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3 years ago

What is the instantaneous velocity v of the particle at t=10.0s?

Physics

2 answers:

algol [13]3 years ago

6 0

The instantenous velocity is just the slope of the graph at a certain instant. Since the graph is a straight line, its instantenous velocity is uniform through out. v = dx / dt = (40 - 10) / (50 - 0) = 0.6 m/s.

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BlackZzzverrR [31]3 years ago

5 0

The instantaneous velocity of the particle is 0.600 m/s. Instantaneous velocity is the velocity of an object at any given time in its path.

<h3>FURTHER EXPLANATION</h3>

The instantaneous velocity of an object can be determined graphically in a position - time graph.

If the velocity of the object is constant, the graph in a position - time graph will be a straight line and the instantaneous velocity will be equal to the slope or gradient of the line.
If the velocity of the object is changing, the graph will be a curve and the instantaneous velocity can be determined from the slope of the curved line at any given moment.

To get the slope of a line, get the ratio of the rise and the run.

In the given problem, the velocity of the object is constant because its position - time graph shows a straight line. As such, the instantaneous velocity will be equal to the average velocity.

To solve for the average velocity, use the equation below:

$average \ velocity \ = \frac{displacement \ ( \Delta y)}{ time \ (\Delta x)}$

Using data from the graph,

$average \ velocity \ = \frac{40\ m - 10 \ m}{50 \ s \ - 0 \ s} \\average \ velocity \ = \frac{30 \ m}{50 \ s}\\ \\ \boxed {average \ velocity = 0.6 \frac{m}{s}}\\$

The instantaneous velocity expressed with 3 significant figures like the given in the problem, therefore, is 0.600 m/s.

<h3>LEARN MORE</h3>

Position - Time Graph brainly.com/question/11533012
Acceleration brainly.com/question/4134594
Velocity - Time Graph brainly.com/question/4710544

Keywords: instantaneous velocity, position - time graph

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A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t

tresset_1 [31]

Answer:

Part a)

$Reading = 2.00 kg$

Part b)

$Reading = 2.00 kg$

Part c)

$Reading = 4.04 kg$

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part b)

When elevator is decending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

$F_{net} = ma$

$T - mg = ma$

$T = mg + ma$

Reading is given as

$Reading = \frac{mg + ma}{g}$

$Reading = 2.00\frac{9.81 + 10}{9.81}$

$Reading = 4.04 kg$

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

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3 years ago

Two children are riding on a merry-go-round that is rotating with a constant angular speed. Abbie is one meter from the center o

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Answer:

Abbie’s acceleration is (1/2) Zak’s acceleration.

Explanation

1. Data:

a) ω = constant

b) Abbie: r₁ = 1 m

c) Zak: r₂ = 2 m

d) Ac₁ = ? Ac₂

2. Formulae

Ac = ω² r

3. Solution:

a) Abbie:

Ac₁ = ω² r₁ = ω² (1m)

b) Zack:

Ac₂ = ω² r₂ = ω² (2m)

c) Divide Ac₁ / Ac₂

Ac₁ / Ac₂ = ω² (1m) / [ω² (2m) ] = 1/2

⇒ Ac₁ = (1/2) Ac₂ = Ac₂ / 2 = 0.5 Ac₂

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2 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.