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3 years ago

What are the three ways of answering a scientific question

Physics

1 answer:

My name is Ann [436]3 years ago

5 0

Answer:

Let's start by understanding what exactly a scientific question is. A scientific question is a question that may lead to a hypothesis and help us in answering (or figuring out) the reason for some observation. A good scientific question has certain characteristics. It should have some answers (real answers), should be testable.

Here's examples of a few:

Why is that a star?

What is that star made of?

Hope this can lead you to the answer you're looking for at least!!

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Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

3 0

3 years ago

To have the highest magnification in a telescope, the focal length of the objective lens should be _________ and the focal lengt

Darina [25.2K]

Answer:

Large; small.

Explanation:

A telescope can be defined as an optical instrument or device which comprises of a curved mirror and lenses used for viewing distant objects i.e objects that are very far away such as stars and other planetary bodies. The first telescope was invented by Sir Isaac Newton.

To have the highest magnification in a telescope, the focal length of the objective lens should be large and the focal length of the eyepiece lens should be small.

This ultimately implies that, the eyepiece lens has a small focal length while the objective lens has a large focal length.

6 0

3 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$