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4 years ago

At what angle should the player aim the ball if it is thrown from a height of 8.5 ftft with a speed of 28 ft/sft/s?

Physics

1 answer:

pav-90 [236]4 years ago

7 0

Answer:

θ = 56.09°

Explanation:

Given that

h = 8.5 ft

speed ,u= 28 ft/s

We know that acceleration due to gravity g

g= 32.2 ft/s²

The maximum height h is given as

$h=\dfrac{u^2sin^2\theta}{2g}$

Now by putting the values

$8.5=\dfrac{28^2sin^2\theta}{2\times 32.2}$

$sin^2\theta=0.69$

$sin\theta=0.83$

θ = 56.09°

Therefore the angle will be θ = 56.09°

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3 years ago

An average human weighs about 650 N. If each of two average humans could carry 1.0 C of excess charge, one positive and one nega

bezimeni [28]

Answer:

Distance, r = 3721.04 meters

Explanation:

It is given that,

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3 years ago

A porsche 911 can go from 0 to 60 mph in 3.4 seconds whats ur acceleration?

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3 years ago

Read 2 more answers

A simple series circuit consists of a 130 ? resistor, a 30.0V battery, a switch, and a 2.10 pF parallel-plate capacitor (initial

V125BC [204]

Answer with Explanation:

We are given that

Resistance,R=130 ohm

Potential difference, V=30 V

Capacitor,C=2.1pF= $2.1\times 10^{-12} F$

$1pF=10^{-12} F$

$d=5.0 mm=5\times 10^{-3} m$

$1mm=10^{-3} m$

a.Maximum flux

$\phi=EA=\frac{V}{d}\times \frac{Cd}\epsilon_0}=\frac{CV}{\epsilon_0}$

$\epsilon_0=8.85\times 10^{-12}$

$\phi=\frac{2.1\times 10^{-12}\times 30}{8.85\times 10^{-12}}=7.12Vm$

b.Maximum displacement current, $I=\frac{V}{R}=\frac{30}{130}=0.23 A$

c.We have to find electric flux at t=0.5 ns

$t=0.5ns=0.5\times 10^{-9} s$

$1ns=10^{-9}s$

$q=CV(1-e^{-\frac{t}{RC}})$

$q=30\times 2.1\times 10^{-12}(1-e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-9}})$

$q=52.9\times 10^{-12} C$

$\phi=\frac{q}{\epsilon_0}=\frac{52.9\times 10^{-12}}{8.85\times 10^{-12}}=5.98Vm$

d.Displacement current at t=0.5ns

$I=(\frac{V}{R})e^{-\frac{t}{RC}}=\frac{30}{130}e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-12}}}$

$I=0.037 A$

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3 years ago

Which describes the relation between an electric field and an equipotential surface?

alina1380 [7]

The answer is the field vector is perpendicular to the surface.

The electric field's angle with the equipotential surface is always 90 degree. The electric field is always perpendicular to the equipotential surface.

What is electric field?

The electric field is formally defined as a vector field associated with each location in space, the force per unit charge exerted on a positive test charge at rest at that place.
The electric charge or time-varying magnetic fields create the electric field. At the atomic level, the electric field is responsible for the attractive forces that hold the atomic nucleus and electrons together.
The normal vector to a surface, often known as the "normal," is a vector that is perpendicular to the surface at a particular position. When considering normals on closed surfaces, the inward-pointing normal (pointing towards the surface's interior) and outward-pointing normal are commonly differentiated.

To learn more about electric field visit:

brainly.com/question/8971780

#SPJ4

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