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MAVERICK [17]
3 years ago
5

What is the mass of an object traveling at 20 m/s with a kinetic energy of 4000 J?​

Physics
1 answer:
valentinak56 [21]3 years ago
5 0

Answer:

20 kg

Explanation:

Kinetic energy=½*Mass * velocity²

4000= ½* m*20²

8000=400m

m=8000/400

m=20 kg

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Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds 5.0
Tems11 [23]
A. Average speed is weighted mean (1 × 2 + 2 × 3 + 3 × 5 + 4 × 7 + 3 × 9 + 2 × 12.5)/15 = (2 + 6 + 15 + 28 + 27 + 25)/15 = 103/15 = 6.867 b. RMS is square root of 1/15 times sum of squares of speeds Sum of squares is 4 + 9 + 9 + 25 + 25 + 25 + 49 + 49 + 49 + 49 + 81 + 81 + 81 +156.25 + 156.25 = 848.5 
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5 0
3 years ago
Me ajudem, Por favor!!!!!!
zzz [600]

Answer:

a)  a=4\,\frac{m}{s^2}

b)  V(t)=4\,t\,+3

c)  V(1)=7 \,\frac{m}{s} \\

d)  Displacement = 22 m

e)  Average speed = 11 m/s

Explanation:

a)

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}

Therefore,  acceleration is a=4\,\frac{m}{s^2}

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

y=m\,x+b\\V(t)=4\,t\,+3

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = \frac{(7+15)\,2}{2} = 22\,\,m

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = \frac{22}{2} = 11\, \,\frac{m}{s}

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