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4 years ago

12

A rock is thrown downward from an unknown height above the ground with an initial speed of 10m/s. It strikes the ground 3s later

. Determine the initial

1 answer:

bazaltina [42]4 years ago

6 0

At the beginning of the 3 sec, its speed is 10 m/s.

Over the course of 3 sec, gravity adds (9.8 x 3) = 29.4 m/s
to its speed, so its speed after 3 sec is (10 + 29.4) = 39.4 m/s.

Its average speed during the 3 sec is

   (1/2) (10 + 39.4)

   = (1/2) (49.4) = 24.7 m/s .

In 3 sec, at an average speed of 24.7 m/s,
the rock travels

   (3) x (24.7) = 74.1 meters.

If there's no air resistance in its path, then it was thrown
from 74.1 meters above the ground.

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A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

2 years ago

The Earth's Radius is 6.3710x106 m and mass is 5.9742x1024 kg. What is the acceleration due to gravity at Mount Everest (elevati

Shalnov [3]

Answer is

9.773m/s^2

-----------------------------------------------------------------------------

Given,

h=8848m

The value of sea level is 9.08m/s^2. So, Let g′ be the acceleration due to the gravity on Mount Everest.

g′=g(1 − 2h/h)

=9.8(1 - 6400000/17696)

=9.8(1 − 0.00276)

9.8×0.99724

=9.773m/s^2

Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2

-----------------------------------------------------------------------

hope this helps :)

3 0

3 years ago

At an accident scene on a level road, investigators measure a car’s skid mark to be 88 m long. The accident occurred on a rainy

Oduvanchick [21]

Answer:

The the speed of the car is 26.91 m/s.

Explanation:

Given that,

distance d = 88 m

Kinetic friction = 0.42

We need to calculate the the speed of the car

Using the work-energy principle

work done = change in kinetic energy

$W=\Delta K.E$

$\mu\ mg\times d=\dfrac{1}{2}mv^2$

$v^2=2\mu g d$

Put the value into the formula

$v=\sqrt{2\times0.42\times9.8\times88}$

$v=26.91\ m/s$

Hence, The the speed of the car is 26.91 m/s.

3 0

3 years ago

Read 2 more answers

What is the kinetic energy of a 478 kg object that is moving with a speed of 15 m/s

Tpy6a [65]

The kinetic energy would be 53,775J:)

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3 years ago

How are the components of a heterogeneous mixture distributed?

Lena [83]

Answer:

heterogeneous mixture has components that are not evenly distributed. This means that you can easily distinguish between the different components.

3 0

3 years ago