.99×(1-0.96) = 0.0396

Therefore, probality of stopping = 0.0096+0.0396 = 0.0492

8 0
2 years ago
Look at the graph. What is the slope of the line?
NeTakaya

For the first hour, the slope is zero.

After that, the slope is -2 miles per hour.

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2 years ago
A car goes forward along a level road T a constant velocity the additional force needed to being the car into equilibrium is?
eimsori [14]
The additional force needed to bring the car into equilibrium is frictional force.
8 0
2 years ago
An unknown substance from planet X has a density of 10 g/mL. It occupies a volume of 80 mL. What is the mass of this unknown sub
asambeis [7]

Answer:

800 mL

Explanation:

D*V=M

You pick out the numbers as well as what it is they represent from the word problem/explanation, then from there plug them in to the equations. Once you do that, you get your product and have the answer.

10*80= 800

5 0
3 years ago
A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected
gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0
3 years ago
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