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2 years ago

If the position of a particle on the x-axis at time t is âˆ’5t2, then the average velocity of the particle for 0 â‰¤ t â‰¤ 3 is

Physics

1 answer:

Drupady [299]2 years ago

3 0

Answer:

v = 15 m / s

Explanation:

In this exercise we are given the position function

x = 5 t²

and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval

$v= \frac{v_{f} - v_{o} }{t_{f} - t_{o} }$

let's look for the displacements

t = 0 x₀ = 0 m

t = 3 $x_{f}$ = 5 3 2

x_{f} = 45 m

we substitute

$v = \frac{45 -0}{3 - 0}$

v = 15 m / s

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A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft

elena-s [515]

Answer:

The water level rises more when the cube is located above the raft before submerging.

Explanation:

These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.

Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.

When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.

$Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]$

Density is given by:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyancy force can be calculated using the following equation:

$F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]$

Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.

7 0

3 years ago

If you see a sedimentary rock outcrop and red layers of sand are on top of pale yellow layers of sand, what do you know for sure

neonofarm [45]

Answer: The yellow layer is definitely older than the red layer

Explanation: According to Nicolaus Steno's law of superposition and original horizontality. Older rocks underlie younger rocks.

Sedimentary rocks are usually deposited in horizontal layers in which each stratigraphic layer is laid down before another can be deposited upon it.

The red layer, in addition to being older, is also likely to have undergone intense oxidation due to earlier exposure.

5 0

2 years ago

A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

vitfil [10]

The acceleration of the sled will be 1.30 m/s². Force is defined as the product of mass and acceleration.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Given data;

m(mass of sled)=8 kg

Θ is the inclination of force= 50°

Force of friction,f=2.4 N.

The applied force at the given angle is resolved into the two-component as;

$\rm F_h=F cos \theta \\\\ F_h= 20 cos 50 ^0 \\\\ F_h= 12.85 \ N$

$\rm F_v=F sin \theta \\\\ F_v=20 sin 50^0 \\\\ F_v=15.32 \ N$

The net vertical force is zero;

$\rm F_N=mg-Fsin50^0 \\\\ \rm F_N=8 \times 9.81 -15.32 \\\\ F_N=63.1 \ N \\\\$

From Newton's second law the net force as;

$\rm \sum F_{net}=ma \\\\ Fcos 60^0-f =ma \\\\ a=\frac{12.855-2.4}{8} \\\\a = 1.30 \ m/s^2$

Hence, the acceleration of the sled will be 1.30 m/s².

To learn more about the force refer to the link;

brainly.com/question/26115859

#SPJ1

5 0

1 year ago

Read 2 more answers

1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (H

JulijaS [17]

Answer:

The angle between the electric field lines and the equipotential surface is 90 degree.

Explanation:

The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.

The electric field lines are always perpendicular to the equipotential surface.

$dV = \overrightarrow{E} . d\overrightarrow{r}\\\\$