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3 years ago

Pls help will mark brainliest

Physics

2 answers:

Blababa [14]3 years ago

5 0

Answer: a!! Bc 10x2=20!

Explanation:

Elanso [62]3 years ago

3 0

Answer:

20N

Explanation:

F=m*a

F=10*2

F=20N

.....

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2 years ago

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3 years ago

What distance will a vehicle travel before coming to a complete stop from a speed of 70 mph, (a) When the vehicle is traveling o

OLga [1]

Answer:

(a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be $723.91>SSD>620.2$

(c), The SSD will be 910.5 ft.

Explanation:

Given that,

Speed = 70 mph

Suppose, a perception reaction time of 2.5 sec and the coefficient of friction is 0.35

We need to calculate the stopping sight distance

Using formula of SSD

$SSD=1.47\times v\times t+\dfrac{v^2}{30\times(f\pm g)}$

Where, v = speed of vehicle

t = perception reaction time

f = coefficient of friction

g = gradient of road

(a). If the gradient of road is zero.

Then, the stopping sight distance will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35)}$

$SSD=723.9\ ft$

(b-1). If the gradient of road is 0.1

Then, the stopping sight distance will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35+0.1)}$

$SSD=620.2\ ft$

(b-2). If the grade continuously decrease then the SSD will be increase.

But if the grade is increase then the SSD will be decrease and for flat grade the SSD will be more.

So, The SSD will be $723.91>SSD>620.2$

(c). When the vehicle is traveling downhill on a roadway of constant grade then the vehicle take will be more SSD

So, The SSD will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35-0.1)}$

$SSD=910.5\ ft$

Hence, (a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be $723.91>SSD>620.2$

(c), The SSD will be 910.5 ft.

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4 years ago

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