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Harrizon [31]
3 years ago
9

a student conducts an experiment to determine how the additional of salt to water affects the density of the water. the student

fills 3 beakers with equal amounts of water. he then adds 1 cup of salt to the first beaker, 2 cups of salt to the second beaker and no salt to the third beaker. what is the dependent variable in the student's investigation?
Physics
1 answer:
Lera25 [3.4K]3 years ago
3 0
The water would be because however much salt you add the water rises
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How many nanoseconds does it take light to travel 3.50 ft in vacuum?
Fiesta28 [93]
Answer:3.56 nanosecond

In this case, you are asked the time and given the light distance(3.5ft)
To answer this question you would need to know the velocity of light. Speed of light is <span>299792458m/s. Then the calculation would be:

time= distance/speed
time= 3.5 ft / (</span>299792458m/s) x 0.3048 meter/ 1 ft=  3.56 10^{-9} second or 3.56 nanosecond
6 0
2 years ago
A cyclist going downhill is accelerating at 1. 2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is
mel-nik [20]

Answer:

\boxed {\boxed {\sf v_i= 4 \ m/s}}

Explanation:

We are asked to find the cyclist's initial velocity. We are given the acceleration, final velocity, and time, so we will use the following kinematic equation.

v_f= v_i + at

The cyclist is acceleration at 1.2 meters per second squared. After 10 seconds, the velocity is 16 meters per second.

  • v_f= 16 m/s
  • a= 1.2 m/s²
  • t= 10 s

Substitute the values into the formula.

16 \ m/s = v_i + (1.2 \ m/s^2)(10 \ s)

Multiply.

16 \ m/s = v_i + (1.2 \ m/s^2 * 10 \ s)

16 \ m/s = v_i + 12 \ m/s

We are solving for the initial velocity, so we must isolate the variable v_i. Subtract 12 meters per second from both sides of the equation.

16 \ m/s - 12 \ m/s = v_i + 12 \ m/s -12 \ m/s

4 \ m/s = v_i

The cyclist's initial velocity is <u>4 meters per second.</u>

6 0
2 years ago
In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp
MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

V_{Ax}=(21m/s)cos(-14^{o})

V_{Ax}=20.38 m/s

V_{Ay}=V_{A}sin\theta

V_{Ay}=(21m/s)sin(-14^{o})

V_{Ay}=-5.08 m/s

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0
2 years ago
A string of 26 identical Christmas tree lights are connected in series to a 120 V source. The string dissipates 73 W. What is th
spin [16.1K]

To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,

R_{eq}= \frac{V}{I}

Here,

V = Voltage

I = Current

While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,

P=VI \rightarrow I = \frac{P}{V}

I =0.608 A

Applying Ohm's law

R_{eq} = \frac{120V}{0.608A}

R_{eq} = 197.4\Omega

Therefore the equivalent resistance of the light string is 197.4\Omega

6 0
2 years ago
According to Newton’s Second Law of Motion, an object will accelerate if which kind of force is applied?
Kobotan [32]

Answer:

unbalanced force

Explanation:

this is a guess so just look it up

7 0
2 years ago
Read 2 more answers
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