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2 years ago

What are common in the pictures?

Physics

1 answer:

sesenic [268]2 years ago

8 0

Answer:

the flowers the left have them on there heads the middle on there chest and the right one has them on there shoulders

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What happens to an iron bar when it is placed within the coils of a solenoid

Rina8888 [55]

When the iron bar is placed within the coils of a solenoid it then becomes a magnet.

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2 years ago

Read 2 more answers

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

2 years ago

Which is not a way to accelerate an object?

GREYUIT [131]

Slowing an object down is not a means of accelerating it. It actually decelerates the motion of an object. Speeding it up, changing its direction and applying balanced forces accelerate an object. In order for an object to accelerate, a force must be applied. It follows Newton’s second law of motion where it states that a body at rest remains at rest unless a force is acted upon it. When you move an object, you are exerting a force onto it. By exerting a force on the object, you are actually displacing it from its initial position. You cannot apply force to the object without altering its position. Keep in mind that when you exert work, you are exerting energy too.

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2 years ago

A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory (ignorin

Nat2105 [25]

Answer:

b) Gravity

Explanation:

Gravity acts all of the time, when you apply force to a projectile it has to be more than the forces of the gravity and air resistance together so the projectile can move, when the rock is at the top of its trajectory the force that you applied at the beginning is getting lost, so the other forces (air resistance and gravity) make the rock fall to the floor.

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3 years ago

As you lift an 88 N box straight upward you produce a power of 72 W. What is the speed of the box?

jeka57 [31]

F = force applied to lift the box = weight of the box = 88 N

P = power produced while lifting the box upward = 72 Watt

v = speed of the box = ?

Speed of the box is given as

v = $\frac{P}{F}$