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levacccp [35]
3 years ago
9

How do you determine true displacement value of an object that has moved?

Physics
1 answer:
WITCHER [35]3 years ago
3 0
Diceplacement is the distance an object has traveled in a certain direction
for example, if you were to walk North for 20m, then east for 40m, the <u>distance</u> you have traveled is 60m however your displacement is the distance between your starting position and your end position;
sqrt(20^2+40^2) = 44.7m
and because displacement is a vector, there needs to be a direction;
sin(theta)=40/44.7
theta=63.4 degrees East of North

therefore the true displacement is 44.7m at 63.4 degrees East of North
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¿Cuál de las siguientes magnitudes es derivada?
Sedaia [141]

Answer:

A. El volumen

B. La densidad.

Explanation:

A derived quantity is defined as one that has to be calculated by using two or more other measurements.

Volume is a derived quantity because it requires one to use different measurements to determine it. For instance, in the case of a cube, the length, width and height of the cube are all needed to calculate volume.

Density is also a derived quantity because it needs both volume and mass for it to be calculated.

8 0
3 years ago
Which is most likely the length of a student’s textbook?
UkoKoshka [18]

For this case, what we must do is to rewrite these measurements in the same unit in order to compare them.

By writing the measurements in meters we have:

30 mm = (30) * (\frac{1}{1000}) = 0.030 m

30 cm = (30) * (\frac{1}{100}) = 0.30 m

30 dm = (30) * (\frac{1}{10}) = 3 m\\30 Hm = (30) * (100) = 3000 m

Therefore, physically the correct measure is:

0.30 m = 30 cm

Answer:

the length of a student's textbook most likely is:

30 centimeters

3 0
4 years ago
Read 2 more answers
What is saturns most distinctive feature?
a_sh-v [17]
Its prominent ring system which is composed of primarily ice particles with smaller amounts of rocky detbris. Hope this helped!
6 0
3 years ago
What is the acceleration during the pushing-off phase, in m/27 Express your answer in meters per second squared. A bush baby, an
Sindrei [870]

Answer:

a = 12.78 g's

Explanation:

Height reached by the object after push off is given as

H = 2.3 m

v_f^2 - v_i^2 = 2 a s

now we have

0 - v^2 = 2(-9.81)(2.3)

v = 6.72 m/s

now we know that this push last for total distance of 0.18 m

so during the push we will have

v_f^2 - v_i^2 = 2 a d

6.72^2 - 0 = 2a(0.18)

a = 125.35 m/s^2

now in terms of g = 9.81 m/s/s we have

a = \frac{125.35}{9.81} gs

a = 12.78 g's

6 0
3 years ago
A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s
Zarrin [17]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± \frac{1}{2} at^2               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± \frac{1}{2} at^2               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - \frac{1}{2} at^2

h = 26.54(1.7) - \frac{1}{2} (10)(1.7)^2

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

7 0
3 years ago
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