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4 years ago

The only type of investigation that involves a control group is

Physics

2 answers:

OleMash [197]4 years ago

4 0

<span>experimental? I hope that helps :)</span>

frez [133]4 years ago

3 0

An experimental one. Hope this helped.

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The driver of a car traveling at a speed of 20.0 m/s slams on the brakes and comes to a stop in 5.8 s. If we assume that the car

Degger [83]

Answer:

10 m/s

Explanation:

u = 20 m/s

v = 0 m/s

t = 5.8 s

Let a be the acceleration and s be the distance traveled by the car in time t.

use first equation of motion

v = u + a t

0 = 20 + a x 5.8

a = -3.45 m/s^2

Use third equation of motion

$v^{2}=u^{2}+2as$

$0^{2}=20^{2}-2 \times3.45 \times s$

s = 58 m

The average speed is defined as the ratio of distance traveled to the time taken.

$Average speed = \frac{58}{5.8} = 10 m/s$

thus, the average speed of the car is 10 m/s.

6 0

3 years ago

Mrs. Davis has asked Kyle to retrieve the compound mixture from the chemical cabinet. Which of the following should Kyle retriev

nignag [31]

Answer:

water

Explanation:

coz out of the choices water is the only one that is not an element.

water formula is H2O

3 0

4 years ago

the main difference between batholiths and stocks is that stocks are formed deeper in Earth’s surface

11Alexandr11 [23.1K]

<span>A batholith is an exposed area of rock that covers an area larger than 100 square kilometers. Areas smaller than 100 square kilometers are called stocks.</span>

3 0

3 years ago

A river flows due south with a speed of 2.0 m/s. You steer a motorboat across the river; your velocity relative to the water is

neonofarm [45]

Answer:

a) 25.5°(south of east)

b) 119 s

c) 238 m

Explanation:

solution:

we have river speed $v_{r}$ =2 m/s

velocity of motorboat relative to water is $v_{m/r}$ =4.2 m/s

so speed will be:

a) $v_{m}$ = $v_{r}$ + $v_{m/r}$

solving graphically

$v_{m}=\sqrt{v^2_{r}+v^2_{m/r}}$

=4.7 m/s

Ф= $tan^{-1} (\frac{v_{r}}{v_{m/r}} )$

=25.5°(south of east)

b) time to cross the river: t= $\frac{w}{v_{m/r}}$ = $\frac{500}{4.2}$ =119 s

c) d= $v_{r}t$ =(2)(119)=238 m

note :

pic is attached

6 0

4 years ago

Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s

AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

$E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J$

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0

4 years ago