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3 years ago

What is tranlsatory motion? Give two examples

Physics

1 answer:

Aleksandr [31]3 years ago

6 0

Answer:

The motion in which all particles of a body move through the same distance in the same time.

Examples: A car moving along the road

A ball rolling on the ground

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Two cars start a race with initial velocity 7ms-1 and 4ms-1 respectively.Their acceleration are 0.4ms-2 and 0.5ms-2 respectively

Ronch [10]

The distance of the track is 600 m

Explanation:

The two cars move by uniformly accelerated motion, so we the distance they cover after a time t is described by the following suvat equation

$d=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

For car 1, we have

$d_1 = u_1 t + \frac{1}{2}a_1 t^2$

where

$u_1 = 7 m/s$ is the initial velocity of car 1

$a_1 = 0.4 m/s^2$ is the acceleration of car 1

So the equation can be rewritten as

$d_1 = 7t + 0.2t^2$

For car 2, we have

$d_2 = u_2 t + \frac{1}{2}a_2 t^2$

where

$u_2 = 4 m/s$ is the initial velocity of car 2

$a_2 = 0.5 m/s^2$ is the acceleration of car 2

So the equation can be rewritten as

$d_2= 5t + 0.25t^2$

The two cars finish the race at the same time: this means that they cover the same distance in the same time t, so we can write

$d_1 = d_2\\7t + 0.2t^2=5t + 0.25t^2$

And solving for t, we find

$2t - 0.05t^2= 0\\t(2-0.05t)=0$

Which gives two solutions:

t = 0 (initial instant)

t = 40 s

Therefore, the distance of the track is

$d_1 = 7t +0.2t^2 = 7\cdot 40+0.2\cdot 40^2=600 m$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0

3 years ago

For a molecule of O2 at room temperature (300 K), calculate the average angular velocity for rotations about the x or y axes. Th

Ksivusya [100]

Answer: 6.65 rad/s

Explanation:

Firstly, we need to know that according the kinetic theory of gases, the average kinetic energy $KE$ of a molecule with $i$ degrees of freedom is:

$KE=\frac{i}{2}kT$ (1)

Where:

$i=5$ because oxigen is a diatomic molecule and has 5 degrees of freedom

$k=1.38064852 (10)^{-23} \frac{m^{2}kg}{s^{2}K}$ is the Boltzman constant

$T=300 K$ is the temperature

Then:

$KE=\frac{5}{2}(1.38064852 (10)^{-23} \frac{m^{2}kg}{s^{2}K})(300 K)$ (2)

$KE=1.035 (10)^{-20} \frac{m^{2}kg}{s^{2}}$ (3) With this value of average kinetic energy we can find the average angular velocity, with the following equation:

$KE=\frac{1}{2}I \omega^{2}$ (4)

Where:

$I$ is the moment of inertia

$\omega$ is the angular velocity

Now, $I=m R^{2}$ (5)

Being $m=31.98 g/mol=0.03198 kg/mol$ the molar mass of Oxigen molecule and $R=0.121(10)^{-9}m$ the distance between atoms

$I=(0.03198 kg/mol)(0.121(10)^{-9}m)^{2}$ (6)

$I=4.682(10)^{-22} \frac{kg}{mol} m^{2}$ (7)

Substituting (7) and (3) in (4):

$1.035 (10)^{-20} \frac{m^{2}kg}{s^{2}}=\frac{1}{2} (4.682(10)^{-22} \frac{kg}{mol} m^{2}) \omega^{2}$ (8)

Finding $\omega$ :

$\omega=6.65 rad/s$ (9) This is the average angular velocity for a molecule of O2

7 0

3 years ago

You have two springs. one has a greater spring constant than the other. you also have two objects, one with a greater mass than

neonofarm [45]

The answer should be "The object with the greater mass should be attached to the spring with the smaller <span>spring constant."

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4 0

3 years ago

Traveling at an average rateaverage rate of between 50 and 60 miles per hour for 4 to 7 hours select the best estimate for the d

borishaifa [10]

Answer:

best close estimate distance is 315 miles

Explanation:

given data

speed v1 = 50 miles per hour

speed v2 = 60 miles per hour

time t1 = 4 hr

time t2 = 7 hr

to find out

best estimate for the distance traveled

solution

we know lower end distance

distance = v1 × t1

distance = 50 × 4 = 200 miles

and

upper end distance

distance = v2 × t2

distance = 60 × 7 = 420 miles

so middle value distance is

v = 55 miles per hour and t = 5.5 hours

distance = v × t

distance = 55 × 5.5

distance = 302.5 miles

so best close estimate distance is 315 miles

3 0

3 years ago

The distance between two slits is 1.50 *10-5 m. A beam of coherent light of wavelength 600 nm illuminates these slits, and the d

Fed [463]

Answer: y = 2.4×10^-6m or y= 2.4μm

Explanation: The formulae for the distance between the central bright fringe to any other fringe in pattern is given as

y = R×mλ/d

Where y = distance between nth fringe and Central bright spot fringe.

m = position of fringe = 4

λ = wavelength of light= 600nm = 600×10^-9 m

d = distance between slits = 1.50×10^-5m

R = distance between slit and screen = 2m

y = 2 × 4 × 600×10^-9/2

y = 4800×10^-9/2

y = 2400 × 10^-9

y = 2.4×10^-6m or y= 2.4μm

8 0

3 years ago

What is tranlsatory motion? Give two examples​

What is tranlsatory motion? Give two examples