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denpristay [2]
1 year ago
9

three different-mass projectiles are launched at different angles of elevation from the top of a building. each particle has the

same initial kinetic energy. which particle has the greatest kinetic energy just as it impacts with the ground?
Physics
1 answer:
MrRa [10]1 year ago
3 0

The correct option is b. The one with the lowest mass.

An object's kinetic energy is determined by

k=1/2mv^2

where

m is the object's mass.

v is the object's speed.

The three missiles in this puzzle have varying masses but the same beginning kinetic energy.

The three projectiles will all have the same kinetic energy when they hit the ground because mechanical energy is conserved, assuming there is no air resistance (because the potential energy that they have lost is the same, since they have been launched from the same height, and they reach the same final altitude, the ground).

hence,

K1=k2=k3

To know more about kinetic energy refer to brainly.com/question/14604194

#SPJ4

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A car is behind a truck going 25 m/s on the highway. The car’s driver looks for an opportunity to pass, guessing that his car ca
user100 [1]

Answer:

No he should not attempt the pass

Explanation:

Let t be the time it takes for the car to pass the truck. The driver should ONLY attempt to pass when the distance covered by himself plus the distance covered by the oncoming car is less than or equal 400 m (a near miss)

At acceleration of 1m/s2 and a clear distance of 10 + 20 + 10 = 40 m, we can use the following equation of motion to estimate the time t in seconds

s = at^2/2

40 = 1t^2/2

t^2 = 80

t = \sqrt{80} = 8.94 s

Within this time frame, the first car would have traveled a total distance of the clear distance (40m) plus the distance run by the truck, which is

8.94 * 25 = 223.6m

So the total distance traveled by the first car is 223.6 + 40 = 263.6m

The distance traveled by the 2nd car within 8.94 s at rate of 25m/s is

8.94 * 25 = 223.6 m

So the total distance covered by both cars within this time frame

223.6 + 263.6 = 487.2m > 400 m

So no, he should not attempt the pass as we will not clear it in time.

8 0
3 years ago
Calculate kp at 298.15 k for the reactions (a), (b), and (c) using δg°f values.
vivado [14]
(a) 2NO(g) + O₂(g) ⇄2NO₂(g)kp
(b)  2N₂O(g)⇄2NO(g) + N₂(g) kp
(c)  N₂(g) + O₂(g)⇄ 2NO(g) kp
Now A is
2NO +O₂⇄2NO₂
ΔG° =ΔG° products - ΔG reactants
=2× 51.3-(256.6)
-70.6kJ/mol.
ΔG° = -RT Inkp
-70.6 = -8.314 ×10⁻³ ˣ 298.15 ˣInkJ
InkJ = 28.48
kp=2.34 ˣ 10¹²

B is 
ΔG° = 2× 86.6 - 2 × 104.2 = -35.2
-35.2 = 8.314 × 10⁻³ ˣ 298.15 ˣInkJ
InkJ = 14.2
kp = 1.47ˣ 10⁶

C is
It is also similar
kp = 4.62 ˣ 10⁻³I
6 0
3 years ago
Two identical resistors are connected first in series and second in parallel. The equivalent resistances of the two types of con
trasher [3.6K]

Answer:

\frac{R_{s} }{R_{p} } =\frac{R_{1} }{R_{2} }+\frac{R_{2} }{R_{1} } +2

Explanation:

We have series and parallel combination of two resisters R_{1} and R_{2}.

Series combination is

R_{s}= R_{1}+R_{2} and Parallel is R_{p} = (\frac{1}{R_{1}}+\frac{1}{R_{2}}  )^-1

Now dividing series equivalent resistance by parallel resistance gives us

\frac{R_{s} }{R_{p} } =\frac{R_{1} }{R_{2} }+\frac{R_{2} }{R_{1} } +2.

Note! series Combination is simply superposition of two elements (resisters in this case ) in a circuit.

4 0
3 years ago
A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal
PilotLPTM [1.2K]
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline 
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
3 0
3 years ago
Set the radius to 2.0 m and the velocity to 1.0 m/s. Keeping the radius the same, record the magnitude of centripetal accelerati
jek_recluse [69]

Answer:

a=4\ m/s^2

Explanation:

Given that,

Radius, r = 2 m

Velocity, v = 1 m/s

We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{1}\\\\=4\ m/s^2

So, the magnitude of centripetal acceleration is 4\ m/s^2.

5 0
2 years ago
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