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3 years ago

If

Physics

1 answer:

nadya68 [22]3 years ago

8 0

Answer:

hhjjkkkksksksjskkskakakkskskksksksoao

Explanation:

hiiii look forward but I don't know how to do it

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Help me please i dont understand

Mamont248 [21]

Answer:

0.58

Explanation:

Sinẞ = opposite ÷ hypotenuse

Sinẞ = 5 ÷ 8.6

Sinẞ = 0.5814

Sinẞ ≈ 0.58

5 0

2 years ago

Read 2 more answers

A power supply is connected to a 59 Ohm resistor and a 53 Ohm resistor in series. The total current is found to be 0.15 A. What

kirill [66]

the answer is 47265.dug

Explanation:

im bot sure

4 0

2 years ago

The plate near the Artic Ridge is among those with the slower growth rates, moving slightly less than 2.5 cm/year. How much will

Aleks04 [339]

Greater than 25 . 2.5 x 10 = 30

4 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

A 14 kg tennis ball moves at a velocity of 11 m/s. The ball is struck by a racket, causing it to rebound in the opposite directi

Charra [1.4K]

To solve this problem we will apply the concepts related to the momentum.

This is defined as the product between the change in velocity and the mass of the object, that is

$p = m\Delta v$

$p = m (v_f-v_i)$

Where,

m = mass

$v_f =$ Final velocity

$v_i =$ Initial velocity

Our values are given as,

m = 14kg

$v_i$ = 11m/s

$v_f= - 5m/s \rightarrow$ the negative Symbol implies that the direction is opposite to the initial one and therefore there is also a change in the sense of magnitude

$p = (14)(-5-11)$

$p = 14(-16)$

$p = -224 kg \cdot m/s$

The negative symbol indicates that the momentum has a direction opposite to that of the initial velocity. Or failing that, it has the same direction of the final speed

7 0

3 years ago