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2 years ago

1. Straight-in spaces can leave you a safer out if

Physics

1 answer:

Juliette [100K]2 years ago

3 0

Answer:

C. you're able to reverse out of the parking spot

Explanation:

Straight-in parking is an approach of parking that allows a more flexible traffic layout where a driver can approach the spot from either direction and still safely park within the lines. It thus helps to prevent blockage of cars. Each car can move in and out freely preventing it from congestion.

This way of parking can leave you safe when you able to reverse out of the parking spot. It gives you greater control and makes it easier to maneuver out space. The benefits of Straight-in parking are,

- Allows for two-way traffic

- Drivers can line up the vehicle from multiple angles

- Saves time for drivers

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How often must you check the temperature of food that is being held with temperature control?.

loris [4]

You need to check the temperature of food being stored in a temperature-controlled environment every four hours. The process of changing a space's temperature is called temperature control.

Cooking food alone may not be enough to avoid food poisoning, though, if the bacteria in food are allowed to grow to large numbers. When the temperature is between 5°C and 63°C, bacteria can grow. The risk zone is the range between 5°C and 63°C.

Temperature control is a process where the passage of heat energy into or out of a space or substance is adjusted to achieve the desired temperature. This process involves measuring or otherwise detecting changes in the temperature of the space (and all of the objects contained therein) or of the substance.

Learn more about temperature here

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6 0

10 months ago

Rosný bod závisí od ?

WITCHER [35]

Answer:

what did u say and what language are you speaking in

8 0

3 years ago

In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.

Fiesta28 [93]

Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

$KE_i+PE_i =KE_f+PE_f$

$\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh$

where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

$u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s$

4 0

3 years ago

A glass lens that has an index of refraction equal to 1.57 is coated with a thin layer of transparent material that has an index

Bogdan [553]

Answer:

Find the given attachment

5 0

3 years ago

Read 2 more answers

A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1

pav-90 [236]

I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
$F_a=F_cf-F_g$
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
$F_a=m\frac{v^2}{r}-mg=179 $N$
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
$F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N$
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
$F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
v=\sqrt{gr}=8.29\frac{m}{s}$

6 0

3 years ago