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uranmaximum [27]
3 years ago
15

A kicker punts a football from the very center of the field to the sideline 39 yards downfield.

Physics
1 answer:
MrRa [10]3 years ago
8 0

Answer:

L = 47.15 yards

∅ = 55.80°

Explanation:

Given

y = 39 yards

x = 53/2 yards

L = ?

We apply the formula

L = √(x²+y²)

L = √((53/2 yards)²+(39 yards)²)

L = 47.15 yards

The angle between the direction of the net displacement of the ball and the central line of the feild is obtained as follows

∅ = tg⁻¹(y/x)

∅ = tg⁻¹(39/(53/2)) = 55.80°

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The change in internal energy during one complete cycle of a heat engine A. equals the net heat flow into the engine. B. equals
Stels [109]

Answer:

B. equals zero

Explanation:

Given data

one complete cycle = heat flow

solution

we have given that when heat engine complete 1 cycle change in energy = net heat flow

that is always equal to zero

from first law of thermodynamics that

ΔU = Q + W

we know ΔU is the change internal energy in system and Q is net heat transfer in system and W is  net work done in system

therefore change of internal energy during one cycle

ΔU = Ufinal -  Uinitial

ΔU  = Uinitial  -  Uinitial  = 0

7 0
3 years ago
You are looking at yourself in a plane mirror, a distance of 3 meters from the mirror. your brain interprets what you are seeing
AleksandrR [38]

You are looking at yourself in a plane mirror, a distance of 3 meters from the mirror. your brain interprets what you are seeing in the mirror as being a person standing 6 meters from you.

<h3>Calculation</h3>

The plane mirror shows an exact replica of the real world. that means the distance of you from the mirror is the same distance as your reflection form the mirror at the opposite side of the mirror.

Thus, distance of image from the plane mirror is same as the distance of object (person) from the plane mirror but the image is formed behind the mirror.

Thereby we have v=u=3 m

Thus, distance between image and the person

is d = v + u = 3 + 3 = 6 m

Thus, the person is 6 meters away from the image.

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8 0
1 year ago
a system of four particles moves along one dimension. the center of mass of the system is at rest, and the particles do not inte
Zinaida [17]

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

m1 = 1.45 kg, v1(t) = (6.09m/s) + (0.299m/s^2) × t

m2 = 2.81 kg, v2(t) = (7.83m/s) + (0.357m/s^2) × t

m3 = 3.89 kg, v3(t) = (8.09m/s) + (0.405m/s^2) × t

m4 = 5.03kg

The velocity of the center mass of the particles is calculated as;

McmVcm = m1v1 + m2v2 +m3v3+m4v4

Vcm= m1v1 + m2v2 +m3v3 +m4v4/ Mcm

0 = m1v1 + m2v2 +m3v3 +m4v4/ Mcm

m1v1 + m2v2 +m3v3+m4v4 = 0

m4v4 = -(m1v1 + m2v2 +m3v3)

v4 =-(m1v1 + m2v2 +m3v3)/ m4

The velocity of particle 1 at time, t = 2.83 s;

vi = 6.09 + 0.299× 2.83

v1 = 6.94 m/s

The velocity of particle 3 at time, t = 2.83 s;

v2 = 7.83 + 0.357 × 2.83

v2 = 8.84 m/s

The velocity of particle 3 at time, t = 2.83 s;

v3 = 8.09 + 0.405 × 2.83

v3 = 9.24 m/s

The velocity of particle 3 at time, t = 2.83 s;

v4 = - (m1v1 + m2v2 + m3v3)/m4

v4 = -(1.45×6.94 + 2.81×8.84 + 3.89×9.24)/5.03

v4 = -14.4 m/s

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

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5 0
1 year ago
What is the potential energy at the origin due to an electric field of 5 x 10^6 N/C located at x=43cm,y=28cm?
Fudgin [204]

Answer:

potential energy at origin is 2.57*10^{6} volt

Explanation:

given data:

electric field E = 5*10^{6} N/C

at x = 43 cm, y = 28 cm

distance btween E and origin

\Delta r = \sqrt{43^2 +28^2}

\Delta r = 51.313 cm

potential energy per unit charge \Delta V = - Edr

\Delta V = 5*10^6*51.313*10^{-2} J/C

\Delta V  =  2.57*10^{6} volt

potential energy at origin is 2.57*10^{6} volt

3 0
3 years ago
Please help me out !!
blagie [28]
I think the answer would be letter a
4 0
2 years ago
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