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2 years ago

What does it mean to investigate science?

Physics

1 answer:

lapo4ka [179]2 years ago

7 0

Answer:

Scientific investigation is a quest to find the answer to a question using the scientific method. In turn, the scientific method is a systematic process that involves using measurable observations to formulate, test or modify a hypothesis.

Explanation:

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Stan is driving north on his scooter at 8m/s, accelerates 11m/s (North) in 4s, drives a constant velocity for the next 15s, and

kow [346]

A) Acceleration: $a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2$

B) The total displacement is 209.5 m north

C) The average velocity is 8.06 m/s north

Explanation:

Acceleration is defined as:

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

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we have:

- In the first segment,

u = 8 m/s north

v = 11 m/s north

t = 4 s

So the acceleration is

$a_1 = \frac{11-8}{4}=0.75 m/s^2$ (north)

- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,

u = v

Therefore, the acceleration is zero: $a_2 = 0$

- In the third segment,

u = 11 m/s (north)

v = 0 (he comes to a stop)

t = 7 s

So the acceleration is

$a=\frac{0-11}{7}=-1.57 m/s^2$

And the negative sign means the acceleration is south, opposite to the direction of motion.

In a uniformly accelerated motion, the displacement can be calculated as:

$s=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

t is the time

- For the first segment, we have

$u = 0\\a = 0.75 m/s^2\\t=4 s$

So the displacement is

$s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m$

- For the second segment, we have

$u = 11 m/s\\a = 0\\t=15 s$

So the displacement is

$s_2 = (11)(15)+0=165 m$

- For the third segment, we have

$u = 11\\a = -1.57 m/s^2\\t=7 s$

So the displacement is

$s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m$

So the total displacement is:

s = 6 m + 165 m + 38.5 m = 209.5 m

In the north direction (positive direction)

The average velocity is given by:

$v=\frac{d}{t}$

where

d is the total displacement

t is the total time

Here we have:

d = 209.5 m

t = 26 s

Therefore, the average velocity is

$v=\frac{209.5}{26}=8.06 m/s$ (north)

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0

3 years ago

A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the

kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂ : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F : Force ( N)

d : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the bar

mm = 1.5 kg : mass of the monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances from forces the point 2

d₁₂ = (3-0.6) m = 2.4m : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium of moments at the point 2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0

2 years ago

The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p

leva [86]

1) $1.86\cdot 10^6 rad/s^2$

2) 2418 rad/s

3) $27000 m/s^2$

4) 36.3 m/s

Explanation:

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

$\theta=\omega_i t +\frac{1}{2}\alpha t^2$

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

Solving for $\alpha$ , we find:

$\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2$

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

where

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

$v=u+at$

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

5 0

3 years ago

Your car's speedometer works in much the same way as its odometer, except that it converts the angular speed of the wheels to a

brilliants [131]

Answer:

Speed will be 30810 rpm

Explanation:

We have given diameter of the tire d = 24 inch

So radius $r=\frac{d}{2}=\frac{24}{2}=12imch$

We have given linear velocity v = 35 mph

We know that linear velocity is given by $v=\omega r$

$35=\omega \times 12$

$\omega =\frac{35}{12}\times \frac{63360}{60}=3080rad/min$

As we know that 1 mile = 63360 inch and 1 hour = 60 min

3 0

3 years ago

a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra

LenaWriter [7]

Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

$a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}$

Now, we have that

${v_f}^2-{v_0}^2=2a\Delta x$

so we end up with a distance traveled of

$\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$

$\implies\Delta x=79.6\,\mathrm m$

6 0

3 years ago