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4 years ago

What step of the scientific method do we record data

1 answer:

7 0

1. Ask a question
2. Do Background Research
3. Construct a Hypothesis
4. Test Your Hypothesis by Doing an Experiment
5. Analyze Your Data and Draw a Conclusion
6. Communicate Your Results

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What is the difference between mass and weight?

I am Lyosha [343]

Mass is the actual amount of material contained in a body and is measured by kg, gm, etc. Whereas weight is the force extorted by the gravity on that object mg. Note that mass is independent of everything but weight is different on earth, moon, etc.

5 0

3 years ago

One kind of baseball pitching machine works by rotating light and stiff rigid rod about a horizontal axis until the ball is movi

erastova [34]

Answer:

(a). The ball's centripetal acceleration is $16.17\times10^{2}\ m/s^2$

(b). The magnitude of the net force is 232.9 N.

Explanation:

Given that,

Mass of baseball = 144 g

Speed = 81 mph = 36.2 m/s

Distance = 81 cm

(a). We need top calculate the ball's centripetal acceleration just before it is released

Using formula of centripetal acceleration

$a=\dfrac{v^2}{r}$

Where, v = speed

r = radius

Put the value into the formula

$a=\dfrac{(36.2)^2}{81\times10^{-2}}$

$a=1617.82\ m/s^2$

$a=16.17\times10^{2}\ m/s^2$

(b). We need to calculate the magnitude of the net force that is acting on the ball just before it is released

Using formula of force

$F=\dfrac{mv^2}{r}$

Put the value into the formula

$F=\dfrac{144\times10^{-3}\times(36.2)^2}{81\times10^{-2}}$

$F=232.9\ N$

Hence, (a). The ball's centripetal acceleration is $16.17\times10^{2}\ m/s^2$

(b). The magnitude of the net force is 232.9 N.

4 0

3 years ago

A horse pulls a barge along a canal using a rope 10m

marta [7]

Answer:

Explanation:

Here the rope is 10m long and making an angle of θ with the direction of canal or direction of flow of water in the canal .

sinθ = 2 / 10 = .2

θ = 11.5⁰

a )

Component of 500 N along the line perpendicular to the canal

= 500 sinθ

= 500 sin 11.5

= 100 N

b )

component of 500 N along the canal

= 500 cos 11.5

= 490 N approx .

8 0

4 years ago

Can someone help me figure out how to find tension in a rope? The question is:

Simora [160]

You are welcome.......

8 0

3 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

3 years ago