Have I answered this question correctly?

An ore sample weighs 17.50 N in air. When the sample

zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T= $W-F_b$

$F_b$ =buoyancy force

T=Tension force

W=Weight

By using the formula

$11.2=17.5-F_b$

$F_b=17.5-11.2=6.3$ N

$F_b=V_{object}\times \rho_{water}\cdot g$

Where

$V_{object}$ =Volume of object

$\rho_{water}=1000 kgm^{-3}$ =Density of water

$g=9.8 ms^{-2}$ =Acceleration due to gravity

Substitute the values then we get

$6.3=9.8\times 1000\times V_{object}$

$V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3$

Volume of sample= $6.43\times 10^{-4} m^3$

Density of sample, $\rho_{object}=\frac{Mass}{volume_{object}}$

Where mass of ore sample=1.79 kg

Substitute the values then, we get

$\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3$

Density of the sample= $2.78\times 10^{3} kgm^{-3}$

7 0

4 years ago

Wha is the frequency of a wave having a period equal to 18 seconds?

Rainbow [258]

Answer:

5.5 × 10-2 hertz

Explanation:

The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.

= 0.055 per second (1 cycle per second = 1 Hertz)

Thus, we can conclude that the frequency of the wave is 5.5 X 10^{-2} hertz.

Hopes this helps, love <3

5 0

3 years ago

Taste and smell would be considered

natka813 [3]

One of your 5 senses because your 5senses are hearing, smelling, seeing, taste, touching

8 0

4 years ago

A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary

Archy [21]

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

m = 10 kg
g = 9.8 m/s²
T = ?

Use formula:

$\boxed{\bold{T=m*g}}$

Replace and solve:

$\boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}$

$\boxed{\boxed{\bold{T=98\ N}}}$

The tension in the rope is <u>98 Newtons.</u>

Greetings.

5 0

3 years ago

A horizontal pipe 18.0 cm in diameter has a smooth reduction to a pipe 9.00 cm in diameter. If the pressure of the water in the

Rzqust [24]

Answer:

The rate of flow of water is 71.28 kg/s

Solution:

As per the question:

Diameter, d = 18.0 cm

Diameter, d' = 9.0 cm

Pressure in larger pipe, P = $9.40\times 10^{4}\ Pa$

Pressure in the smaller pipe, P' = $2.80\times 10^{4}\ Pa$

Now,

To calculate the rate of flow of water:

We know that:

Av = A'v'

where

A = Cross sectional area of larger pipe

A' = Cross sectional area of larger pipe

v = velocity of water in larger pipe

v' = velocity of water in larger pipe

Thus

$\pi \frac{d^{2}}{4}v = \pi \frac{d'^{2}}{4}v'$

$18^{2}v = 9^v'$

v' = 4v

Now,

By using Bernoulli's eqn:

$P + \frac{1}{2}\rho v^{2} + \rho gh = P' + \frac{1}{2}\rho v'^{2} + \rho gh'$

where

h = h'

$\rho = 10^{3}\ kg/m^{3}$

$9.40\times 10^{4} + \frac{1}{2}\rho v^{2} = 2.80\times 10^{4} + \frac{1}{2}\rho (4v)^{2}$

$6.6\times 10^{4} = \frac{1}{2}\rho 15v^{2}$

$v = \sqrt{\frac{2\times 6.6\times 10^{4}}{15\times 10^{3}}} = 8.8\ m/s$

Now, the rate of flow is given by:

$\frac{dm}{dt} = \frac{d}{dt}\rho Al = \rho Av$

$\frac{dm}{dt} = 10^{3}\times \pi (\frac{18}{2}\times 10^{- 2})^{2}\times 8.8 = 71.28\ kg/s$

3 0

3 years ago

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