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2 years ago

Which of the following is a helpful process in evaluating scientific claims?

2 answers:

8 0

Answer: D. If you look at the claims with critical thinking then you will be less likely to be fooled by something false

Agata [3.3K]2 years ago

7 0

Answer: d. critical thinking

Explanation:

A scientific claim is a statement which is based on evidences obtained after the experimental trials and direct observatory approach of the process occurring in the nature.

Critical thinking based on the approach of verification and analysis can be used to evaluate the scientific claims. The result of evaluation will be reliable and acceptable. The evaluation should not be biased or presumed it will yield inappropriate results.

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Which of the following statements is FALSE?

Levart [38]

A thrust fault is a reverse fault with an extremely high dip (close to 90°). This is the false statement.

Answer: Option D

<u>Explanation:</u>

Faults are the fracture or fracture zone occurring on the rocks. These fractures can travel through the rocks leading to massive destruction. So, depending upon the direction of their travel, the faults can be classified as normal, reverse and strike slip fault. Also, the angle of dip along the fault is one of the important criteria for determining the type of faults.

There is dip-slip fault which has its movement along the vertical fault plane while the strike slip fault will be in horizontal direction. Similarly, an oblique fault will be acting in both vertical and the horizontal direction. So, the fourth statement related to thrust fault is false as in reverse fault or thrust fault the dip will be shallow and not high.

5 0

2 years ago

During the construction of an office building, a hammer is accidentally dropped from a height of 784 ft. the distance (in feet)

Serga [27]

T= 24.5 feet per second. That is the velocity it reaches at the end of its fall

7 0

2 years ago

Read 2 more answers

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

2 years ago

A steel aircraft carrier is 370 m long when moving through the icy North Atlantic at a temperature of 2.0 °C. By how much does t

34kurt

Answer:

The carrier lengthen is 0.08436 m.

Explanation:

Given that,

Length = 370 m

Initial temperature = 2.0°C

Final temperature = 21°C

We need to calculate the change temperature

Using formula of change of temperature

$\Delta T=T_{f}-T_{i}$

$\Delta T=21-2.0$

$\Delta T=19^{\circ}C$

We need to calculate the carrier lengthen

Using formula of length

$\Delta L=\alpha_{steel}\times L_{0}\times\Delta T$

Put the value into the formula

$\Delta L=1.2\times10^{-5}\times370\times19$

$\Delta L=0.08436\ m$

Hence, The carrier lengthen is 0.08436 m.

8 0

2 years ago

A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text

antoniya [11.8K]

Answer:

The energy of the system is 15 J.

Explanation:

Given that,

Energy E = 2.5 J

Amplitude = 10 cm

We need to calculate the spring constant

Using formula of mechanical energy of the system

$E=\dfrac{1}{2}kA^2$

Put the value into the formula

$2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2$

$k=\dfrac{2.5\times2}{(10\times10^{-2})^2}$

$k=500\ N/m$

If the block is replaced by a block with twice the mass of the original block

Amplitude = 6 cm

We need to calculate the energy

Using formula of mechanical energy

$E=\dfrac{1}{2}kA^2$

Put the value into the formula

$E=\dfrac{1}{2}\times500\times(6\times10^{-2})$

$E=15\ J$

Hence, The energy of the system is 15 J.

8 0

2 years ago