All categories

3 years ago

How many liters are equivalent to 500 centiliters

Physics

2 answers:

andreev551 [17]3 years ago

4 0

5 Liters are equal to 500 centiliters

PilotLPTM [1.2K]3 years ago

4 0

1000 centiliters = 1 liter.

500 centiliters = x liter

1000 / 500 = 1/x Cross multiply

1000x = 500 Divide by 1000

x = 500/1000

x = 1/2

500 centiliters = 1/2 or 0.5 of a liter.

You might be interested in

Steve want to catch up to Cami, a girl he likes. If he is jogging at 8 m/s and she is walking at 1 m/s. How long will it take hi

Aleks04 [339]

Answer:

1) if they are moving away from each other it will take 1.43 secs

2) if they are moving towards each other then it will take 1.11 secs

Explanation:

Distance between them is 10 m

Speed ( if they are moving towards each other)= distance/time

time = 10/8+1

time = distance / speed= 10/9= 1.11 secs

if they are moving away from each other than it will take

time = 10/8-1= 10/7= 1.43 secs

5 0

2 years ago

A stone is dropped from a certain height, distance covered by it in one second is:

Alinara [238K]

Answer:

d = 4.9 m

Explanation:

It is mentioned that a stone is dropped from a certain height. It is required to find the distance covered by it in one second.

The initial speed of the stone is equal to 0 as it was at rest. Let d is the distance covered by the stone.

Using second equation of motion :

$d=ut+\dfrac{1}{2}at^2$

Put u = 0 and a = g

$d=\dfrac{1}{2}gt^2\\\\d=\dfrac{1}{2}\times 9.8\times 1^2\\\\d=4.9\ m$

So, the distance covered by it in one second is 4.9 m.

5 0

3 years ago

11 kg is a familiar weight for a bag of flour. You are baking cookies for a Save The Rain Forest fund drive. It takes 500 g of f

arlik [135]

Answer: 22 batches.

Explanation:

Given that 11 kg is a familiar weight for a bag of flour. Also, it is given that It takes 500 g of flour to make one batch of cookies.

How many batches of cookies can you make with one bag of flour

Let's first convert 11 kg into grams (g) by multiplying it by 1000

11 × 1000 = 11000 g

Divide 11000 by 500

11000/500 = 22

Therefore, 22 batches of cookies can be made with one bag of flour.

8 0

3 years ago

The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is th

Rom4ik [11]

Answer:

Approximate escape speed = 45.3 km/s

Explanation:

Escape speed

$v=\sqrt{\frac{2GM}{R}}$

Here we have

Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

R = 1 AU = 1.496 × 10¹¹ m

M = 2.3 × 10³⁰ kg

Substituting

$v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.3\times 10^{30}}{1.496\times 10^{11}}}=4.53\times 10^4m/s=45.3km/s$

Approximate escape speed = 45.3 km/s

6 0

3 years ago

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

3 years ago