Assuming the power delivered by the horse does not change, the speed of the cart will decrease.
In fact, the power delivered by the horse is the work done by the horse (W) per unit time (t):

<span>If several bags are added to the cart, the horse must do more work to transport them. Therefore, W in the fraction increases. But if the power P of the horse is constant, then it means that the time t must increase as well. So, the horse will take more time to transport the car, and this means that the speed of the cart has decreased.</span>
> Non-zero numbers (like 1,2,3,4...) are always significant
> A zero sandwiched between two non-zero numbers is always significant
> Trailing zeros in a decimal (not whole number like million) are always significant.
<span>0,020170 = 2.0170 × 10^-2
5 sig-figs
</span>
Answer:
a) 19.4 m/s
b) 19 m/s
Explanation:
a) In the given question,
the potential energy at the initial point = Ui = 0
the potential energy at the final point = Uf = mgh
the kinetic energy at the initial point = Ki = 1/2 mv₀².
the kinetic energy at the final point = Kf = 0
work done by air= Ea= fh = 0.262 N
Now, using the law of conservation of energy
initial energy= final energy
Ki +Ui = Kf + Uf +Ea
1/2 mv₀² + 0 = 0 + mgh + fh
1/2 mv₀² = mgh + fh
h = v₀²/ 2g (1 +f/w)
calculate m
m= w/g = 5.29 /9.8
= 0.54 kg
h = 20 ²/ (2 x9.80) x (1 0.265/5.29)
h = 19.4 m.
b) 1/2 mv² + 2fh = 1/2 mv₀²
Vg = 19 m/s
Answer:
a= - 6.667 m/s²
Explanation:
Given that
The initial speed of the box ,u= 20 m/s
The final speed of the box ,v= 0 m/s
The distance cover by box ,s= 30 m
Lets take the acceleration of the box = a
We know that
v²= u ² + 2 a s
Now by putting the values in the above equation we get
0²=20² + 2 a x 30

a= - 6.667 m/s²
Negative sign indicates that velocity and acceleration are in opposite direction.
Therefore the acceleration of the box will be - 6.667 m/s² .
The answer is A: can change