Question 4 (2 points)

When developing an experiment design, which could a scientist take to improve the quality of the results?

Bad White [126]

The options missing are;

A. Make only a few changes to the manipulated variable.

B. Identify any biases about the answer to the scientific question.

C. Include as many details as possible when writing the conclusion.

D. Share her ideas through peer review.

Answer:

B. Identify any biases about the answer to the scientific question.

Explanation:

Since the question is asking what can be done to improve the quality of the results, it means what can the scientist do to make sure that the result answers the question properly and has little or no biases.

Thus, looking at the options, the only one that comes close to actually improving on the quality of the results before finalizing is option B where the scientist is to identify any biases.

3 0

3 years ago

Which of the following statements is true about winds from the ocean? (2 points)

aleksley [76]

The correct answers are:

Winds from the ocean reduce the number of clouds.

Winds from the ocean are faster.

HOpe this helps!

7 0

3 years ago

Read 2 more answers

A 2.0 kg block, initially moving at 10.0 m/s, slides 50.0 m across a sheet of ice beforecoming to rest. What is the magnitude of

inna [77]

Answer:

The magnitude of the average frictional force on the block is 2 N.

Explanation:

Given that.

Mass of the block, m = 2 kg

Initial velocity of the block, u = 10 m/s

Distance, d = 50 m

Finally, it stops, v = 0

Let a is the acceleration of the block. It can be calculated using third equation of motion. It can be given by :

$v^2-u^2=2ad$

$-u^2=2ad$

$a=\dfrac{-u^2}{2d}\\\\a=\dfrac{-(10\ m/s)^2}{2\times 50\ m}\\\\a=-1\ m/s^2$

The frictional force on the block is given by the formula as :

F = ma

$F=2\ kg\times (-1)\ m/s^2\\\\F=-2\ N$

|F| = 2 N

So, the magnitude of the average frictional force on the block is 2 N. Hence, this is the required solution.

8 0

4 years ago

A 3.00-kg object has a velocity 16.00 i ^ 2 2.00 j ^2 m/s.

trapecia [35]

Answer:

390 J

Explanation:

m = 3 kg

u = 16 i + 2 j

(a) Magnitude of velocity = $\sqrt{16^{2}+2^{2}}$ = 16.1245 m/s

KEi = 1/2 m v^2 = 0.5 x 3 x 16.1245 = 390 J

(b) v = 18 i + 14 j

Magnitude of velocity = $\sqrt{18^{2}+14^{2}}$ = 22.804 m/s

KEf = 1/2 m v^2 = 0.5 x 3 x 22.804 = 780 J

According to the work energy theorem

Work done = change in KE = KEf - KEi = 780 - 390 = 390 J

8 0

4 years ago

A car moving at 10 m/s crashes into a large bush and stops in 1.3 m. Using the Work-Energy theorem, calculate the average force

ad-work [718]

Answer:

The magnitude of the average force the seat belt exerts on the passenger is 2692.3 N.

It takes 0.26 s to bring the passenger in the car to a halt.

Explanation:

Hi there!

The negative work (W) needed to bring a moving object to stop is equal to its kinetic energy (KE):

W = KE

F · s = 1/2 · m · v²

Where:

F = applied force on the passenger.

s = displacement of the passenger.

m = mass of the passenger.

v = velocity of the passenger.

Solving the equation for F:

F = 1/2 · m · v² / s

Replacing with the data:

F = 1/2 · 70 kg · (10 m/s)² / 1.3 m

F = 2692.3 N

The magnitude of the average force the seat belt exerts on the passenger is 2692.3 N.

According to the second law of Newton:

F = m · a

Where "a" is the acceleration of the passenger.

We also know from kinematics that the velocity of an object can ve calculated as follows:

v = v0 + a · t

Where:

v = velocity of the passenger at time t.

v0 = initial velocity.

t = time.

a = acceleration.

When the passenger stops, its velocity is zero. So replacing a = F/m, let´s solve the equation for the time it takes the passenger to stop:

v = v0 + a · t

0 = 10 m/s + (-2692.3 N/ 70 kg) · t

-10 m/s / (-2692.3 N/ 70 kg) = t

t =0.26 s

It takes 0.26 s to bring the passenger in the car to a halt.

5 0

3 years ago