A cardio kickboxing class challenges 3 things. List them.

Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration

LenaWriter [7]

Answer:

c. $5.6m/s^2$

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration, $a=\frac{v-u}{t}$

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a= $\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2$

Acceleration= $a=5.6 m/s^2$

Hence, the acceleration of cheetah= $5.6m/s^2$

5 0

3 years ago

The radial velocity method preferentially detects: Choose one: A. all of the above described planets equally well. B. small plan

Zina [86]

The radial velocity method preferentially detects large planets close to the central star

what is the Radial velocity:

The radial velocity technique is able to detect planets around low-mass stars, such as M-type (red dwarf) stars.

This is due to the fact that low mass stars are more affected by the gravitational tug of planets.

When a planet orbits around a star, the star wobbles a little.

From this, we can determine the mass of the planet and its distance from the star.

hence we can say that,

option D is correct.

The radial velocity method preferentially detects large planets close to the central star

Learn more about radial velocity here:

brainly.com/question/13117597

#SPJ4

5 0

2 years ago

Football player 1 has a mass of 80 kg and a velocity of 2 m/s east while player 2 has a mass of 70 kg and a velocity of 3 m/s we

sukhopar [10]

The total momentum of the system is equal to 50 Kgm/s.

Given the following data:

Mass 1 = 80 kg

Velocity 1 = 2 m/s east.

Mass 2 = 70 kg

Velocity 2 = 3 m/s west.

To determine the total momentum of the system:

Mathematically, momentum is given by the formula;

$Momentum = mass \times velocity$

For Football player 1:

$Momentum = 80 \times 2$

Momentum 1 = 160 Kgm/s.

For Football player 2:

$Momentum = 70 \times 3$

Momentum 1 = 210 Kgm/s.

Now, we can calculate the total momentum of the system:

$Total\;momentum = momentum \;2 - momentum \;1\\\\Total\;momentum = 210-160$

Total momentum = 50 Kgm/s.

Note: We subtracted because the football players were moving in opposite directions.

6 0

3 years ago

Solve this physics for me please with steps

Mars2501 [29]

Answer:

The answers are located in each of the explanations showed below

Explanation:

a)

(i) Surface Tension: The tensile force that causes this tension acts parallel to the surface and is due to the forces of attraction between the molecules of the liquid. The magnitude of this force per unit of length is called surface tension.

σ = F/l [N/m]

where:

F = force [N]

l = length [m]

σ = Surface Tension [N/m]

(ii) Frequency is the number of repetitions per unit of time of any periodic event.

f = 1/T [1/s] or [s^-1] or [Hz]

where:

T = period [s] or [seconds]

f = frecuency [Hz] or [hertz]

(iii) Each of the units will be shown for each variable

v = velocity [m/s]

a = accelertion [m/s^2]

s = displacement [m]

$[\frac{m}{s} ]^{2} =[\frac{m}{s} ]^{2} + 2* [\frac{m}{s^{2} } ]*[m]\\$

$[\frac{m^2}{s^2} ] =[\frac{m^2}{s^2} ] + [\frac{m^{2} }{s^{2} } ]$

$[\frac{m^2}{s^2} ]$

b) To find the velocity we must derivate the function X with respect to t because this derivate will give us the equation for the velocity, it means:

$v=\frac{dx}{dt} \\v = 0.75*2*t+5*t$

$(i) X = 0.75*t^{2} +5*t+1\\X = 0.75*(4)^{2} +5*(4)+1\\X = 33 [m]$

ii) replacing in the derivated equation.

$v=1.5*(4)+5\\v=11[m/s]$

iii) the average velocity is defined by the expresion v = x/t

$v = \frac{x-x_{0} }{t-t_{0} } \\$

$x_{0}=0.75(2)^{2}+5(2)+1 \\ x_{0}=14[m]\\x=0.75(7)^{2}+5(7)+1\\x=72.75[m]\\t = 7 [s]t0= 2[s]Now replacing:[tex]v_{prom} = \frac{72.75-14}{7-2} \\v_{prom} = 11.75 [m/s]$

2

a) Pascal's principle or Pascal's law, where the pressure exerted on an incompressible fluid and in balance within a container of indeformable walls is transmitted with equal intensity in all directions and at all points of the fluid.

Therefore:

P1 = pressure at point 1.

P2 = pressure at point 2.

P1 = F1/A1

P2= F2/A2

$\frac{F_{1} }{A_{1} }=\frac{F_{2}}{A_{2} } \\F_{1}=A_{1}*(\frac{F_{2}}{A_{2} })$

b) One of the applications of the surface tension is the capillarity this is a property of liquids that depends on their surface tension (which, in turn, depends on the cohesion or intermolecular force of the liquid), which gives them the ability to climb or descend through a capillary tube.

Other examples of surface tension:

The mosquitoes that can sit on the water.

A clip on the water.

Some leaves that remain floating on the surface.

Some soaps and detergents on the water.

5 0

3 years ago

If a simple truss member is known to carry a tensile force, then the internal force drawn on a free-body at a section cut when u

mote1985 [20]

Answer: b) pointed toward and parallel to the member.

Explanation:

It is shown in the picture attached

6 0

3 years ago