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3 years ago

What is the potential energy of a 0.5 kg ball sitting on a table 1 meter

Physics

2 answers:

arsen [322]3 years ago

6 0

Hey!!

Mass=0.5kg

H=1 meter

Gravity=9.8m/s^2

Now,

Potential energy=m*g*h

 = 0.5*9.8*1

The right answer is 4.9 Joule.

Hope it helps

Good luck on your assignment

nevsk [136]3 years ago

3 0

The answer is 4.9 j above ground

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Find the sum 5.24 g, 43.261 g, and 7.3458 g. Write your answer with the correct amount of significant figures.

lapo4ka [179]

Answer:

This is how I do it:

5.24 rounded to one significant figure is 5
43.261 rounded to one significant figure is 40.
7.3458 rounded to one significant figure is 7.
5 + 40 + 7 is 52 g

Hope this helps you.

Explanation:

7 0

3 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

3 years ago

Five properties of magnet

Olenka [21]

Answer:

1. The magnet is magnetic and can attract iron articles.

2. The magnet has magnetic poles. Each magnet has two kinds of poles: N pole and S pole. They are in pairs.

3. Temporary magnet and permanent magnet: when the ferromagnetic material is magnetized, it is easy to lose the magnetic property, which is called temporary magnet (for example: iron); when the ferromagnetic material is magnetized, it is not easy to lose the magnetic property, which is called permanent magnet (for example: steel).

4. When two magnets are close to each other, the same poles will repel and push away from each other, and the different poles will attract and stick to each other. Therefore: the same pole repels each other, the different pole attracts each other.

5. The attraction of a magnetic object is called magnetism. An object is surrounded by a magnetic material. The area affected by the magnetic force is called the magnetic field.

3 0

3 years ago

The direction of deflection of an electron beam in a magnetic field can be determined by which of the following?

amid [387]

Answer:

Right Hand Rule

Explanation:

When a charged particle travels in a magnetic field, it experiences a force whose magnitude is given by:

$F=qvBsin\theta$

where

q is the charge of the particle

v is the velocity

B is the magnetic field strength

$\theta$ is the angle between the directions of v and B

The direction of the force can be determined by using the Right Hand Rule, as follows:

- index finger: this should be put in the direction of the velocity

- middle finger: this should be put in the direction of the magnetic field

- thumb: this will give the direction of the force -> however, for a negative charge (as the electron) the direction must be reversed, so it will be opposite.

4 0

3 years ago

An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value

Dimas [21]

Answer:

$n_f=2$

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$