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2 years ago

What does the scientific method help test in environmental science?

Physics

1 answer:

horsena [70]2 years ago

5 0

Answer:

The scientific method is a method of research with defined steps that include experiments and careful observation. One of the most important aspects of this method is the testing of hypotheses by means of repeatable experiments. A hypothesis is a suggested explanation for an event, which can be tested.

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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 59 kg physics student is riding her 220 kg Harley at 12 m/s when she has a head-on collision with a 2.1 kg pigeon flying the o

IgorLugansk [536]

Answer:

Explanation:

Using conservation of momentum

$m_1=59+220=279kg$

$m_2=2.1kg$

$U_1=12m/s$

$U_2=-44m/s$

$m_1U_1+m_2U_2=(m_1+m_2)V\\\\\frac{278\times 12-2.1\times 44}{279+2.1}=V\\\\V=11.58m/s$

6 0

2 years ago

How much work is done when a 5 N force moves a block 4 m?

katen-ka-za [31]

20N•m or 20J. Work is equal to force•distance, and 5N•4m is 20N•m, or J

4 0

3 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

2 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

8 0

2 years ago