A car is making a 50 mi trip. It travels the first half of the total distance 25.0 mi at 7.00 mph and the last half of the total distance 25.0 mi at 52.00 mph. (a) What is the total time in hours of the trip? Keep two decimal places. 4.05 Correct (100,0%) (b) What is the car's average speed in mph for the entire trip? Keep two decimal places. 12 35 Correct (100,0%) Submit The car travels the same distance again, but this time, in the first half of the time its speed is 7.00 mph and in the second half of the time its speed is 52.00 mph. c) What is the total time in hours of the trip? Keep two decimal places.
Answers
Answer:
a) The total time of the trip is 4.05 h.
b) The average speed of the car is 12.35 mi/h.
c) The total time of the trip is 1.69 h.
Explanation:
Hi there!
a) The equation of traveled distance for a car traveling at constant speed is the following:
x= v · t
Where:
x = traveled distance.
v = velocity.
t = time.
Solving the equation for t, we can find the time it takes to travel a given distance "x" at a velocity "v":
x/v = t
So, the time it takes the car to travel the first half of the distance will be:
t1 = 25.0 mi / 7.00 mi/h
And for the second half of the distance:
t2= 25.0 mi / 52.00 mi / h
The total time will be:
total time = t1 + t2 = 25.0 mi / 7.00 mi/h + 25.0 mi / 52.00 mi / h
total time = 4.05 h
The total time of the trip is 4.05 h.
b) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it takes to travel that distance (t). In this case, the traveled distance is 50 mi and the time is 4.05 h. Then:
a.s = d/t
a.s = 50 mi / 4.05 h
a.s = 12.35 mi/h
The average speed of the car is 12.35 mi/h
c) Let's write the equations of traveled distance for both halves of the trip:
For the first half, you traveled a distance d1 in a time t1 at 7.00 mph:
7.00 mi/h = d1/t1
Solving for d1:
7.00 mi/h · t1 = d1
For the second half, you traveled a distance d2 in a time t2 at 52.00 mph.
52.00 mi/h = d2/t2
52.00 mi/h · t2 = d2
We know that d1 + d2 = 50 mi and that t1 and t2 are equal to t/2 where t is the total time:
d1 + d2 = 50 mi
52.00 mi/h · t/2 + 7.00 mi/h · t/2 = 50 mi
Solving for t:
29.5 mi/h · t = 50 mi
t = 50 mi / 29.5 mi/h
t = 1.69 h
The total time of the trip is 1.69 h.
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Thank you!!
Answers
Answer:
(a) 22 kN
(b) 36 kN, 29 kN
(c) left will decrease, right will increase
(d) 43 kN
Explanation:
(a) When the truck is off the bridge, there are 3 forces on the bridge.
Reaction force F₁ pushing up at the first support,
reaction force F₂ pushing up at the second support,
and weight force Mg pulling down at the middle of the bridge.
Sum the torques about the second support. (Remember that the magnitude of torque is force times the perpendicular distance. Take counterclockwise to be positive.)
∑τ = Iα
(Mg) (0.3 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L)
F₁ = ½ Mg
F₁ = ½ (44.0 kN)
F₁ = 22.0 kN
(b) This time, we have the added force of the truck's weight.
Using the same logic as part (a), we sum the torques about the second support:
∑τ = Iα
(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)
F₁ = ½ Mg + ⅔ mg
F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)
F₁ = 36.0 kN
Now sum the torques about the first support:
∑τ = Iα
-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)
F₂ = ½ Mg + ⅓ mg
F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)
F₂ = 29.0 kN
Alternatively, sum the forces in the y direction.
∑F = ma
F₁ + F₂ − Mg − mg = 0
F₂ = Mg + mg − F₁
F₂ = 44.0 kN + 21.0 kN − 36.0 kN
F₂ = 29.0 kN
(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)
As x increases, F₁ decreases and F₂ increases.
(d) Using our equation from part (c), when x = 0.6 L, F₂ is:
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)
F₂ = ½ Mg + mg
F₂ = ½ (44.0 kN) + 21.0 kN
F₂ = 43.0 kN
Answer:
a. Left support = Right support = 22 kN
b. Left support = 36 kN
Right support = 29 kN
c. Left support force will decrease
Right support force will increase.
d. Right support = 43 kN
Explanation:
given:
weight of bridge = 44 kN
weight of truck = 21 kN
a) truck is off the bridge
since the bridge is symmetrical, left support is equal to right support.
Left support = Right support = 44/2
Left support = Right support = 22 kN
b) truck is positioned as shown.
to get the reaction at left support, take moment from right support = 0
∑M at Right support = 0
Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0
Left support = 44 (0.3) + 21 (0.4)
0.6
Left support = 36 kN
Right support = weight of bridge + weight of truck - Left support
Right support = 44 + 21 - 36
Right support = 29 kN
c)
as the truck continues to drive to the right, Left support will decrease
as the truck get closer to the right support, Right support will increase.
d) truck is directly under the right support, find reaction at Right support?
∑M at Left support = 0
Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0
Right support = 44 (0.3) + 21 (0.6)
0.6
Right support = 43 kN
Answers
Answer:
The length is 5.2 mm.
Explanation:
Given that,
Time period T²= 0.021 s²
Gravity due to acceleration = 9.78 m/s²
We need to calculate the length
Using formula of time period of pendulum
Where, l = length
g = acceleration due to gravity
T = time period
Put the value into the formula
Hence, The length is 5.2 mm.
Answers
E = hv, where v = the frequency and h = planck's constant.
So E = 5.20 x10^14 x 6.63 x 10^-34
= 3.45 x 10^-19 Joules
Answers
Explanation:
a scientific theory is a well substantiated explanation of some aspect of the nature world, based on a body of facts that have been repeatedly confirmed through observation and experiment. search fact-supported theories are not "guesses" but reliable account of the real world .
2)(b) If you throw your 1.08-kg boot with an average force of 391 N, and the throw takes 0.576 s (the time interval over which you apply the force), what is the magnitude of the force that the boot exerts on you? (Assume constant acceleration.)
391 N
3)(c) How long does it take you to reach shore, including the short time in which you were throwing the boot?
Just number 3
Answers
Answer:
1a) The direction to throw the boot is directly away from the closest shore.
2b) The magnitude of the force that the thrown boot exerts on the engineer = 391 N
3c) Time taken to reach shore = 8.414 s
Explanation:
1a) Newton's third law of motion explains that for every action, there is an equal and opposite reaction.
The force generated by throwing the boot in one direction is exerted back on the engineer as recoil in the opposite direction.
Hence, the best direction to throw the boot is opposite the direction that the engineer intends to move towards.
2b) Just as explained in (1a) above, the force exerted in one direction always has a reaction of the same magnitude in the opposite direction.
Hence, the force exerted by the boot on the engineer is equal to the force exerted by the engineer on the boot = 391 N.
3c) For this part, we analyze the total motion of the engineer.
The force exerted by the boot on the engineer initially accelerates the engineer until the engineer reaches a constant velocity dictated the impulse of the initial force (since impulse is equal to change in momentum), this constant velocity then takes the engineer all the way to shore, since the ice surface is frictionless.
The weight of the engineer = W = 588 N
W = mg
Mass of the engineer = (W/g) = (588/9.8) = 60 kg
Force exerted on the engineer by the thrown boot = F = 391 N
F = ma
Initial acceleration of the engineer = (F/m) = (391/60) = 6.52 m/s²
We can then calculate the distance covered during this acceleration
X₁ = ut + ½at₁²
u = initial velocity of the engineer = 0 m/s (the engineer was initially at rest)
t₁ = time during which the force acts = 0.576 s
a = acceleration during this period = 6.52 m/s²
X₁ = 0 + 0.5×6.52×0.576² = 1.08 m
For the second part of the engineer's motion, the velocity becomes constant.
So, we first calculate this constant velocity
Impulse = Change in momentum
F×t = mv - mu
F = Force causing motion = 391 N
t = time during which the force acts = 0.576 s
m = mass of the engineer = 60 kg
v = final constant velocity of the engineer = ?
u = initial velocity of the engineer = 0 m/s
391 × 0.576 = 60v
v = (391×0.576/60) = 3.7536 m/s.
The distance from the engineer's initial position to shore is given as 30.5 m
The engineer covers 1.08 m during the time the force causing motion was acting.
The remaining distance = X₂ = 30.5 - 1.08 = 29.42 m
We can then calculate the time taken to cover the remaining distance, 29.42 m at constant velocity of 3.7536 m/s
X₂ = vt₂
t₂ = (X₂/v) = (29.42/3.7536) = 7.838 s
Time taken to reach shore = t₁ + t₂ = 0.576 + 7.838 = 8.414 s
Hope this Helps!!!
Answers
Answer: The volume is decreasing at a rate of 80 cm3/min
Explanation: Please see the attachments below
Answer: 80 cm³/min
Explanation:
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