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vo = 25 m/sec
<span>vf = 0 m/sec </span>
<span>Fμ = 7100 N (Force due to friction) </span>
<span>Fg = 14700 N </span>
<span>With the force due to gravity, you can find the mass of the car: </span>
<span>F = ma </span>
<span>14700 N = m (9.8 m/sec²) </span>
<span>m = 1500 kg </span>
<span>Now, we can use the equation again to find the deacceleration due to friction: </span>
<span>F = ma </span>
<span>7100 N = (1500 kg) a </span>
<span>a = 4.73333333333 m/sec² </span>
<span>And now, we can use a velocity formula to find the distance traveled: </span>
<span>vf² = vo² + 2a∆d </span>
<span>0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d </span>
<span>0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d </span>
<span>-625 m²/sec² = (-9.466666666667 m/sec²) ∆d </span>
<span>∆d = 66.0211267605634 m </span>
<span>∆d = 66.02 m</span>
Answer:
Distance = 30 m
Explanation:
Since the position of the bus is at zero, and distance is a scalar quantity. That is, we are only concerned about the magnitude
If Angela ran to the bus and back to where she started, the distance travelled will be:
Distance = 15 + 15 = 30 m
But her displacement will be:
Displacement = 15 - 15 = 0.
Answer:
x=0.154kg
Explanation:
(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0
(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0
Answer:
Part a)

Part b)

Explanation:
Part a)
In order to have same range for same initial speed we can say


so after comparing above we will have

so we have


Part b)
Time of flight for the first ball is given as



Now for other angle of projection time is given as


So here the time lag between two is given as



Answer:
Heat would flow from a hot cup of coffee to a room temperature table.
Explanation:
Thermal energy will flow from a hot cup of coffee to a room temperature table.
Heat is a form of thermal energy which is the average kinetic energy of the particles of system.
- Typically heat flows from a place system at high temperature to one with a low temperature.
- Heat never flows from a cold to a hot body
- Heat moves from a hot to a cold body in a system.