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3 years ago

6

To test the hypothesis above, you will observe the changes during the experiment. To do this, you will use these observations to

compare the of the substances to the of the substances.

2 answers:

julsineya [31]3 years ago

8 0

^^ correct

Answer:

For the second part, select all of them

Explanation:

100%

wariber [46]3 years ago

4 0

The answer to this is initial appearance and the second is final appearance.

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A miner develops cancer of the esophagus. Ten years before, he had been exposed to radiation when he worked for a year in a mine

slamgirl [31]

D.) It is unlikely that a specific cause can be determined, but the treatment would likely be the same in either case.

3 0

3 years ago

Read 2 more answers

Suppose you throw a ball vertically upward with a speed of 49 m/s. Neglecting air friction, what would be the height of the ball

vlabodo [156]

Answer:

78.4 m

Explanation:

Using newton's equation of motion,

S = ut + 1/2gt²......................... Equation 1

Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.

Note: Taking upward to be negative, and down ward positive

Given: u = 49 m/s, t = 2.0 s, g = -9.8 m/s²

Substitute into equation 1

S = 49(2) - 1/2(9.8)(2)²

S = 98 - 19.6

S = 78.4 m

Hence the height of the ball two seconds later = 78.4 m

6 0

3 years ago

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

xz_007 [3.2K]

Answer:

a) $\Delta{t} = 5.39s$

b) the motorcycle travels 155 m

Explanation:

Let $t_2-t_1 = \Delta{t}$ , then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

where:

$v_{m2}$ is the speed of the motorcycle at time 2

$v_{c}$ is the velocity of the car (constant)

$v_{0}$ is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

$\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]$

$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

$x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933$

3 0

3 years ago

Based on the data table,which car won the race?

kobusy [5.1K]

Answer:

car one

Explanation:

3 0

2 years ago

A piano string having a mass per unit length of 5.00 g/m is under a tension of 1350 N. Determine the speed of transverse waves i

padilas [110]

Answer:

The speed of transverse waves in this string is 519.61 m/s.

Explanation:

Given that,

Mass per unit length = 5.00 g/m

Tension = 1350 N

We need to calculate the speed of transverse waves in this string

Using formula of speed of the transverse waves

$v=\sqrt{\dfrac{T}{\mu}}$

Where, $\mu$ = mass per unit length

T = tension

Put the value into the formula

$v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}$

$v =519.61\ m/s$

Hence, The speed of transverse waves in this string is 519.61 m/s.

6 0

3 years ago