A car covers 100 km in 2 h , what is the speed of the car.

All of the waves in the electromagnetic spectrum are _______ waves.

crimeas [40]

All of the waves in the electromagnetic spectrum are transverse waves.

5 0

3 years ago

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Why don't atoms get too close?

Lera25 [3.4K]

Answer:

I would have to go with A, or maybe....yea A

3 0

2 years ago

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What is a non-example of gravitational force?

qaws [65]

A non-example of force would be something that stay sill like a balloon in the air..... taste, smell, feel, texture, color, opinion, faith, hope, sincerity, honest, speed, momentum, altitude, volume, loudness, area, length, acidity, obesity, nationalism, current, resistance, viscosity, wavelength, flow, rate, frequency, albedo, diameter, age, temperature, acceleration, body mass index, salinity, specific, specific gravity, consciousness, intelligence, refraction index, mass, time, date rate, switching speed, libido, focal length, and latency are not force. And even there are many other things that also are not force, too.

4 0

3 years ago

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What the kinetic Energy of 80kg object this is Moving with a Speed of 23 m/s

Katyanochek1 [597]

Explanation:

Kinetic Energy Formula = ½mv²

½×800kg×(23m/s)²
400kg×529m²/s²
211600 kg•m²/s²
211700 joule Answer

4 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

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

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

3 years ago

A car covers 100 km in 2 h , what is the speed of the car.​

A car covers 100 km in 2 h , what is the speed of the car.