Ask question

Ask question

All categories

andrew-mc [135]

3 years ago

12

3. What was a constraint (a limitation or a condition) in this version?

1 answer:

pav-90 [236]3 years ago

7 0

Answer: limitations or constraints in the vision of science is that the hypothesis must be testable and falsifiable.

Explanation:

You might be interested in

What is a longitudinal wave?

Vedmedyk [2.9K]

The answer is 'A', they travel parallel to the direction of motion

4 0

3 years ago

Read 2 more answers

What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Marianna [84]

Answer:

The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

When two point are at same height so ,

$P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2$ ....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

$Q=A_{1}v_{1}$

$v_{1}=\dfrac{Q}{A_{1}}$

$v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}$

$v_{1}=56.58\ m/s$

For output velocity,

$v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}$

$v_{2}=6.28\ m/s$

Put the value into the formula

$P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)$

$\Delta P=1.58\times10^{6}\ N/m^2$

(b). We need to calculate the maximum height

Using formula of height

$\Delta P=\rho g h$

Put the value into the formula

$1.58\times10^{6}=1000\times9.8\times h$

$h=\dfrac{1.58\times10^{6}}{1000\times9.8}$

$h=161.22\ m$

Hence, The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

8 0

4 years ago

Do you have to use significant figures in converting temperatures

schepotkina [342]

Not only in converting temperatures I think, we always use significant figures every time we want to make some experimental report using the exact data from our experiment. If that isn't experimental report ( some exercise/questions from your textbook) it's not necessary

4 0

3 years ago

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

6 0

4 years ago

Can somebody explain it to me please?

Volgvan

Answer:

45°.

It is a property of the parabolas.. When the angle between a parabola and the x-axis is 45° the range is maximum.

6 0

3 years ago

Read 2 more answers