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bazaltina [42]
3 years ago
11

You perform an experiment in which you measure the distance traveled by a projectile fired with different initial velocities. Wh

ich piece of technology is appropriate for gathering the experimental data you need?
A. Scale
B. Computer
C. Tape measure
D. Thermometer
Physics
2 answers:
aleksklad [387]3 years ago
5 0

The correct answer is C. Tape measure

Explanation:

In an experiment, the technology or instrument that is appropriate for gathering experimental data mainly depends on the factors or variables to be studied. In this way, a scale would be useful for gathering information related to mass; a thermometer for gathering information related to temperature; a computer for analyzing numerical information or large amounts of information and a tape measure in the case of gathering information about distance. This implies if you are measuring the distance of a projective with different initial velocities the best is to use a tape measure as this measures distance in millimeter, centimeters, and meters and therefore it is useful to know in numbers the difference of each projectile.

Vlad [161]3 years ago
4 0

During the motion of projectile we know that the distance traveled by the object is in the range of meter

So the object will move to certain distance at different velocity which we have to measure by a device which has range of distance from 0 meter to several meters

So the correct device to measure it is

<em>C. Tape Measure</em>

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A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in t
HACTEHA [7]

Answer:

(a) 2.85 m

(b) 16.5 m

(c) 21.7 m

(d) 22.7 m

Explanation:

Given:

v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

(c) Find Δx when t = 3.5 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

7 0
3 years ago
Relate the structure of water to its ability to act as a solvent
pishuonlain [190]

Explanation:

Water (H2O) as a polar covalent molecule has its arrangement of oxygen and hydrogen atoms where one end (hydrogen) has a partially positive charge while the other side (oxygen) had a partially negative charge.

It is also capable of forming hydrogen bonds with polar molecules. Each water molecule can form two hydrogen bonds involving their hydrogen atoms and two further hydrogen bonds using the hydrogen atoms attached to neighboring water molecules.

5 0
3 years ago
The amount of potential energy depends on the objects _________ and position
disa [49]
Mass would be the answer
4 0
3 years ago
Read 2 more answers
starting at a position of 0 m, if the man is moving at a constant velocity of 2 m/s, it will take seconds for him to reach a pos
Harman [31]

The calculated time is 6 seconds.

Time is defined by physicists as the flow of events from the past through the present and into the future. In essence, a system is timeless if it is unchanging. When describing events that take place in three-dimensional space, time can be thought of as the fourth dimension of reality. Even if time isn't directly connected to energy, it is undoubtedly connected to many other components of energy. For instance, the movement of energy across the universe can affect the direction of time (from the past to the future).

V= 2 m/s

d=12 m

t=?

we know that velocity=displacement / time

time= displacement / velocity

= 12/2

=6 seconds

the calculated time is 6 seconds.

Learn more about time here-

brainly.com/question/28050940

#SPJ4

4 0
1 year ago
A 75kg Tibetan is trekking along flat, but icy ledge with his 450kg yak when he slips over the edge. Luckily, he is holding the
Tcecarenko [31]

Answer:

minimal coefficient of static friction: \mu_s=0.1667

Explanation:

Once the Tibetan is hanging from the strap, he is exerting a horizontal force on the yak equal to his weight which is the product of his mass times the acceleration of gravity (g) as written below:

w = m\,*\,g= 75\,kg\,*\,g

The other forces acting on the yak are (see attached diagram):

* the force of gravity on the yak (identified in blue color in the image as F_g,

* the normal force (indicated in green in the image and identified by the letter "n") of the ledge on the yak as reaction to the yak's weight

* the force of static friction between the yak's hooves and the ledge (pictured in red in the image and identified with f_s)

Since the normal force and the force of gravity on the yak cancel each other (balance - the yak is not moving vertically), the only forces we need to analyse are the force of the Tibetan's weight via the strap, and the force of static friction which should at least be equal in magnitude so the Tibetan doesn't fall. We assume these two forces are acting horizontally (one to the right: the Tibetan's weight, and one to the left: the static friction).

As we said, we want them to be at least equal so thy are in balance.

We recall that the force of static friction is the product of the normal force (n) times the coefficient of static friction (\mu_s), such that: f_s=\mu_s\,*\,n

In our case these are the forces at play:

F_g= M\,*\,g=450\, kg \,*\,g\\n=F_g=450\,kg\,*\,g\\f_s=\mu_s\,*\,n=\mu_s\,*450\,kg\,*\,g\\w=m\,*\,g=75\,kg\,*\,g

So we need to find what is the minimum coefficient of static friction that precludes the Tibetan from falling. We therefore proceed to make an equality between the force of static friction on the yak and the weight of the Tibetan:

f_s=w\\\mu_s\,*450\,kg\,*g=75\,kg\,*\,g

and proceed to solve for the coefficient of friction by dividing both sides by "g" (which by the way cancels out), and by the yak's mass:

\mu_s\,*450\,kg\,*g=75\,kg\,*\,g\\\mu_s=\frac{75}{450} \\\mu_s=0.1667

where we have rounded to four decimal places the periodic number that the quotient generates. Notice that as expected, the coefficient of friction has no units (they all cancelled out in the division).

5 0
3 years ago
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