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2 years ago

If 400g is 1kg find the ratio in the simplest form

Physics

1 answer:

uranmaximum [27]2 years ago

6 0

2:5

Explanation:

400g : 1kg

400g: 1000g

4 : 10

2 : 5

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Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho

77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge $q{t}= 0.65 q_{0}$

We need to calculate the time constant

Using expression for charging in a RC circuit

$q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

Where, $\dfrac{t}{RC}$ = time constant

Put the value into the formula

$0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

$1-e^{-(\dfrac{t}{RC})}=0.65$

$e^{-(\dfrac{t}{RC})}=0.35$

$-\dfrac{t}{RC}=ln (0.35)$

$-\dfrac{t}{RC}=-1.049$

$\dfrac{t}{RC}=1.049$

Hence, The time constant is 1.049.

6 0

2 years ago

A swimming pool has the shape of a right circular cylinder with radius 21 feet and height 10 feet. Suppose that the pool is full

AysviL [449]

Answer:

The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

Explanation:

Given that,

Radius = 21 feet

Height = 10 feet

Weighing = 62.5 pounds/cubic

Work = 4329507.37572

Height = 2 feet

Let's look at a horizontal slice of water at a height of h from bottom of pool

We need to calculate the area of slice

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times21^2$

$A=441\pi\ feet^2$

Thickness of slice $t=\Delta h\ ft$

The volume is,

$V=(441\pi\times\Delta h)\ ft^3$

We need to calculate the force

Using formula of force

$F=W\times V$

Where, W = water weight

V = volume

Put the value into the formula

$F=62.5\times(441\pi\times\Delta h)$

$F=27562.5\pi\times\Delta h\ lbs$

We need to calculate the work done

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs$

We do this by integrating from h = 0 to h = 10

We need to find the total work,

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh$

$W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(10\times10-\dfrac{100}{2}-0)$

$W=4329507.37572$

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh$

$W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(12\times10-\dfrac{100}{2}-0)$

$W=1929375\pi$

$W=6061310.32\ foot- pound$

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

8 0

2 years ago

What is the speed of a skater who travels a distance of 210 m in a time of 10 seconds?

Klio2033 [76]

21m per second. Take 210 divided by 10. Hope this helps!

7 0

3 years ago

A model rocket is launched straight upward with an initial speed of 56.5 m/s. It accelerates with a constant upward acceleration

marta [7]

Answer:

Maximum height reached by the rocket, h = 202.62 meters

Explanation:

It is given that,

Initial speed of the model rocket, u = 56.5 m/s

Constant upward acceleration, $a=1.96\ m/s^2$

Distance traveled by the engine until it stops, d = 198.8 m

Let v is the speed of the rocket when the engine stops. It can be calculated using the third equation of motion as :

$v=\sqrt{u^2+2ad}$

$v=\sqrt{(56.5)^2+2\times 1.96\times 198.8}$

v = 63.02 m/s

At the maximum height, v = 0 and the engine now decelerate under the action of gravity, a = -g. Let h is the maximum height reached by the rocket.

Again using third equation of motion as :

$v^2-u^2=-2gh$

$h=\dfrac{v^2-u^2}{-2g}$

$h=\dfrac{u^2}{2g}$

$h=\dfrac{(63.02)^2}{2\times 9.8}$

h = 202.62 meters

So, the maximum height reached by the rocket is 202.62 meters. Hence, this is the required solution.

3 0

2 years ago

What frequency is received by the ambulance after reflecting from a wall near the person watching the oncoming ambulance (as in

Zepler [3.9K]

Answer:

900.48925 Hz

979.9785 Hz

Explanation:

$v_a$ = Relative velocity of ambulance = $109\ km/h=\dfrac{109}{3.6}$

$v_w$ = Velocity of wall = 0

v = Velocity of sound in air = 343 m/s

From doppler effect we have

$f=f'\dfrac{v+v_w}{v-v_a}\\\Rightarrow f=821\dfrac{343+0}{343-\dfrac{109}{3.6}}\\\Rightarrow f=900.48925\ Hz$

The frequency of sound is 900.48925 Hz

When the wall acts like a source

$f=f'\dfrac{v+v_a}{v-v_w}\\\Rightarrow f=900.48925\dfrac{343+\dfrac{109}{3.6}}{343-0}\\\Rightarrow f=979.9785\ Hz$

The frequency of sound is 979.9785 Hz

4 0

3 years ago

If 400g is 1kg find the ratio in the simplest form​

If 400g is 1kg find the ratio in the simplest form