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taurus [48]
3 years ago
14

A car travels 3 km north, then turns and travels 4 km west. What is the car's displacement?

Physics
2 answers:
Olin [163]3 years ago
5 0

Explanation:

Displacement = Shortest path Travelled from starting point to final point..

Displacement formula = <em>Final</em><em> </em><em>position</em><em> </em><em>-</em><em> </em><em> </em><em>Initial</em><em> </em><em>position</em>

= 7km - 3km

= 4km

^Displacement is 4km

<em>If</em><em> </em><em>this</em><em> answer</em><em> helps</em><em> you</em><em> plz</em><em> mark</em><em> as</em><em> brainlist</em><em>.</em><em>.</em>

<em>Tq</em><em>.</em><em>.</em>

Karolina [17]3 years ago
4 0

"Displacement" is the straight-line distance and direction between the start-point and the end-point, regardless of what route was taken to get there.

If you draw a picture of how this car traveled, you'll see that it drove along two legs of a right triangle. I'm guessing the driver probably decided to stay on the streets, instead of driving the STRAIGHT line across people's lawns to get to the grocery store.  if a BIRD wanted to make the same trip, it would just fly in a straight line, straight over houses and yards.

The bird would fly along the hypotenuse of the right triangle.  The distance would be  √(3² + 4²).

That's √(9 + 16), which is √(25), and that's 5 km .

'Displacement' is a vector, and it also needs a direction.

After driving 3km north and 4km west, you are 5km roughly Northwest of where you started from.  (actually 53.1° west of north)

The car's displacement is 5 km , 53.1° west of north,

or <em>5 km at an azimuth of 306.9°</em> .

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3 years ago
Two bodies fall freely from different heights and reach the ground simultaneously. The time of descent for the first body is 1s
jok3333 [9.3K]
The initial height of the first body is given by:
h_1 =  \frac{1}{2}gt^2
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
h_1 =  \frac{1}{2}(9.81 m/s^2)(1 s)^2=4.9 m

The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
h_2 =  \frac{1}{2}(9.81 m/s^2)(2 s)^2=19.6 m

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6 0
3 years ago
A train travels 85 kilometers in 5 hours, and then 63 kilometers in 5 hours what is its average speed?
Lesechka [4]
We know average speed =total distance/time taken
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3 0
3 years ago
One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel
Gekata [30.6K]

Answer:

a) v_{U-235} = 2.68 \cdot 10^{5} m/s

v_{He-4} = -1.57 \cdot 10^{7} m/s  

b) E_{He-4} = 8.23 \cdot 10^{-13} J

E_{U-235} = 1.41 \cdot 10^{-14} J

 

Explanation:

Searching the missed information we have:                                        

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg  

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg            

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

p_{i} = p_{f}

m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}

Since the plutonium nucleus is originally at rest, v_{Pu-239} = 0:

0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}  

v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}    (1)

Kinetic Energy:

E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}

2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}    

1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}   (2)    

By entering equation (1) into (2) we have:

1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}  

1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}  

Solving the above equation for v_{U-235} we have:

v_{U-235} = 2.68 \cdot 10^{5} m/s

And by entering that value into equation (1):

v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s                        

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J

For U-235:

E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J

 

I hope it helps you!                                                                                    

3 0
3 years ago
Two blocks of masses 20 kg and 8.0 kg are connected togetherby
OleMash [197]

Answer:

14 N

Explanation:

The tension in the second string is puling both the masses of 20 kg and 8 kg with acceleration of 0.5 m s⁻²

So tension in the second string = total mass x acceleration

= 28 x .5 = 14 N . Ans..

4 0
3 years ago
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