All categories

3 years ago

Which is an example of making a quantitative observation?

Physics

2 answers:

MAXImum [283]3 years ago

5 0

Answer:

Quantitative observation is a data collection method that belongs to quantitative research, and refers to observation that uses precise and computational methods, often statistical in human sciences. The term is commonly used in the human sciences, such as social sciences and education, but broadly it can also refer to natural sciences.

For example, sociology focuses on human-made communities and phenomena resulting from the activities of human communities. In quantitative observation, the researcher seeks to collect empirical evidence for himself. By looking at the evidence, the researcher seeks to understand a social phenomenon. By examining the observational data, the researcher seeks to generalize the observational data he has collected through a quantitative (numerical) hypothesis.

A clear example of quantitative observation occurs when the researcher records the number of people who possess a certain characteristic in a massive event (for example, the number of white people in a Republican rally).

malfutka [58]3 years ago

3 0

An example of a quantitative observation is measuring the surface of an oil painting and finding its dimensions to be 12 inches by 12 inches. A quantitative observation occurs when a researcher takes a measurement that is recorded in an objective number of units.

You might be interested in

A cart of mass 20 kg, initially at rest, is pushed with a net force of 40 N on a flat surface. If the cart is pushed 5 m, what i

bixtya [17]

Using the theorem of kinetic energy
1/2mVf² - 1/2mVi²= WF + Wp, Wp=0
WF = F. AB, AB=5m and F= 40N, m=20kg
so the final kinetic is KEf= 1/2mVf² = WF =<span>F. AB= 40*5=200J
</span>
the final velocity is 1/2mVf² <span>=200, implies Vf= sqrt(20)=2sqrt(5)m/s</span>

3 0

3 years ago

What statement BEST describes a solid?

g100num [7]

C. Solids have a definite shape and definite volume.
is the correct one.

5 0

4 years ago

Read 2 more answers

SOMEONE PLEASE HELP ME ASAP PLEASE!!!!!

Inessa [10]

The correct answer is 2.479

5 0

4 years ago

Read 2 more answers

A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0

Xelga [282]

Answer:

$v_m \approx -4.38\; \rm m \cdot s^{-1}$ (moving toward the incline.)

$v_M \approx 4.02\; \rm m \cdot s^{-1}$ (moving away from the incline.)

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

If $g = 9.81\; \rm m \cdot s^{-2}$ , the potential energy of the block of $m = 2.20\; \rm kg$ would be $m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J$ when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this $m = 2.20\; \rm kg\!$ block right before the collision would also be approximately $77.695\; \rm J$ .

Calculate the velocity of that $m = 2.20\; \rm kg$ based on its kinetic energy:

$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

Sum of the energy of each block right after the collision:

$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

Combine to obtain two equations about $v_m$ and $v_M$ (given that $m = 2.20\; \rm kg$ whereas $M = 7.00\; \rm kg$ .)

$\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.$ .

Solve for $v_m$ and $v_M$ (ignore the root where $v_M = 0$ .)

$\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.$ .

The collision flipped the sign of the velocity of the $m = 2.20\; \rm kg$ block. In other words, this block is moving backwards towards the incline after the collision.

6 0

3 years ago

A 20-ton truck collides with a 1500-lb car and causes a lot of damage to the car. Since a lot of damage is done on the car : Sel

Anni [7]

Answer:

Explanation:

Newton's 3rd Law tells us that for every action we will have a reaction equal in magnitude and opposite in direction. In a collision between obects A and B this means that the force on A applied by B will be equal in magnitude than the force on B applied by A, which leaves d as the true answer and the rest as false.

8 0

3 years ago