All categories

3 years ago

What is the control group used for

Physics

1 answer:

LUCKY_DIMON [66]3 years ago

6 0

Answer:

What is a Control Group? The control group (sometimes called a comparison group) is used in an experiment as a way to ensure that your experiment actually works. It's a way to make sure that the treatment you are giving is causing the experimental results, and not something outside the experiment

Explanation:

You might be interested in

A person is riding a motorized tricycle. They weigh 180kg and are moving at 3 m/s over a distance of 300 m. How much work is don

agasfer [191]

If I am to understand this question correctly this is what asks you:

If a person is riding a motorized tricycle how much work do they do?

You may ask yourself, why did I only use part of the question. Simple, the rest is not relevant to what is being asked. The weight, speed, and distance wont affect the person riding any motorized vehicle other than the time it takes to get from one place to another.

So to answer this question I would say:

Not much, all they really have to do is to steer and set the motorized tricycle to cruise control. Just like any rode certified vehicle.

If you have any questions about my answer please let me know and I will be happy to clarify any misunderstandings. Thanks and have a great day!

3 0

3 years ago

Help me find the acceleration

ANEK [815]

a = 3.09 m/s²

<h3>Explanation</h3>

This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the timeless suvat equation is likely what you need for this question.

In the timeless suvat equation,

$a = \dfrac{v^2 - u^2}{2\; x}$

where

$a$ is the acceleration of the car;
$v$ is the final velocity of the car;
$u$ is the initial velocity of the car; and
$x$ is the displacement of the car.

Note that v and u are velocities. Make sure that you include their signs in the calculation.

In this question,

$a$ is the unknown;
$v = -10.9 \; \text{m} \cdot \text{s}^{-2}$ ;
$u = -27.7 \; \text{m} \cdot \text{s}^{-2}$ ; and
$x = - 105 \; \text{m}$ .

Apply the timeless suvat equation:

$a = \dfrac{v^{2} - u^{2}}{2\; x}\\\phantom{a} = \dfrac{(-10.9)^{2} - (-27.7)^{2}}{2 \times (-105)}\\\phantom{a} = 3.09 \; \text{m} \cdot \text{s}^{-2}$ .

The value of $a$ is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.

7 0

3 years ago

Through which of these media do sound waves travel most slowly? A.iron. B.Wood. C.Water. D.Cold Air

EleoNora [17]

The answer should be D) Cold air because even though its true sound can travel through all types of matter, air which is a gas, can travel but it travels SLOWLY while sound travels quickly in SOLIDS.

5 0

3 years ago

A Bullet Off mass 100 gm is fired From A Gun Off mass 5 Kg. If the backward velocity of the gun's 5 m / s, what is forward veloc

Elena L [17]

Answer:

250 m/s

Explanation:

The mass of the bullet, m₁ = 100 g = 0.1 kg

The mass of the gun, m₂ = 5 kg

The backward velocity of the gun, v₂ = -5 m/s

Given that the momentum is conserved, we have;

The total initial momentum = The total final momentum

The gun and the bullet are at rest, therefore, we have;

The initial momentum = 0

The total final momentum = m₁·v₁ + m₂·v₂

Where;

v₁ = The forward velocity of the bullet

Therefore, we get;

m₁·v₁ + m₂·v₂ = 0

0.1 kg × v₁ + 5 kg × (-5 m/s) = 0

0.1 kg × v₁ = 5 kg × 5 m/s

v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s

The forward velocity of the bullet, v₁ = 250 m/s

6 0

3 years ago

A weightlifter lifted 800 Newtons to a height of 2.5 meters. What is the power output if this was

stiv31 [10]

The power required is 1.6 kW.

Answer:

Explanation:

Power is defined as the amount of work done on any object for a given time interval. In other words, power is the amount of force required to move an object in a given period of time.

Power = Work done / time taken for that work done.

Here the force is given as 800 N and the displacement is given as 2.5 m, while the time required for the displacement is given as 1.22 seconds.

So the power will be ratio of the product of force acting on the weightlifter, displacement of the weight to the time taken for that displacement.

Power = (800×2.5)/1.22 =1639 W

So the power required is 1.6 kW.

5 0

3 years ago