Ask question

Ask question

All categories

2 years ago

9

You kick a soccer ball with a mass of 2 kg. The ball leaves your foot with a speed of 30 m/s. How much kinetic energy does the b

all have?

2 answers:

____ [38]2 years ago

8 0

900 J is your answer :)

Lyrx [107]2 years ago

7 0

Answer:

KE= 900 (I think)

Explanation:

KE=½mv²

KE= ½(2)(30)²

KE=½(60)²

KE=30²

KE=900

Hope this helps!

You might be interested in

Elsa has a jar of honey that was left outside on a hot summer day. She notices that there are now sugar crystals at the bottom o

gavmur [86]

The solution is supersaturated. The answer is letter C.

A supersaturated solution has more dissolved solute in it than what it would normally have. This could be achieved if you dissolve more solute while heating the solution. However, once the solution cools, crystallization would occur. As a result, you see crystals settling at the bottom.

7 0

3 years ago

Read 2 more answers

A piece of copper wire is formed into a single circular loop of radius 9.1 cm. A magnetic field is oriented parallel to the norm

bazaltina [42]

Given Information:

Radius of circular loop = r = 9.1 cm = 0.091 m

Change in time = Δt = 0.66 seconds

Change in magnetic field = ΔB = 0.90 T

Resistance of wire per unit length = R = 2.9x10⁻² Ω/m

Number of turns = N = 1

Required Information:

Electrical energy dissipated = E = ?

Answer:

Electrical energy dissipated = 50.09x10⁻³ Joules

Step-by-step explanation:

We know that energy is given by

E = Pt

Where power is given by

P = ξ²/R

Where ξ is the induced EMF in the wire and is given by

ξ = -NΔΦ/Δt

Where ΔΦ is the change in flux and is given by

ΔΦ = ΔBAcosφ

Where φ is the angle between magnetic field and circular loop

A = πr² and R = 2.9x10⁻²*2πr

Substituting the above relations into the energy equation and simplifying yields,

E = [-Nπr²cosφ(ΔB/Δt)²]*t/R

E = [-1*π(0.091)²*cos(0)(0.90/0.66)²*0.66]/2.9x10⁻²*2π*(0.091)

E = 0.050094 Joules

E = 50.09x10⁻³ Joules

Therefore, the average electrical energy dissipated in the circular loop of the wire is 50.09x10⁻³ Joules.

6 0

3 years ago

All of the following are factors affecting flow rate except what?

faust18 [17]

Answer:

c. Vessel side holes

Explanation:

The "Poiseuille formula" which is given by $\\\begin{aligned} \small Q& = \small \frac{\pi r^4}{8 \eta}.\frac{\Delta P}{\Delta L}\\\end{aligned}$ describes the volumetric flow rate ( $\small Q$ ) through tubular sections.
Here, $\Delta P,\,\, \Delta L,\,\, r,\,\, \eta$ represent the injection pressure difference, the length of the section, the radius of the section and the viscosity index of the fluid that flows through the section respectively.
With this, one can confirm that all the factors except the vessel side holes affect the flow rate.
Side holes, however, are a factor that could give a measure of how much volume would flow to a particular location. In such a situation the flow rate remains unchanged and one location would receive a lower volume (not the whole) as some volume would spill out at the side holes.

#SPJ4

5 0

2 years ago

A hoop rolls down a 4.75 m high hill without slipping. What is the final speed of the hoop, in meters per second?

german

Answer:6.82 m/s

Explanation:

Given

Height of hill $h=4.75\ m$

Suppose r is the radius of hoop and m be its mass

So moment of inertia is given by

$I=mr^2$

Using conservation of energy we get

Potential energy at top=Kinetic+Rotational energy at bottom

$mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2$

where $\omega =\frac{v}{r}$

v=velocity of hoop

$2gh=v^2+r^2\times \frac{v^2}{r^2}$ (without slipping)

$v^2=gh$

$v=\sqrt{gh}$

$v=\sqrt{98\times 4.75}$

$v=6.822\ m/s$

4 0

3 years ago

The Concorde's exterior is made of aluminum, which has acoefficient of linear expansion of At 15 degrees celsius, the Concorde m

Rom4ik [11]

Answer: D) 28 cm

Explanation:

This problem can be solved by the following equation:

$\frac{\Delta L}{L_{o}}= \alpha \Delta T$ (1)

Where:

$\Delta L$ is the change in the length of the outer skin of the Concorde

$L_{o}=62.1 m$ is the initial length of the outer skin of the Concorde

$\alpha=2.40(10)^{-5} \°C^{-1}$ is the coefficient of linear thermal expansion for aluminium

$\Delta T=T_{f}-T_{o}=200\°C - 15\°C=185\°C$ is the variation in temperature

Isolating $\Delta L$ :

$\Delta L=\alpha \Delta T L_{o}$ (2)

$\Delta L=(2.40(10)^{-5} \°C^{-1})(185\°C)(62.1 m)$ (3)

$\Delta L=0.275 m \frac{100 cm}{1m}=27.5 cm$ (4)

Finally:

$\Delta L=27.5 cm \approx 28 cm$

3 0

4 years ago