All categories

3 years ago

During which interval is the cart accelerating?

Physics

1 answer:

Brilliant_brown [7]3 years ago

7 0

the barriers if the interval are Aand B the of displacement goes up.

from B to C it just stays the same, then from C to D no Movement, and from time D to E movement, but steady

You might be interested in

12. Suppose a person is covered head to foot by wool clothing with average thickness of 2.00 cm and is transferring energy by co

Vikki [24]

27 degrees because yea ik the answer

7 0

2 years ago

All steroids are illegal.<br> A. True<br> B. False

TiliK225 [7]

False noth all steroids are legal

7 0

3 years ago

Read 2 more answers

One method of determining the location of the center of gravity of a person is to weigh the person as he/she lies on a board of

andrey2020 [161]

Answer:

x = 1.018 m

Explanation:

given,

height of man = 190 cm

= 1.9 m

scale reading on left = 450 N

scale reading on the right = 390 N

Let center of gravity of man be x distance from feet, feet is on right side.

For system to be in equilibrium moment about center should be equal to zero.

∑M = 0

now,

450(1.9 - x ) - 390 × x = 0

450(1.9 - x ) = 390 × x

855 - 450 x = 390 x

840 x = 855

$x = \dfrac{855}{840}$

x = 1.018 m

hence, point of center of gravity from feet is equal to x = 1.018 m

7 0

4 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.

7 0

4 years ago

ANSWER PLEASE FAST I NEED IT

Deffense [45]

14.0.5 cm 13. wind speed surface area temperature

6 0

3 years ago

During which interval is the cart accelerating? ​

During which interval is the cart accelerating?