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3 years ago

6

Which of the following statements concerning the scientific method is true?

1 answer:

Bezzdna [24]3 years ago

7 0

Answer:

Los científicos formulan una hipótesis antes de realizar experimentos y luego los

experimentos se utilizan para sacar una conclusión

Explanation:

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An unbalanced force of 20N acts on a 4.0kg mass what is it's acceleration

tankabanditka [31]

Hi there!

According to Newton's second law:

∑F = m · a, where:

∑F = net force (N = kgm/s²)

m = mass (kg)

a = acceleration (m/s²)

Rearrange to solve for acceleration:

F/m = a

20N / 4.0kg = 5 m/s²

4 0

3 years ago

A 60 g ball of clay is thrown horizontally at 40 m/s toward a 1.5 kg block sitting at rest on a frictionless surface. the clay h

Bingel [31]

The solution for this problem is:
Let u denote speed.

Equating momentum before and after collision:
= 0.060 * 40 = (1.5 + 0.060) u
= 2.4 = 1.56 u
= 2.4 / 1.56 = 1.56 u / 1.56
= 1.6 m / s is the answer for this question. This is the speed after the collision.

7 0

3 years ago

.A coin rolls off the edge of a table. The coin

geniusboy [140]

Answer:

Apply the following formulae horizontally And get A value for time

Remember horizontal acceleration is zero

$s = ut + \frac{1}{2}a {t}^{2} \\ 0.8 = 1.7 \times t \\ \frac{0.8}{1.7} = t \\ t = 0.47s$

and then to find the height apply the same above equation vertically...remember vertical initial velocity is zero

$s = ut + \frac{1}{2} a {t}^{2} \\ s = \frac{1}{2} \times 10 \times (0.47) ^{2} \\ s = 1.1045m$

5 0

2 years ago

One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at

Zolol [24]

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature $(T_i)=300 K$

$P_i=10 atm$

$P_f=2 atm$

Work done in iso-thermal process $=P_iV_iln\frac{P_i}{P_f}$

$P_i$ =initial pressure

$P_f$ =Final Pressure

$W=10\times 2.463\times ln\frac{10}{2}=39.64 J$

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is $=P\Delta V$

$V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L$

$V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L$

$\Delta W=1\times (12.315-2.463)=9.852 J$

$\Delta q=\Delta W=9.852 J$

$\Delta U=0$

8 0

3 years ago

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne

Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

$F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz$

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0

3 years ago