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2 years ago

11

Item 4 A student conducts an experiment to determine how the temperature of water affects the time for sugar to dissolve. In eac

h trial, the student uses a different amount of sugar and a different temperature of water. What is wrong with this experimental design? A.The student did not change enough variables. b.The student does not have a dependent variable. C.The student changed too many variables. D. The student needs to perform more trials.

1 answer:

expeople1 [14]2 years ago

3 0

Answer:

The Answer is C. the student changed too many variables.

Explanation:

For the quiz in K12

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The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th

worty [1.4K]

Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

$\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg$

where $m_a$ is the mass of the arrow, $m_b$ is the mass of the block, $\Delta H$ of the change in height of the block after the collision, and $v$ is the velocity of the arrow before it hit the block.

Solving for the velocity $v$ , we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} }$

and we put in the numerical values

$m_a = 0.045kg$ ,

$m_b = 1.40kg,$

$\Delta H = 0.4m,$

$g= 9.8m/s^2$

and simplify to get:

$\boxed{ v= 15.9m/s}$

The arrow was moving at 15.9 m/s

6 0

3 years ago

As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

marshall27 [118]

Answer:

The resultant velocity is $v_t=10 knots$

Explanation:

Apply the law of conservation of momentum

$M_L *v_L + M_f * V_f = (M_L + M_f) v_t$

Where $M_L$ is the mass of the Luxury Liner = 40,000 ton

$v_L$ is the velocity of Luxury Liner = 20 knots due west

$M_f$ mass of freighter = 60,000

$v_f$ is the velocity of freighter = 10 knots due north

Apply the law of conservation of momentum toward the the west direction

$v_f = 0 \ knots$

So the equation would be

$M_L *v_L = (M_L + M_f) v_t$

Substituting values

$40000*20 = (40000+ 60000)v_t_w$

Where $v_t_w$ the final velocity due west

Making $v_t_w$ the subject

$v_t_w = \frac{40,000* 20}{(40000 + 60000)}$

$= 8 \ knots$

Apply the law of conservation of momentum toward the the north direction

$v_L = 0 \ knots$

So the equation would be

$M_f *v_f = (M_L + M_f) v_t_n$

Where $v_t_n$ the final velocity due north

Making $v_t_n$ the subject

$v_t_n = \frac{60,000* 10}{(40000 + 60000)}$

$= 6 \ knots$

The resultant velocity is

$v_t = \sqrt{v_t_w^2 + v_t_n^2}$

$= \sqrt{8^2 +6^2}$

$v_t=10 knots$

8 0

3 years ago

A symbolic model for learning is a model that is observed in person.

kap26 [50]

Answer:

True

Explanation:

3 0

3 years ago

Read 2 more answers

A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,

shusha [124]

Answer:

$v=d\sqrt{\frac{k}{m}}$

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

$U_{0}+K_{0}=U_{f}+K_{f}$

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

$U_{0}=K_{f}$

The initial potential energy of the spring is given by the equation:

$U_{0}=\frac{1}{2}kd^{2}$

the Kinetic energy of the block is then given by the equation:

$K_{f}=\frac{1}{2}mv_{f}^{2}$

so we can now set them both equal to each other, so we get:

$=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}$

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

$kd^{2}=mv_{f}^{2}$

so now we can solve this for the final velocity, so we get:

$v=d\sqrt{\frac{k}{m}}$

6 0

3 years ago

A 5.00 kilogram block slides along a horizontal,frictionless surface at 10.0 meters per second. for 4.00 seconds. The magnitude

finlep [7]

In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
= (5 * 10) Kg m/s
= 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".

5 0

3 years ago