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polet [3.4K]
3 years ago
11

12–139. Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is posted at 60 km>h, d

etermine the minimum acceleration experienced by the passengers.
(60)2



y

2

0)2 = 1


*12–140. Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is posted at 60 km>h, determine the maximum acceleration experienced by the passengers.

12–142. The race car has an initial speed vA = 15 m> s at
a. If it increases its speed along the circular track at the rate at = (0.4s) m> s2, where s is in meters, determine the time needed for the car to travel 20 m. Take r = 150 m.




x


Probs. 12–139/140

Physics
1 answer:
Sergio039 [100]3 years ago
5 0
Check the attached file for the answer.

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A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of
sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height \Delta h depends on the square of the time:

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t,

where

  • \Delta h is the change in the ball's height.
  • g is the acceleration due to gravity.
  • t is the time for which the ball is in the air.
  • v_0 is the initial vertical velocity of the ball.
  • The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. \Delta h = -0.950\;\text{m}.
  • Gravity pulls objects toward the earth, so g is also negative. g \approx -9.81\;\text{m}\cdot\text{s}^{-2} near the surface of the earth.
  • Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, v_0 = 0.

Solve for t.

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t;

\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2};

\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)};

t \approx 0.440315\;\text{s}.

What's the initial horizontal velocity of the ball?

  • Horizontal displacement of the ball: \Delta x = 0.352\;\text{m};
  • Time taken: \Delta t = 0.440315\;\text{s}

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}.

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0
3 years ago
two negative charges that are both -3.0 C push each other apart with a force of 19.2 N how far apart are the charges
Hoochie [10]
The electrostatic force between two charges q1 and q2 is given by
F=k_e  \frac{q_1 q_2}{r^2}
where k_e =8.99 \cdot 10^9 N m^2 C^{-2} is the Coulomb's constant and r is the distance between the two charges.

If we use F=19.2 N and q1=q2=-3.0 C, we can find the value of r, the  distance between the two charges by re-arranging the previous formula:
r= \sqrt{k_e \frac{q_1 q_2}{F} }= \sqrt{ 8.99 \cdot 10^9 N m^2 C^{-2} \frac{(-3.0C)^2}{19.2 N} } =6.49 \cdot 10^4 m=64.9 km
5 0
3 years ago
A 7.75-l flask contains 0.482 g of hydrogen gas and 4.98 g of oxygen gas at 65°c. What is the partial pressure of oxygen in the
dimulka [17.4K]

Answer:

0.558 atm

Explanation:

We must first consider that both gases behaves like ideal gases, so we can use the following formula: PV=nRT

Then, we should consider that, whithin a mixture of gases, the total pressure is the sum of the partial pressure of each gas:

P₀ = P₁ + P₂ + ....

P₀= total pressure

P₁=P₂= is the partial pressure of each gass

If we can consider that each gas is an ideal gas, then:

P₀= (nRT/V)₁ + (nRT/V)₂ +..

Considering the molecular mass of O₂:

M O₂= 32 g/mol

And also:

R= ideal gas constant= 0.082 Lt*atm/K*mol

T= 65°C=338 K

4.98 g O₂ = 0.156 moles O₂

V= 7.75 Lt

Then:

P°O₂=partial pressure of oxygen gas=  (0.156x0.082x338)/7.75

P°O₂= 0.558 atm

3 0
3 years ago
A zebra starts from rest and accelerates at 1.9 m/s how far has a zebra gone in five seconds
Maksim231197 [3]
To that multiply 1.9 by 5 to get 9.5.  The zebra has gone 9.5 m.
5 0
3 years ago
2- A car on a straight highway goes in the positive direction for 8 km and then backs up for 3.6 km. What are the distance and d
IRISSAK [1]

Answer:

11.6km

4.4km in the negative direction

Explanation:

Distance is the total length of path covered and traveled by a body.

So, for this car on a straight line;

  Total distance  = 8km + 3.6km  = 11.6km

Displacement is the distance traveled along a path and the direction it takes.

It is a vector quantity with magnitude and directional attributes.

For this journey;

 Displacement  = 8km  - 3.6km  = 4.4km in the negative direction.

4 0
2 years ago
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