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3 years ago

Can you help me pls

Physics

1 answer:

Kruka [31]3 years ago

7 0

Answer:

The student can conclude that as the bulb is not lighting up so the object can be insulator

Explanation:

PLEASE MARK ME BRAINLIEST IF MY ANSWER IS CORRECT PLEASE

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A student pushes on a crate with a force of 100 Newtons directed to the right.

romanna [79]

Answer:

100N

Explanation:

Newton's third law of motion

For every action, there is an equal and opposite reaction.

Therefore 100N of force is exerted by the crate on student as a reaction to his action

6 0

3 years ago

The Moon is attracted to the Earth. The mass of the Earth is 6.0x1024 kg and the mass of the Moon is 7.4x1022 kg. If the Earth a

xxTIMURxx [149]

Sorry I don't know the answer

7 0

3 years ago

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

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

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

For downgrade of 4%, Substituting values

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

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

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

3 0

3 years ago

A 8 g object displaces 50ml of water, what is its specific gravity?

Igoryamba

Answer:

8.3

Explanation:

thats what i think sorry if its wrong im ok at doing this stuff

4 0

3 years ago

Read 2 more answers

If the electric field of an electromagnetic wave is in the x-direction and the magnetic field of the wave is in the y-direction,

Ulleksa [173]

Answer:

Positive z-direction

Explanation:

According to Fleming's right hand rule, point the fingers of your right hand in the direction of Electric field E ( positive x-direction), and curl your fingers toward magnetic field B (positive y-direction), and your thumb points in the direction of propagation of wave (positive z-direction).

Therefore, the correct option will be positive z-direction.

3 0

3 years ago