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3 years ago

The information about groundwater recorded by a student is valid when

2 answers:

7 0

Such information is valid when A. it is not biased.
There is no bias in science - the results are either correct or incorrect; personal opinions have nothing to do with that. B is incorrect because the point of experiments is that they can be repeated to achieve the same results. C and D are likewise incorrect as primary sources are important, and you can share the results with others if you want to.

Alik [6]3 years ago

4 0

Answer:

a. it is not biased

Explanation:

Scientific information can not be biased to be credible and valid, and it is only possible to be sure that this scientific information is true if the results are repeated in different types of experiments and tests. We can not forget that scientific information gains credibility with primary sources that give scientific background to the statements, moreover, the purpose of formulating scientific information is to share information and let everyone know about it.

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A 1.0 kg cart moving right at 5.0 m/s on a frictionless track collides with a second cart moving at 2.0m/s. The 1.0 kg cart has

____ [38]

Answer:

m₂ = 3kg

Explanation:

The question wasn't clear about what direction the initial velocity of the second cart was, so I'll assume it was going left at 2.0m/s.

Anyway, this is a conservation of momentum problem. The equation you need to use is the one written in blue. They want you to solve for the mass of the second cart, so do some algebra and rearrange that blue equation in term of m₂.

Now that you have the equation for m₂, plug in all the values given from the question and you'll get 3kg.

3 0

4 years ago

A soccer player attempting to steal the ball from an opponent was extending her knee at 50 deg/s when her foot struck the oppone

denpristay [2]

Answer:

$\alpha = -4.36 rad/s^2$

Explanation:

Initial angular speed of the player is given as

$\omega = 50 deg/s$

here we know that

$1 deg = \frac{\pi}{180} rad$

now we know that

$\omega = 50\frac{\pi}{180}$

$\omega = 0.87 rad/s$

now its speed comes to zero in 0.2 s

so angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

$\alpha = \frac{0 - 0.87 rad/s}{0.2 s}$

$\alpha = -4.36 rad/s^2$

7 0

3 years ago

The 200-mm test tube also contained some water (besides the metal) that was subsequently added to the calorimeter (in Part A.4.)

lbvjy [14]

Answer:

The temperature change in the calorimeter will be lower

Explanation:

Water is an example of a molecular substance. They have relatively low melting points and boiling points usually below 300° C . Water reacts with metals to a degree varying with their position in the electrochemical series.

The specific heat of water is 4179.6 Joules which is relatively high . This typically implies that water absorbs a larger amount of heat but the increase in temperature of its boiling points is relatively low. Thus; in the 200-mm test tube that contains water and was subsequently added to the calorimeter , the heat present was initially absorbed by the water and that does not result to an increase in the temperature change in the calorimeter. Thus the temperature change in the calorimeter will be lower.

5 0

4 years ago

I'M GIVING 95POINTS PLZ HELP!!

Travka [436]

initial height = 25000 ft

as we know that

1 ft = 0.3048 m

so we have

H = 25000(0.3048) = 7620 m

If mass of the man = 90 kg

now the initial potential energy is given as

$U = mgh$

$U = (90)(9.8)(7620)$

$U = 6.72 \times 10^6 J$

Gravitational force that is acting on it is given by

$F = mg$

$F = 90(9.8) = 882 N$

so this is the force by which earth is attracting him towards it as we can see due to this force the man is accelerating towards the earth

As we know that gravitational force is given by

$F = \frac{GMm}{r^2}$

here we know that

r = R + h

R = radius of earth ( $6.37 \times 10^6 m$ )

h = 7620 m

now we have acceleration at that point is

$a = \frac{F}{m}$

$a = \frac{GM}{r^2}$

$a = \frac{(6.67\times 10^{-11})(5.98 \times 10^{24})}{(6.37 \times 10^6 + 7620)^}$

$a = 9.806 m/s^2$

by energy conservation we have

$KE = PE$

$\frac{1}{2}mv^2 = mgh$

$v^2 = 2gh$

$v^2 = 2(9.8)(7620)$

$v = 386.5 m/s$

final speed due to wind resistance is 150 mph

now we know that

1 mile = 1609 meter

so this speed in m/s is given as

$150 mph = 150(\frac{1609 m}{3600 s})$

$v = 67 m/s$

No Luke is not accelerating when his speed is 150 mph

because at this speed his velocity will become constant

and since there is no change in velocity so his acceleration will become zero

since at 150 mph the acceleration of Luke is zero

so net force on him must be zero

so we will have

wind force = weight or force due to earth

$F_{wind} = 882 N$

If there is no wind resistance then there will be energy conservation

so KE = PE

$KE = 6.72 \times 10^6 J$

#10

Net will reduce the velocity of luke to be zero by taking long time

As we know that force on a system is given by

$F = \frac{\Delta P}{\Delta t}$

so if we increase the time interval here then we will have less force

so Luke will unhurt due to more time to decrease his speed to be zero

5 0

3 years ago

Find the magnitude of the sum of these two vectors:

sukhopar [10]

Answer:

Explanation:

You can decompose those vectors into their components in x and y direction. For the first vector:

$\vec{r}_{1}=r_{1}\cos 30\hat{i}+r_{1}\sin 30\hat{j}=3.14(\frac{1}{2}\sqrt{3}\hat{i}+\frac{1}{2}\hat{j})=1.57\sqrt{3}\hat{i}+1.57\hat{j}$

For the second vector:

$\vec{r}_{2}=r_{2}\cos 60\hat{i}-r_{2}\sin60 \hat{j}=1.355\hat{i}-1.355\sqrt{3}\hat{j}$

The sum of two vectors will be:

$\vec{r}=\vec{r}_{1}+\vec{r}_{2}=(1.57\sqrt{3}+1.355)\hat{i}+(1.57-1.355\sqrt{3})\hat{j}\approx 4.0711\hat{i}-0.77415\hat{j}$

The magnitude of the sum of two vectors is:

$r=\sqrt{(4.0711)^{2}+(-0.77415)^{2}}\approx 4.14$ meter

7 0

2 years ago