All categories

3 years ago

Which of the following statements is true about the scientific process?

1 answer:

4 0

The best and most correct answer among the choices provided by your question is the third choice or letter C.
The statement "Your hypothesis must be testable." is true about the scientific process.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

You might be interested in

An object weighs 32 newtons. What is its mass if a gravitometer indicates that g = 8.25 m/s?

Elan Coil [88]

Explanation:

weight=mass×g

32=mass×8.25

mass=

$\frac{128}{33} = 3.878$ kg

8 0

3 years ago

* CRIMINOLOGY*

Setler [38]

I think the answer is B.

Hope this helps.

5 0

3 years ago

Read 2 more answers

A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat

Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is 1215 N

Explanation:

$\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}$

$\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}$

$\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}$

Using Newton's second law, it is established that F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object $\left(m_{1}+m_{2}\right)$

Now consider both $\mathrm{m}_{1} \text { and } \mathrm{m}_{2}$ as a system, so net force acting on the system is $\text { Force }=\left(m_{1}+m_{2}\right) a$

Substitute the given values in the above formula,

$\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

$\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

Force = 1215 N

1215 N is the force required to move the quarterback with linebacker.

5 0

3 years ago

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid

Tanzania [10]

A) 18.4 m

a) mass of the load

b) mass of the truck

Explanation:

In order for the oad not to slide, its acceleration must be the same as the acceleration of the truck.

Since there is only one force acting on the load (the force of static friction), the acceleration of the load will be equal to the force of friction divided by the mass of the load (Newton's second law of motion):

$a=\frac{F_f}{m}=\frac{-\mu mg}{m}=-\mu g$ (1)

where

m is the mass of the load

$\mu=0.400$ is the coefficient of static friction

$g=9.8 m/s^2$ is the acceleration due to gravity

The acceleration of the truck (and the load) is also related to the stopping distance of the truck by the suvat equation:

$v^2-u^2=2as$ (2)

where

v = 0 is the final velocity of the car

u = 12.0 m/s is the initial velocity

a is the acceleration

s is the stopping distance

Since the acceleration must be the same, we can substitute (1) into (2), and solving for s we find:

$v^2-u^2=-2\mu g s\\s=\frac{v^2-u^2}{-2\mu g}=\frac{0^2-12.0^2}{-2(0.400)(9.8)}=18.4 m$

From part A, we see that the data that we have not used in the calculation are:

- The mass of the load

- The mass of the truck

Therefore, the two pieces of data unnecessary for the solution are

a) mass of the load

b) mass of the truck

8 0

3 years ago

An object is moving with an initial velocity of 3.3 m/s. It is then subject to a constant acceleration of 3.7 m/s2 for 10 s. How

Llana [10]

Answer:

Explanation:

s = s₀ + v₀t + ½at²

ASSUMING the acceleration is in the direction of initial motion.

s = 0 + 3.3(10) = ½(3.7)(10²)

s = 218 m

5 0

3 years ago