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Margarita [4]
4 years ago
6

A 55.0-kg skydiver drew falls for a period of time before opening his parachute. what is his kinetic energy when he reaches a ve

locity of 16.0 meters/second?
Physics
1 answer:
Jlenok [28]4 years ago
3 0
Mass (m)=55kg

acceleration (a)=9.81 m/s^2, this is the acceleration due to gravity.

initial velocity=0m/s. The skydiver doesn’t start with any speed because she is on the plane or helicopter.

final velocity=16m/s This is the velocity (speed) the skydiver reaches

The equation we use is KE=.5mv^2
Kinetic energy=.5 mass x velocity^2

KE=.5(55kg)(16m/s)^2
KE=.5(55kg)(256m/s)
KE=.5(14080J)

J=Joules

KE=7040J

Kinetic energy is 7040 Joules (J)

Hope this helps
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Tennis balls experience a large drag force. A tennis ball is hit so that it goes up and then comes back straight down.
Effectus [21]

A tennis ball is hit by a large force so that it goes up into the air and then it comes back straight down because of gravity.

<h3>How object move upward and downward?</h3>

We know that objects move upward due to application of force on it while on the other hand, object comes to the ground because of the attraction of earth which we called gravity.

So we can conclude that a tennis ball is hit by a large force so that it goes up into the air and then it comes back straight down because of gravity.

Learn more about force here: brainly.com/question/12970081

#SPJ1

8 0
2 years ago
The dimensions of a cylinder are changing, but the height is always equal to the diameter of the base of the cylinder. If the he
VashaNatasha [74]

Answer:

dV/dt  = 9 cubic inches per second

Explanation:

Let the height of the cylinder is h

Diameter of cylinder = height of the cylinder = h

Radius of cylinder, r = h/2

dh/dt = 3 inches /s

Volume of cylinder is given by

V = \pi r^{2}h

put r = h/2 so,

V = \pi \frac{h^{3}}{4}

Differentiate both sides with respect to t.

\frac{dV}{dt}=\frac{3h^{2}}{4}\times \frac{dh}{dt}

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

\frac{dV}{dt}=\frac{3\times 2\times 2}{4}\times 3

dV/dt  = 9 cubic inches per second

Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.

6 0
3 years ago
Levi and Clara are trying to move a very heavy box. Levi is pushing the box with a force of 30 N, and Clara is pulling the box w
musickatia [10]
The net force on the box is:
50 + 30 - 65
= 15 Newtons
This can be used in conjunction with
F = ma
to calculate the acceleration of the box if its mass is known.
3 0
3 years ago
You measure an electric field of 1.36×106 N/C at a distance of 0.158 m from a point charge. There is no other source of electric
marysya [2.9K]

Answer:

The Electric flux will be 0.42\times10^6\ \rm N.m^2/C

Explanation:

Given

Strength of the Electric Field at a distance of 0.158 m from the point charge is

E=1.36\times10^6\ \rm N/C

We know that the flux of the Electric Field can be calculated by using Gauss Law which is given by

\int E.dA=\dfrac{q_{in}}{\epsilon_0}\\

Let consider a  sphere of radius 0.158 m as Gaussian Surface at a distance of 0.158 m from the point charge and Let \phi be the flux of the Electric Field coming out\passing through it which is given  by

\phi=\int E.dA=1.36\times10^6 \times4\pi \times 0.158^2\\\\=0.42\times10^6\ \rm N.m^2/C

It can be observed that same amount of  flux which is passing through the Gaussian sphere of radius 0.158 is also passing through the Gaussian sphere of radius 0.142 m at a distance of 0.142 m from its centre.

Also it can be observed that the charge inside the two Gaussian Sphere mentioned have same value so the Flux of electric field through them will also be same.

So the electric flux through the surface of sphere that has given charge at its centre and that has radius 0.142 m is  0.42\times10^6\ \rm N.m^2/C

8 0
3 years ago
A compass originally points North; at this location the horizontal component of the Earth's magnetic field has a magnitude of 2e
astraxan [27]

Answer:

μ =5.40 A-m²

Explanation:

The components of the net magnetic field are the magnetic field of the dipole and the magnetic field of Earth, then from the right triangle, the deflection angle is computed by

tan θ = Bdipole / Bearth     ⇒   Bdipole = Bearth* tan θ  

Bdipole = 2e-5 T*tan 70° = 5.49e-5 T

The magnetic field at the location of the compass due to the dipole has a magnitude

Bdipole = (μ₀/4π)(2μ/r³)    ⇒    μ = Bdipole r³ / 2(μ₀/4π)

μ = (5.49e-5 T)(0.27m)³ / 2(1 × 10−7 T m² /(C m/s)) = 5.40 A-m²

8 0
3 years ago
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