What is the approximate potential energy of a 65 kg person standing on a step that is 0.15 m above the ground?

Need help with formula: H(t) = -16^2 + vt + s

Rom4ik [11]

<span> The equation is h(t) = at^2 + vt + d where a = acceleration of gravity = - 32.174 ft/sec^2 v = 25 feet/sec d = starting height = 0 and h(t) = 0 when the ball hits the ground. So, 0 = - 32.174t^2 + 25t + 0 You can use the quadratic formula on that if you want, or you can solve like this: 0 = - 32.174t^2 + 25t 0 = t ( -32.174t + 25) So, one solution of that is t = 0, corresponding to the initial time when the ball is kicked. The other time is: 25 = 32.174t t = 25/32.174 = 0.777 seconds or approximately 0.8 seconds.</span>

4 0

4 years ago

Encontrar la distancia y desplazamiento de las dos trayectorias si se mueve el móvil Desde A hasta B

Lerok [7]

Answer:

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Explanation:

6 0

3 years ago

A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment. (a) calculate the average force

iragen [17]

(a) $-1.5\cdot 10^6 N$

First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:

$v^2 -u ^2 = 2ad$

where

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d = 1.00 cm = 0.01 m is the displacement of the person

Solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.01)}=-20000 m/s^2$

And the average force on the person is given by

$F=ma$

with m = 75.0 kg being the mass of the person. Substituting,

$F=(75)(-20000)=-1.5\cdot 10^6 N$

where the negative sign means the force is opposite to the direction of motion of the person.

b) $-1.0\cdot 10^5 N$

In this case,

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d = 15.00 cm = 0.15 m is the displacement of the person with the air bag

So the acceleration is

$a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.15)}=-1333 m/s^2$

So the average force on the person is

$F=ma=(75)(-1333)=-1.0\cdot 10^5 N$

7 0

4 years ago

• The average length is ________ cm. This is the mean or average. b)• Subtract the highest value from the lowest value: ________

coldgirl [10]

16, 5 , 3 = 16+5+3= 24 + 3

So at the end put 24 + 3 cm
And put 16 for the lengths
For the value 5 and for the diving thingy 3

7 0

4 years ago

15) A 328-kg car moving at 19.1 m/s in the +x direction hits from behind a second car moving at 13 m/s in the same direction. If

kolbaska11 [484]

Answer:

Explanation:

Given that,

Mass of first car

M1= 328kg

The car is moving in positive direction of x axis with velocity

U1 = 19.1m/s

Velocity of second car

U2 = 13m/s, in the same direction as the first car..

Mass of second car

M2 = 790kg

Velocity of second car after collision

V2 = 15.1 m/s

Velocity of first car after collision

V1 =?

This is an elastic collision,

And using the conservation of momentum principle

Momentum before collision is equal to momentum after collision

P(before) = P(after)

M1•U1 + M2•U2 = M1•V1 + M2•V2

328 × 19.1 + 790 × 13 = 328 × V1 + 790 × 15.1

16534.8 = 328•V1 + 11929

328•V1 = 16534.8—11929

328•V1 = 4605.8

V1 = 4605.8/328

V1 = 14.04 m/s

The velocity of the first car after collision is 14.04 m/s

5 0

3 years ago