What is the Force in Newtons exerted by an object with a mass of 15 kg and an Acceleration of 12 m/s2?

A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out

anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

$0 + mgH = \frac{1}{2}mv^{2} + 0$

$2gH = v^{2}$

$v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s$

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

$mv = (m + m')v'$

$65.0\times 9.89 = (65.0 + 20.0)v'$

$v' = 7.56\ m/s$

Now, time taken for the fall:

$h = \frac{1}{2}gt^{2}$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s$

Now, the horizontal distance is given by:

x = v't = $7.56\times 0.638 = 4.823\ m$

7 0

3 years ago

Read 2 more answers

A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time

saul85 [17]

Answer:

t = 0.714 s and x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

vₓ = 7.0 m / s

Let's find the time it takes to get to the river

y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

0 = y₀ + 0 - ½ g t²

t = $\sqrt{\frac{2y_o}{g} }$

t = ra 2 2.5 / 9.8

t = 0.714 s

the distance traveled is

x = vₓ t

x = 7.0 0.714

x = 5.0 m

3 0

3 years ago

What happens to centripetal acceleration as the radius of curvature decreases and the speed is constant, and why

lutik1710 [3]

It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature.

Hopr it helps :)

4 0

3 years ago

A circus act involves a trapeze artist and a baby elephant. They are going to balance on a teeter-totter that is 10 meters long

elena-s [515]

Answer:

Explanation:

Balance point will be achieved as soon as the weight of the baby elephant creates torque equal to torque created by weight of woman about the pivot. torque by weight of woman

weight x distance from pivot

= 500x 5

= 2500 Nm

torque by weight of baby woman , d be distance of baby elephant from pivot at the time of balance

= 2500x d

for equilibrium

2500 d = 2500

d = 1 m

So elephant will have to walk up to 1 m close to pivot or middle point.

6 0

3 years ago

Kissa and Lily ran a race in gym class Lily was off balance so she did not start until 2 seconds had passed

Paul [167]

Average speed = (distance covered) / (time to cover the distance)

Tissa covered 60 meters in 10 seconds. Her average speed was

(60 m) / (10 sec) = 6 m/s.

That's the slope of the dotted line.

Lilly covered 60 meters in 8 seconds. Her average speed was

(60 m) / (8 sec) = 7.5 m/s .

That's the slope of the solid line.

Lilly covered the same distance in less time, and both girls
arrived at the finish line together. Technically, in science talk,
we would say that Lilly ran "faster", and her average speed
was "greater".

We can detect that by looking at the graph, because Lilly's line
has the characteristic of being "steeper", and we know that the
slope of the line on a distance/time graph is "speed".

3 0

3 years ago