An athlete of mass 70kg runs at a velocity of 6ms5-1 calculate the kinetic energy of the athlete.

The radius of our clock face is 9.2cm. It is 8:44; we are done at 9:11. How far will the minute hand (the larger one) travel?

gogolik [260]

Answer:

9:36 and how far it will travel is 26 minutes

8 0

2 years ago

A ball is shot straight up into the air from the ground with initial velocity of 44ft/sec44ft/sec. assuming that the air resista

solniwko [45]

Convert the given in SI units.

(44 ft/sec)(1 m/ 3.28 ft) = 13.41 m/sec

The distance traveled and the initial velocity can be related through the equation,

d = (Vf)² - (Vi)²/ 2a

where d is the distance, Vf is the final velocity, Vi is the initial velocity, a is the acceleration due to gravity. Substituting the known values from the given above,

d = ((0 m/s)² - (13.41 m/s)²)/ 2(-9.8 m/s²)

The value of d from the equation,

d = 9.17 meters

Convert this to feet,

d = (9.17 m)(3.28 ft / 1 m) = 30 ft

Answer: 30 ft

3 0

3 years ago

The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared

velikii [3]

Answer:

$\displaystyle a(5)=-32$

Explanation:

<u>Instant Acceleration</u>

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

$\displaystyle v(t)=\frac{df}{dt}$