Answer:
The box will move to the right with increasing speed
Explanation:
According to Newton's second law:
(1)
where
F is the net force on an object
m is the mass of the object
a is its acceleration
For the box in this problem, there is only one force acting on it: a force of 5 N from the left (so, to the right). This means that according to (1) the box will experience an acceleration: the direction of the acceleration is the same as the direction of the net force (so, to the right), therefore the box will move to the right and its speed will continue increasing due to fact that it is being accelerated.
vf = 10 m/s. A ball with mass of 4kg and a impulse given of 28N.s with a intial velocity of 3m/s would have a final velocity of 10 m/s.
The key to solve this problem is using the equation I = F.Δt = m.Δv, Δv = vf - vi.
The impulse given to the ball with mass 4Kg is 28 N.s. If the ball were already moving at 3 m/s, to calculate its final velocity:
I = m(vf - vi) -------> I = m.vf - m.vi ------> vf = (I + m.vi)/m ------> vf = I/m + vi
Where I 28 N.s, m = 4 Kg, and vi = 3 m/s
vf = (28N.s/4kg) + 3m/s = 7m/s + 3m/s
vf = 10 m/s.
.
Answer:
The answer to your question is:
a) t1 = 2.99 s ≈ 3 s
b) vf = 39.43 m/s
Explanation:
Data
vo = 10 m/s
h = 74 m
g = 9.81 m/s
t = ? time to reach the ground
vf = ? final speed
a) h = vot + (1/2)gt²
74 = 10t + (1/2)9.81t²
4.9t² + 10t -74 = 0 solve by using quadratic formula
t = (-b ± √ (b² -4ac) / 2a
t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)
t = (-10 ± √ 1550.4 ) / 9.81
t1 = (-10 + √ 1550.4 ) / 9.81 t2 = (-10 - √ 1550.4 ) / 9.81
t1 = (-10 ± 39.38 ) / 9.81 t2 = (-10 - 39.38) / 9.81
t1 = 2.99 s ≈ 3 s t2 = is negative then is wrong there are
no negative times.
b) Formula vf = vo + gt
vf = 10 + (9.81)(3)
vf = 10 + 29.43
vf = 39.43 m/s
Answer:
= 85.7 ° C
Explanation:
For this exercise we will use the calorimetry heat ratios, let's start with the heat lost by the evaporation of coffee, since it changes from liquid to vapor state
Q₁ = m L
Where m is the evaporated mass (m = 2.00 103-3kg) and L is 2.26 106 J / kg, where we use the latent heat of the water
Q₁ = 2.00 10⁻³ 2.26 10⁶
Q1 = 4.52 10³ J
Now the heat of coffee in the cup, which does not change state is
Q coffee = M ( -)
Since the only form of energy transfer is terminated, the heat transferred is equal to the evaporated heat
Qc = - Q₁
M ce ( -) = - Q₁
The coffee dough left in the cup after evaporation is
M = 250 -2 = 248 g = 0.248 kg
-Ti = -Q1 / M
= Ti - Q1 / M
Since coffee is essentially water, let's use the specific heat of water,
= 4186 J / kg ºC
Let's calculate
= 90.0 - 4.52 103 / (0.248 4.186 103)
= 90- 4.35
= 85.65 ° C
= 85.7 ° C