All categories

3 years ago

When does the velocity and speed of a moving body becomes identical?

Physics

1 answer:

d1i1m1o1n [39]3 years ago

6 0

The speed and velocity of a moving body become identical when it tends to move in a straight line.

You might be interested in

Planet x has a mass of 4x1022 kg and a radius of 6x105 m What lise the grav ritational field strength in the surface of planet X

SCORPION-xisa [38]

Answer:

g ≈ 7.4 m/s²

Explanation:

The acceleration due to gravity on planet XX is ...

g = GM/r² = (6.67·10^-11 × 4·10^22)/(6·10^5)^2

g ≈ 7.4 m/s²

6 0

2 years ago

Choose the correct association for: dense bushes savanna rain forest or jungle

Gemiola [76]

Answer:

Climate is determined by averaging the seasonal weather conditions for a region over a period of many ______ years

Choose the correct association for: dense bushes rain forest or jungle

Choose the correct association for: plains savanna

3 0

2 years ago

A stack of 55 business cards is 1.85 cm tall. Use this information to determine the thickness of one business

Sindrei [870]

To find the answer, take 55 and divide it by 1.85 to get the thickness of one card. In this case the answer would be 29.72973 cm. each.

8 0

3 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

3 years ago

The fastest man alive can run the 100 meter dash in 9.58 seconds, calculate average speed in meters per second

densk [106]

Answer:

100 ÷ 9.58 = 10.44 (approximate answer)

3 0

3 years ago