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3 years ago

Hello how do you find density

Physics

1 answer:

Alekssandra [29.7K]3 years ago

4 0

Explanation:

First find your mass

Second determine your volume

Then plug into formula

The formula is p = m/v

So basically divide your mass by your volume to get the density.

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What happen to the frequency of transverse vibration of a stretched string if its tension is halved and the area of cross sectio

Stolb23 [73]

Answer:

The fundamental frequency of the stretched string is:

$f=$ $\frac{1}{2} \sqrt{\frac{T}{L} }$ [ T = Tension and μ = mass per unit length]

Here,

μ = $\frac{m}{L} = \frac{Vp}{L} = Ap$

$f= \frac{1}{2} \sqrt{\frac{T}{Ap} }$

If T is halved and A is doubled,

$f= \frac{1}{2} \sqrt{\frac{T'}{A'p} } = \sqrt{\frac{1}{2* 2* A* p} } = \frac{1}{2} (\frac{1}{2} \sqrt{\frac{T}{Ap} } = \frac{1}{2} f$

Thus, the frequency is reduced to half if its tension is halved and the area of cross-section of the string is doubled.

7 0

3 years ago

How does the energy from the person's hand get to the last domino?

Illusion [34]

Answer:

woh its really exist?

Explanation:

cakra

8 0

3 years ago

A liquid of density 1390 kg/m 3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the

Aloiza [94]

Answer:

The difference of power is

ΔP = 172.767 kPa

Explanation:

ρ = 1390 kg / m³

v = 9.63 m/s

d₁ = 10.1 cm , d₂ = 15.3 cm

Δz = 8.85 m

To find the difference ΔP between the fluid pressure at locations 2 and the fluid pressure at location 1

ΔP = ρ * g * Z + ¹/₂ * ρ * v² * ( 1 - (d₁ / d₂)⁴ )

ΔP = 1390 kg / m³ * 9.8 m/s² * 8.85 m + 0.5 * 1390 kg / m³ *(9.63 m /s)² * (1 - (0.101 m / 0.153 m )⁴ )

ΔP = 172.767 x 10 ³ Pa

ΔP = 172.767 kPa

6 0

3 years ago

A 650 kg steel beam is being pulled up by a crane with a force of 7020 N. What is the upwards acceleration of the beam?

zimovet [89]

Answer: 1 m/s^2

Explanation:

n= w+ma

n= Mg+Ma

7020 = (650)(9.8)+650a

7020 = 6370+ 650a

650=650a

A= 1 m/s^2

8 0

3 years ago

Given: G = 6.67259 × 10−11 N m2 /kg2 . A 438 kg geosynchronous satellite orbits a planet similar to Earth at a radius 1.94 × 105

photoshop1234 [79]

Answer:

449.37412 N

Explanation:

G = Gravitational constant = 6.67259 × 10⁻¹¹ m³/kgs²

m = Mass of satellite = 438 kg

M = Mass of planet

T = Time period of the satellite = 24 h

r = Radius of planet = $1.94\times 10^8\ m$

The time period of the satellite is given by

$T=2\pi\sqrt{\frac{r^3}{GM}}\\\Rightarrow M=4\pi^2\frac{r^3}{T^2G}\\\Rightarrow M=4\pi^2\times \frac{(1.94\times 10^8)^3}{(24\times 3600)^2\times 6.67259\times 10^{-11}}\\\Rightarrow M=5.78686\times 10^{26}\ kg$

The gravitational force is given by

$F=G\frac{Mm}{r^2}\\\Rightarrow F=6.67259\times 10^{-11}\times \frac{5.78686\times 10^{26}\times 438}{(1.94\times 10^8)^2}\\\Rightarrow F=449.37412\ N$

The force acting on this satellite is 449.37412 N

3 0

3 years ago