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Amiraneli [1.4K]
2 years ago
5

4. Susan observed that different kinds and amounts of fossils were present in a cliff behind her house. She wondered why changes

in fossil content occurred from the top to the bottom of the bank. She marked the bank at five positions: 5, 10, 15, 20 and 25 m from the surface. She removed 1 bucket of soil from each of the positions and determined the kind and number of fossils in each sample.
Hypothesis:
Independent Variable:
Dependent Variable:
Constant (at least one):
Control:
Number of groups:
Number of trials per group:
Physics
2 answers:
8090 [49]2 years ago
8 0

Answer:

A scientific experiment is divided into two sample groups: control and the experimental. The control group lacks the key variable to be studied while the experimental group contains the sample to be studied.

The experiment contains three experiments: the variable to be studied (dependent variable), the variable which can be changed and influence the dependent variable (independent variable) and the variable which do not change in the experiment (constant variable).

In the given experiment,  

1. Hypothesis: when the height changes in soil, the content of the fossil in the soil also changes.

2. Independent Variable: the height of the bank from surface

3. Dependent Variable: the number and type of fossil.

4. Constant variable: the amount of soil in the bucket

5. Control: the soil at the surface (0 cm)

6. Number of groups: 5 groups

7. Number of trials per group: should be 3

posledela2 years ago
3 0
5,10,15,20,25,30, that's how much it should have been
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You want to examine the hairy details of your favorite pet caterpillar, using a lens of focal length 8.9 cm 8.9 cm that you just
Zepler [3.9K]

Answer:

The angular magnification is M = 2.808

Explanation:

From the question we are told

           The focal length is  f = 8.9cm

          The near point is n_p = 25.0cm

The angular magnification is mathematically represented as

                          M = \frac{n_p}{f}

Substituting values

                        M = \frac{25}{8.9}

                           = 2.808

4 0
1 year ago
A truck is traveling down a road with a 4-percent grade at a speed of 75 mi/h when its brakes are applied to slow it down to 22.
kvasek [131]

Answer:

3.964 s

Explanation:

Metric unit conversion:

1 miles = 1.6 km = 1600 m.

1 hour = 60 minutes = 3600 seconds

75 mph = 75 * 1600 / 3600 = 33.3 m/s

22.5 mph = 22.5 * 1600/3600 = 10 m/s

Let g = 9.81 m/s2

Friction is the product of coefficient and normal force, which equals to the gravity

F_f = \mu N = \mu mg

The deceleration caused by friction is friction divided by mass according to Newton 2nd law.

a = F_f / m = \mu mg / m = \mu g = 0.6 *9.81 = 5.886 m/s^2

So the time required to decelerate from 33.3 m/s to 10 m/s so the wheels don't slide, with the rate of 5.886 m/s2 is

t = \frac{\Delta v}{a} = \frac{33.3 - 10}{5.886} = 3.964 s

3 0
1 year ago
A wood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in.2 The
WINSTONCH [101]

Answer:

31.67 in

Explanation:

Given:

Diameter of the pipe, D = 3ft = 36 in

cross-sectional area of the steel = 0.2 in²

Note: Refer to the figure attached

From the free body diagram represented in the figure, we have

ΣFx = 0

or

pressure × projected area = 2 × Force in steel

Now, the projected area = spacing (s) × diameter of the wood pipe

force in steel = stress in steel (σ) × cross-sectional area of the steel

on substituting the values we get

4 ksi × (s × 36 in) = 2 × σ × 0.2 in²

also, allowable hoop stress = 11.4 ksi

thus,

σ = 11.4 ksi = 11.4 × 10³ psi

therefore, we have

4 psi × (s × 36 in) = 2 × 11.4 × 10³ psi × 0.2 in²

thus,

s = 31.67 in

hence the maximum spacing is 31.67 in

3 0
2 years ago
Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is
topjm [15]

Answer:

β2= β1+10*f

Explanation:

comparing β2 and β1, it is said that β2 is increased by a factor of f.

for each factor of f, there is a 10*f dB increase.

therefore if the β1 is increases by an intensity of factor f

the new intensity would be β1+ 10*f

4 0
1 year ago
Constants Periodic Table Suppose the top surface of the vessel in the figure (Figure 1) is subjected to an external gauge pressu
Gnom [1K]

Answer:

a)  v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]

b) Water does not flow,

Explanation:

a) For this exercise we will use Bernoulli's equation, where index 1 is at the exit and index 2 on the surface of the water

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

This case does not indicate at the surface pressure is P₂, the pressure at the outlet is P₁ = P₀, the surface velocity is zero v₂ = 0

          P₀ + ½ ρ v₁² + ρ g y₁ = P₂ + 0 + ρ g y₂

           ½ ρ v₁² = P₂-P₀ + ρ (y₂ -y₁)

          v₁² = 2 (P₂-P₀) /ρ + 2 (y₂ -y₁)

          v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]

b) Reduce the pressure to SI units

         P₂ = 0.86 atm (1.01 10⁵ Pa / 1 atm) = 0.8686 10⁵ Pa

         P₀ = 1.01 10⁵ Pa

         ρ = 1 10³ kg / m³

Let's calculate

         v₁ = √ [2 (0.8686 -1.01) 10⁵/10³ + 2 (2.6)]

         v₁ = √ [-0.2828 10² + 5.2] = Ra [-23]

Water does not flow, this is because the pressure of the inner part is less than atmospheric, so that the water flows the pressure P₂> = P₀

For example if the pressure P₂ = P₀

         v₁ = √ 5.2

          v₁ = 2.28 m / s

5 0
1 year ago
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