Ask question

Ask question

All categories

3 years ago

9

A child and sled with a combined mass of 48.8 kg slide down a frictionless hill that is 7.05 m high. If the sled starts from res

t, what is its speed at the bottom of the hill?
___m/s

1 answer:

nexus9112 [7]3 years ago

7 0

Answer:

Speed at bottom of the hill (v) = 11.74 m/s

Explanation:

Given:

Combined mass = 48.8 kg

Height h = 7.05 m

Find:

Speed at bottom of the hill (v)

Computation:

v² = 2gh

v = √2 x 9.8 x 7.05

v = √138.18

v = 11.74 m/s

Speed at bottom of the hill (v) = 11.74 m/s

You might be interested in

With what minimum speed must you toss a 110 g ball straight up to just touch the 11-m-high roof of the gymnasium if you release

Sholpan [36]

Answer:

v = 13.79 m/s

Explanation:

given,

mass of ball = 110 g

height = 11 m

ball is released from = 1.3 m

minimum speed = ?

using conservation of energy

Potential energy is conserved in the form of kinetic energy

$P E =KE$

$m g h= \dfrac{1}{2}mv^2$

$v^2 = 2 g h$

$v=\sqrt{2 g h}$

$v=\sqrt{2\times 9.8 \times (11-1.3)}$

$v=\sqrt{2\times 9.8 \times 9.7}$

$v=\sqrt{190.12}$

v = 13.79 m/s

7 0

3 years ago

A plant box falls from the windowsill 25.0 m above the sidewalk and hits the cement 3.0 s later. What is the box's velocity when

mamaluj [8]

Answer:

22m/s

Explanation:

To find the velocity we employ the equation of free fall: v²=u²+2gh

where u is initial velocity, g is acceleration due to gravity h is the height, v is the velocity the moment it hits the ground, taking the direction towards gravity as positive.

Substituting for the values in the question we get:

v²=2×9.8m/s²×25m

v²=490m²/s²

v=22.14m/s which can be approximated to 22m/s

8 0

3 years ago

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

$g=-9.8 m/s^2$ is the acceleration of gravity

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

$v'^2-u^2=2gh'$

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s$

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

$\Delta t = 2\cdot 0.04 =0.08 s$

8 0

4 years ago

What is a prediction

faust18 [17]

An educated guess about something. (What might happen in the future)

7 0

3 years ago

Read 2 more answers

Nonmetals in the periodic table have a negative oxidation number. Which statement best explains why?

Tema [17]

They have a negative oxidation number because nonmetals gain electrons, thus making them negatively charged ions (anions). Metals become cations and have a positive charge because they lose electrons.

7 0

3 years ago