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gtnhenbr [62]
4 years ago
5

If a car is moving at 90 km/h and it rounds a corner, also at 90 km/h, does it maintain a constant speed? A constant velocity? I

f
Physics
1 answer:
valentina_108 [34]4 years ago
6 0

Answer:

Constant speed but different velocity

Explanation:

Since the car speed at the corner is 90km/h which is the same as before. The speed, or magnitude of the velocity vector, remains unchanged. The direction of the velocity vector, on the other hand, is changing as the car turns around the corner. Therefore, the car maintains a constant speed, but not a constant velocity.

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_____ electrons are the most important electrons to an atom .
nalin [4]

Answer:

stable is the answer

Explanation:

stable electrons

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3 years ago
Calculate the potential energy of a 5.2 kg object positioned 5.8 m above the ground.
GrogVix [38]

Answer:

295.568J

Explanation:

use P=mgh

plug in givens

P= 5.2*9.8*5.8= 295.568J

4 0
3 years ago
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It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet
TEA [102]

Answer:

v=346.05\ m.s^{-1}

Explanation:

Given:

initial temperature of the lead bullet, T_i=43^{\circ}C

latent heat of fusion of lead, L_f=2.32\times 10^4\ J.kg^{-1}

melting point of lead, T_m=327.3^{\circ}C

We have:

specific heat capacity of lead, c=129\ J.kg^{-1}.K^{-1}

<em>According to question the whole kinetic energy gets converted into heat which establishes the relation:</em>

\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)

\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f

\frac{1}{2} m.v^2=m(c.\Delta T+L_f)

\frac{v^2}{2} =129\times(327.3-43)+23200

v=346.05\ m.s^{-1}

3 0
3 years ago
Predictions about the future based on the position of planets is an example of
tatyana61 [14]
Predictions about the future based on the position of planets is an example of astrology.
Hope this helps! :)
6 0
3 years ago
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During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 4.02 s, how high above the point where i
alukav5142 [94]

Answer:

d = 19.796m

Explanation:

Since the ball is in the air for 4.02 seconds, the ball should reach the maximum point from the ground in half the total time, therefore, t=2.01s to reach maximum height. At the maximum height, the velocity in the y-direction is 0.

So we know t=2.01, vi=0, g=a=9.8m/s and we are solving for d.

Next, you look for a kinematic equation that has these parameters and the one you should choose is:

d=vt+\frac{1}{2}at^2

Now by substituting values in, we get

d=(0\frac{m}{s})*(2.01s)+\frac{1}{2}(9.8\frac{m}{s^2})(2.01)^2

d = 19.796m

7 0
3 years ago
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