Give two example of conditions when two objects (Object A=1 kg and Object B =10 kg) would have the same momentum.

Define very long base line interferometry

chubhunter [2.5K]

Very Long Baseline Interferometry (VLBI) is a technique being used by the United States Naval Observatory (USNO) to determine the reference frames for stars and the Earth, and to predict the variable orientation of the Earth in three-dimensional space.

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5 0

2 years ago

Read 2 more answers

Two people on skateboards, Brian and Amy, push off from each other. Brian has more mass than Amy. Who experiences a larger force

amm1812

Amy because she is lighter, so the more force will be put on her

4 0

3 years ago

The step by step experiment to determine the cubic expansivity of a liquid

Anit [1.1K]

In accordance with the definition of density as r = m/V, in order to determine the density of matter, the mass and the volume of the sample must be known. The determination of mass can be performed directly using a weighing instrument. The determination of volume generally cannot be performed directly. Exceptions to this rule include · cases where the accuracy is not required to be very high, and · measurements performed on geometric bodies, such as cubes, cuboids or cylinders, the volume of which can easily be determined from dimensions such as length, height and diameter. · The volume of a liquid can be measured in a graduated cylinder or in a pipette; the volume of solids can be determined by immersing the sample in a cylinder filled with water and then measuring the rise in the water level. Because of the difficulty of determining volume with precision, especially when the sample has a highly irregular shape, a "detour" is often taken when determining the density, by making use of the Archimedean Principle, which describes the relation between forces (or masses), volumes and densities of solid samples immersed in liquid: From everyday experience, everyone is familiar with the effect that an object or body appears to be lighter than in air – just like your own body in a swimming pool. Figure 3: The force exerted by a body on a spring scale in air (left) and in water (right)

7 0

3 years ago

A car is making a 50 mi trip. It travels the first half of the total distance 25.0 mi at 7.00 mph and the last half of the total

worty [1.4K]

Answer:

a) The total time of the trip is 4.05 h.

b) The average speed of the car is 12.35 mi/h.

c) The total time of the trip is 1.69 h.

Explanation:

Hi there!

a) The equation of traveled distance for a car traveling at constant speed is the following:

x= v · t

Where:

x = traveled distance.

v = velocity.

t = time.

Solving the equation for t, we can find the time it takes to travel a given distance "x" at a velocity "v":

x/v = t

So, the time it takes the car to travel the first half of the distance will be:

t1 = 25.0 mi / 7.00 mi/h

And for the second half of the distance:

t2= 25.0 mi / 52.00 mi / h

The total time will be:

total time = t1 + t2 = 25.0 mi / 7.00 mi/h + 25.0 mi / 52.00 mi / h

total time = 4.05 h

The total time of the trip is 4.05 h.

b) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it takes to travel that distance (t). In this case, the traveled distance is 50 mi and the time is 4.05 h. Then:

a.s = d/t

a.s = 50 mi / 4.05 h

a.s = 12.35 mi/h

The average speed of the car is 12.35 mi/h

c) Let's write the equations of traveled distance for both halves of the trip:

For the first half, you traveled a distance d1 in a time t1 at 7.00 mph:

7.00 mi/h = d1/t1

Solving for d1:

7.00 mi/h · t1 = d1

For the second half, you traveled a distance d2 in a time t2 at 52.00 mph.

52.00 mi/h = d2/t2

52.00 mi/h · t2 = d2

We know that d1 + d2 = 50 mi and that t1 and t2 are equal to t/2 where t is the total time:

d1 + d2 = 50 mi

52.00 mi/h · t/2 + 7.00 mi/h · t/2 = 50 mi

Solving for t:

29.5 mi/h · t = 50 mi

t = 50 mi / 29.5 mi/h

t = 1.69 h

The total time of the trip is 1.69 h.

6 0

3 years ago

The following data were collected during a short race between two friends. Velocity (m/s) 0 0.5 1 1.5 2 2 4 6 2 0 Time (s) 0 2 4

scoundrel [369]

The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:

a) we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s is: a = 0.25 m / s²

c) The maximum acceleration is: a = 1 m / s²

d) The displacement is: i) d₁ = 8m, ii) $d_{total}$ = 16 m

e) maximum speed is: v = 6 m / s

Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.

v = v₀ + a t

y = v₀ t + ½ a t²

where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.

In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:

Positive slope the body is accelerating and the speed is increasing.

Negative slope the body is stopping, the speed decreases.

Let's answer the different questions about the system.

a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s

v = v₀ + a t

$a = \frac{v-v_o}{\Delta t}$

$a = \frac{2-0}{8-0}$

a = 0.25 m / s²

c) The maximum acceleration is when the slope is maximum.

$a = \frac{6-2}{ 14-10}$

a = 1 m / s²

Therefore the acceleration is maximum in the section between 10 and 14 s

d) The total displacement is the sum of the displacements of each section.

$d_{total } = d_1 +d_2 + d_3 +d_4$

We look for every displacement.

d₁ = v₀ + ½ a₁ Δt²

d₁ = 0 + ½ 0.25 8²

d₁ = 8 m

In the second section the velocity is constant

d₂ = v₂ Δt₂

d₂ = 2 (10-8)

d₂ = 4 m

The third section.

d₃ = v₀ + ½ a t²

d₃ = 2 + ½ 1 (14-10) ²

d₃ = 10 m

The distance of the fourth section.

we look for acceleration

a₄ = $\frac{v-v_o}{\Delta t}$

a₄ = $\frac{0-6}{18-14}$

a₄ = -1.5 m / s²

d₄ = 6 + ½ (-1.5) (1814) ²

d₄ = -6 m

The total displacement is;

$d_{total}$ = 8 + 4 + 10 -6

$d_{total}$ = 16 m

e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to

v = 6 m / s

In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:

a) we have four sections;

0 to 8 s The body is accelerating.

8 to 10 s The body goes at a constant speed, the acceleration is zero.

10 to 14 Body accelerating.

14 to 18 Body slowing down.

b) The acceleration is the first 8 s is: a = 0.25 m / s²

c) The maximum acceleration is: a = 1 m / s²

d) The displacement is: i) d₁ = 8m, ii) $d_{total}$ = 16 m

e) maximum speed is: v = 6 m / s

Learn more about kinematics here: brainly.com/question/24783036

3 0

2 years ago