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3 years ago

Rearrange the equation for density to solve for mass.

Physics

1 answer:

Sedaia [141]3 years ago

5 0

Density = mass/volume

To rearrange the equation for density to solve for MASS

you will get
mass = density x volume

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Twist-on connectors without the spring-steel coils (plastic threads only) are suitable for making branch-circuit connections.

Gnoma [55]

Answer:

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Explanation:

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3 years ago

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Travka [436]

Explanation:

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2 years ago

Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11

tiny-mole [99]

Answer:

The line charge density is $1.59\times10^{-4}\ C/m$

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

$V=EA$

$V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2$

$\lambda=\dfrac{V\times2\epsilon_{0}}{r}$

Where, r = radius

V = potential difference

Put the value into the formula

$\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}$

$\lambda=1.59\times10^{-4}\ C/m$

Hence, The line charge density is $1.59\times10^{-4}\ C/m$

4 0

3 years ago

A 3.00 kg mud ball has a perfectly inelastic collision with a second mud ball that is initially at rest. the composite system mo

viva [34]

Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:

$m_{1} * v_{1} + m_{2} * v_{2} = m_{1} * v'_{1} + m_{2} * v'_{2}$

In case of perfectly inelastic collision v'1 and v'2 are same.

We have following information:
m₁=3 kg
m₂=? kg
v₁=x m/s
v₂=0 m/s
v'1 = v'2 = 1/3 * v₁

Now we insert given information and solve for m₂:
3*v₁ + 0*? = 3*1/3*v₁ + m₂*1/3*v₁
3v₁ = v₁ + m₂*1/3*v₁
2v₁ = m₂*1/3*v₁
2 = m₂*1/3
m₂= 6kg

Mass of second mud ball is 6kg.

7 0

3 years ago

A skater has rotational inertia 4.2 kg.m2 with his fists held to his chest and 5.7 kg.m2 with his arms outstretched. The skater

Fynjy0 [20]

Answer:

6.13428 rev/s

Explanation:

$I_f$ = Final moment of inertia = 4.2 kgm²

I = Moment of inertia with fists close to chest = 5.7 kgm²

$\omega_i$ = Initial angular speed = 3 rev/s

$\omega_f$ = Final angular speed

r = Radius = 76 cm

m = Mass = 2.5 kg

Moment of inertia of the skater is given by

$I_i=I+2mr^2$

In this system the angular momentum is conserved

$L_f=L_i\\\Rightarrow I_f\omega_f=I_i\omega_i\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{(5.7+2\times 2.5\times 0.76^2)3}{4.2}\\\Rightarrow \omega_f=6.13428\ rev/s$

The rotational speed will be 6.13428 rev/s

5 0

3 years ago