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2 years ago

How old do I have to be?

Physics

2 answers:

BartSMP [9]2 years ago

8 0

Answer:

For what?

Specification pls.

melamori03 [73]2 years ago

5 0

What are you asking? What subject do you need help with?

You might be interested in

How can a heavy moving van have the same momentum as a small motorcycle?

vladimir1956 [14]

Hello there!

The formula for momentum is $p=mv$ .
p is momentum, m is mass, and v is velocity.

The mass of a heavy moving van will be much greater than the mass of a motorcycle, so how can they have the same momentum?

They can have the same momentum if the velocity of the moving van is less than the velocity of the motorcycle, or if the velocity of the motorcycle is greater than the moving van (same thing but differently worded)

Let the mass of the moving van be greater than the mass of the motorcycle by a factor of k. In order for the two to have the same momentum, the velocity of the motorcycle must be greater than the velocity of the moving van by the same factor, k.

Hope this helps! :)

3 0

3 years ago

The absolute pressure in water at a depth of 5m is read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the

irga5000 [103]

Answer:

a) 95950 pascals

b) 137642.5 pascals

Explanation:

The absolute pressure (Pabs) on a fluid is:

$P_{abs}=P_{gauge}+P_{atm}$ (1)

With Pgauge the pressure due depth on the fluid and Patm the atmospheric pressure. Pgauge is equal to:

$P_{gauge}=\rho gh$ (2)

with ρ the fluid density, g the gravitational acceleration and h the depth on the fluid. Using (2) on (1) and solving for Patm:

$P_{atm}=P_{abs}-P_{gauge}=P_{abs}-\rho_{water} gh$

$P_{atm}=(145000Pa)-(1000\frac{kg}{m^{3}})(9.81\frac{m}{s^{2}})(5m)$

$P_{atm}=95950Pa$

b) Here we're going to use again (1) but now we have another value of density because it's other liquid, to know that value we should use the fact that specific gravity (S.G) for liquids is the ratio between fluid density and water density:

$S.G=\frac{\rho_{fluid}}{\rho_{water}}$

$\rho_{liquid}=S.G*\rho_{water}$

$\rho_{liquid}=(0.85)*(1000\frac{kg}{m^{3}})=850\frac{kg}{m^{3}}$

so:

$P_{abs}=\rho_{liquid} gh+P_{atm}=(850\frac{kg}{m^{3}})(9.81\frac{m}{s})(5m)+95950Pa$

$P_{abs}=137642.5 Pa$

3 0

4 years ago

E4.Suppose Galileo’s pulse rate was 75 beats per minute. How many beats per second is this? What is the time in seconds between

Umnica [9.8K]

Answer

given,

Galileo's pulse rate = 75 beats per minutes

Beats per second

1 minutes = 60 s

pulse rate = $\dfrac{75}{60}$

= 1.25 beats/s

times in second

$t = \dfrac{1}{1.25}$

t = 0.8 s/pulse

distance of the object

initial velocity = 0 m/s

$s = ut + \dfrac{1}{2}gt^2$

$s = \dfrac{1}{2}\times 9.8\times 0.8^2$

s = 3.14 m

distance in feet

1 m = 3.28 ft

3.14 m = 3.14 x 3.28 ft

= 10.3 ft

distance in feet is equal to 10.3 ft.

3 0

4 years ago

A telephone line hangs between two poles 14 m apart in the shape of the catenary , where and are measured in meters.

Masja [62]

Answer:

a) At x=14 the slope will be given by:

$\frac{dy}{dx}(14)=a\sinh \left({\frac {14-C_{1}}{a}}\right)$ .

b) Then, the angle between the line and the pole will be:

$\phi=\pi - \theta$

where $\theta$ is the angle between the tangent to the catenary and the x-axis.

Explanation:

The catenary has the following general form:

$y(x)==a\cosh \left({\frac {x-C_{1}}{a}}\right)+C_{2}$

a) The slope at any point will be given by the derivative of y.

$\frac{dy}{dx}(x)=a\sinh \left({\frac {x-C_{1}}{a}}\right)$

At x=14:

$\frac{dy}{dx}(14)=a\sinh \left({\frac {14-C_{1}}{a}}\right)$ .

b) The angle between the tangent to the catenary and the x-axis at a given point will be given by:

$\frac{dy}{dx}(x)=tan(\theta)$ ⇒ $\theta=tan^{-1} (\frac{dy}{dx}(x))$

Then, the angle between the line and the pole will be:

$\phi=\pi - \theta$ .

5 0

4 years ago

How do I solve this

Svetradugi [14.3K]

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:

W = q*V

where:

q = charge = 1,602 × 10^(-19) [C]

V = voltage = 500 [V]

W = work [J]

W = 1,602 × 10^(-19) * 500

W = 8.01 × 10^(-17) [J]

8 0

4 years ago