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Artemon [7]
2 years ago
15

At room temperature, an oxygen molecule, with mass of 5.31x10-26kg , typically has a kinetic energy of about 6.21x10-21J.How fas

t is it moving?
Physics
1 answer:
alex41 [277]2 years ago
3 0
<h2>Oxygen is moving at 483.63 m/s</h2>

Explanation:

Kinetic energy = 0.5 x Mass x Velocity²

Kinetic energy = 6.21 x 10⁻²¹ J

Mass = 5.31 x 10⁻²⁶ kg

Substituting

        Kinetic energy = 0.5 x Mass x Velocity²

        6.21 x 10⁻²¹ = 0.5 x 5.31 x 10⁻²⁶ x Velocity²

        Velocity² = 2.34 x 10⁵

        Velocity = 483.63 m/s

Oxygen is moving at 483.63 m/s  

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A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m
Marrrta [24]

Answer:

-1.47 m/s^2

Explanation:

We can use the following SUVAT equation to solve the problem:

v^2 - u^2 = 2ad

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2

4 0
2 years ago
A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod
ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = 8.0\times 10^{- 5}\ T

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

e = - l(vec{v}\times \vec{B})

Here, velocity v is perpendicular to the rod

Thus

e = lvB           (1)

For the orbital velocity of the satellite at an altitude, H:

v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}

where

G = Gravitational constant

m_{e} = 5.972\times 10^{24}\ kg = mass of earth

R_{E} = 6371\ km = radius of earth

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s

Using this value value in eqn (1):

e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V

5 0
2 years ago
At 5000-kg freight car runs into 10000-kg freight car at rest. they couple upon collision and move with a speed of 2 m/s. what w
aliina [53]

Solution for the problem is:

Total momentum before collision is always equal to total momentum after collision. So note that:
Momentum of car A = 5000 x Xm/s 
Momentum of car A + B = 15,000 x 2m/s 

So combining the two, will give us the equation:
15,000/5,000 = 3 
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6 0
2 years ago
A car enters a 300-m radius horizontal curve on a rainy day when the coefficient of static friction between its tires and the ro
Vlada [557]

To solve this problem it is necessary to take into account the concepts related to Centripetal Force and Friction Force.

In the case of the centripetal force, we know that it is defined as

F_c = \frac{mv^2}{R}

Where,

m=mass

v= velocity

r= Radius

In the case of the Force of Friction we have to,

F_f = \mu m*g

Where,

\mu =Friction Constant

m= mass

g= gravity

According to the information given, the centripetal force must be less than or equal to the friction force to stay on the road, in this way

\frac{mv^2}{R} \leq \mu m*g

Re-arrange to find the velocity,

v \leq \sqrt{\mu gR}

v \leq \sqrt{(0.6)(9.8)(300)}

v \leq 42m/s

Therefore la velocidad del carro debe ser igual o menor a 42m/s para mantenerse en el camino

6 0
2 years ago
The first law of thermodymaic stae the <br> 5ktiopheuithkjhuiguihaoitg
pentagon [3]

I'm assuming you want the first law of thermodynamics.

The First Law of Thermodynamics states that heat is a form of energy and cannot be created or destroyed. It can, however, be transferred from one location to another and can be converted into other forms of energy.

3 0
2 years ago
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