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3 years ago

Calculate the volume of 19 kilograms of petrol if the density of petrol is 800 kg/m?

Physics

1 answer:

earnstyle [38]3 years ago

6 0

Answer:

i hope this will help you :)

Explanation:

mass=19kg

density=800kg/m³

volume=?

as we know that

density=mass/volume

density×volume=mass

volume=mass/density

putting the values

volume=19kg/800kg/m³

so volume=0.02375≈0.02m³

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2 years ago

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A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel

trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2$

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s$

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

$1\ mile=1609.34\ m$

$1\ h=3600\ seconds$

$18.03\times \frac{3600}{1609.34}=40.33\ mph$

Speed of the boat by when it will hit the dock is 40.33 mph

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m$

The distance at which the boat will have to start decelerating is 312.5 m

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2 years ago

In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so

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Answer:

(a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

Explanation:

Given that,

Radius of the circle = 0.681 m

Angular acceleration = 67.7 rad/s²

Angular speed =18.6 rad/s

We need to calculate the centripetal acceleration of the ball

Using formula of centripetal acceleration

$a_{c}=\omega^2\times r$

Put the value into the formula

$a_{c}=(18.6)^2\times0.681$

$a_{c}=235.5\ m/s^2$

We need to calculate the tangential acceleration of the ball

Using formula of tangential acceleration

$a_{t}=r\alpha$

Put the value into the formula

$a_{t}=0.681\times67.7$

$a_{t}=46.104\ m/s^2$

(a). We need to calculate the magnitude of the total acceleration of the ball

Using formula of total acceleration

$a=\sqrt{a_{c}^2+a_{t}^2}$

Put the value into the formula

$a=\sqrt{(235.5)^2+(46.104)^2}$

$a=239.97\ m/s^2$

(b). We need to calculate the angle of the total acceleration relative to the radial direction

Using formula of the direction

$\theat=\tan^{-1}(\dfrac{a_{t}}{a_{c}})$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{46.104}{235.5})$

$\theta=11.0^{\circ}$

Hence, (a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

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2 years ago

Calculate the volume of 19 kilograms of petrol if the density of petrol is 800 kg/m?​

Calculate the volume of 19 kilograms of petrol if the density of petrol is 800 kg/m?