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3 years ago

Two scientists interpret the same set of data differently. How would the scientific community deal with this problem?

1 answer:

7 0

The correct answer would be D. A new experiment would be needed to be done in order to test the conclusions. In science there is no authority, data is the only thing that matters. So if we have two different conclusions from the same date the only solution is to perform more tests and more experiments to see what is correct.

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One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo

yuradex [85]

Answer:

μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

W₁ x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

cos 60 = x / L

where L is the length of the ladder

x = L cos 60

sin 60 = y / L

y = L sin60

the horizontal distance of man is

cos 60 = x2 / 7.0

x2 = 7 cos 60

we substitute

m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

fr = (735 + 2450) / 8.66

fr = 367.78 N

the friction force has the expression

fr = μ N

write the translational equilibrium equation

N - W₁ -W₂ = 0

N = m₁ g + W₂

N = 30 9.8 + 700

N = 994 N

we clear the friction force from the eucacion

μ = fr / N

μ = 367.78 / 994

μ = 0.37

3 0

3 years ago

Please help with 4 and 5, thank you :)

allochka39001 [22]

Answer: #4

Sally is faster.

Explanation:

If you multiply Sallies it is going to be less than Jessica's.

6 0

3 years ago

A ball is dropped from 8.5 meters above the ground. If it begins at rest, how long does it take to hit the ground?

Anna35 [415]

Answer:

Explanation:

Givens

d = 8.5 meters

vi = 0

a = 9.81

t = ?

Formula

d = vi * t + 1/2 a t^2

Solution

8.5 = 0 + 1/2 9.81 * t^2 multiply both sides by 2

8.5 = 4.095 t^2 Divide both sides by 4.095

8.5/4.095 = t^2

1.7329 = t^2 Take the square root of both sides

t = 1.316

It takes 1.316 seconds to hit the ground.

6 0

3 years ago

Calculate the equivalent resistance for both circuits. Series circuit: 2 Ω and 4 Ω Parallel circuit: 2 Ω and 4 Ω Which circuit h

goldenfox [79]

Equivalent resistance is also known as the overall resistance.

For resistors in a series circuit, the total resistance is computed using the formula:

$R_{T} = R_{1}+ R_{2}+ R_{3}... R_{n}$

In other words, you just add up the resistance of each resistor in the series circuit. In your case you only have two resistors. You have 2Ω and 4Ω. So all you need to do is add that up.

$R_{T} = R_{1}+ R_{2}$
$R_{T} = 2 + 4=6$

The total resistance of the series circuit is 6Ω

In a parallel circuit you get the total resistance using the formula:
$\frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}...+\frac{1}{R_{n}}$

First you get the sum of all fractions and at the end take the reciprocal of the resulting fraction and divide. So let us take your problem into consideration where you have two resistors that have a resistance of 2Ω and 4Ω.

$\frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}$
$\frac{1}{R_{T}} = \frac{1}{2}+\frac{1}{4}$
$\frac{1}{R_{T}} = \frac{2}{4}+\frac{1}{4}$
$\frac{1}{R_{T}} = \frac{3}{4}$

Get the reciprocal of the resulting fraction 3/4 and then divide. The reciprocal of 3/4 is 4/3.

4/3 = 1. 33Ω

So if you compare the equivalent resistance of the two circuits, the series circuit has a higher equivalent resistance.

3 0

3 years ago

Read 2 more answers

A rocket accelerates upward from rest, due to the first stage, with a constant acceleration of a1 = 67 m/s2 for t1 = 39 s. The f

Igoryamba

Answer:

(a) $v_1= a_1t_1$

(b) $v_2 =a_1t_1+a_2t_2$

(c) 44133.5 m

Explanation:

<u>Given:</u>

$u$ = initial speed of the rocket in the first stage = 0 m/s
$v_1$ = final speed of the rocket in the first stage
$v_2$ = final speed of the rocket in the second stage
$t_1$ = time interval of the first stage
$t_2$ = time interval of the second stage
$s_1$ = distance traveled by the rocket in the first stage

$s_2$ = distance traveled by the rocket in the second stage
$s$ = distance traveled by the rocket in whole time interval

Part (a):

Since the rocket travels at constant acceleration.

$\therefore v_1 = u+a_1t_1\\\Rightarrow v_1 = a_1t_1$

Hence, the expression of the rocket's speed at time $t_1\ is\ v_1 = a_1t_1$ .

Part (b):

In this part also, the rocket moves with a constant acceleration motion.

$\therefore v_2 = v_1+a_2t_2\\\Rightarrow v_2 = a_1t_1+a_2t_2$

Hence, the expression of the rocket's speed in the time interval $t_2$ is $v_2 = a_1t_1+a_2t_2$ .

Part (c):

For the constant acceleration of rocket, let us first calculate the distance traveled by the rocket in both the time intervals.

$s_1 = u+\dfrac{1}{2}a_1t_1^2\\\Rightarrow s_1 = 0+\dfrac{1}{2}67\times(1)^2\\\Rightarrow s_1 =33.5\ m$

Similarly,

$s_2 = v_1t_2+\dfrac{1}{2}a_2t_2^2\\\Rightarrow s_2 = a_1t_1t_2+\dfrac{1}{2}a_2t_2^2\\\Rightarrow s_2 = 67\times1\times49+\dfrac{1}{2}\times 34\times(49)^2\\\Rightarrow s_2 =44100\ m\\\therefore s = s_1+s_2\\\Rightarrow s = 33.5\ m+44100\ m\\\Rightarrow s =44133.5\ m$

Hence, the rocket moves a total distance of 44133.5 m until the end of the second period of acceleration.

5 0

3 years ago

Read 2 more answers