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Greeley [361]
3 years ago
13

Choose the situation below in which the force applied is the greatest.

Physics
1 answer:
Gnesinka [82]3 years ago
6 0

Answer:

D

Explanation:

We know the formula for Work to be:

W = f * d

Where W is work done

f is force

d is the distance

A)

Work = 50

Distance = 50

So, Force is:

Force = 50/50 = 1

B)

Work = 400

Distance = 80

Force = 400/80 = 5

C)

Work = 365

Distance = 73

Force = 365/73 = 5

D)

Work = 144

Distance = 16

Force = 144/16 = 9

Hence, D is the situation in which the force applied is the greatest.

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A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai
shutvik [7]

Answer:

d) None of the above

Explanation:

v_{o} = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

v_{oy} = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

a_{y} = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m  

t = time of travel

using the equation

y=v_{oy} t+(0.5)a_{y} t^{2}

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

v_{ox} = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

a_{x} = acceleration along the horizontal direction = 0 m/s²

t = time of travel =  2.1 s

Using the kinematics equation

x =v_{ox} t+(0.5)a_{x} t^{2}

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

v_{oy} = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

a_{y} = acceleration along the vertical direction = - 9.8 m/s²

y_{o} =initial vertical position at the time of launch = 20 m  

y = vertical position at the maximum height = 20 m

v_{fy} = final velocity along vertical direction at highest point = 0 m/s

using the equation

{v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})

0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)

y = 20.02 m

h = height above the starting height

h = y - y_{o}

h = 20.02 - 20

h = 0.02 m

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The early workers in spectroscopy (Fraunhofer with the solar spectrum, Bunsen and Kirchhoff with laboratory spectra) discovered
Anestetic [448]

The hot gases produce their own characteristic pattern of spectral lines, which remain fixed as the temperature increases moderately.

<h3><u>Explanation: </u></h3>

A continuous light spectrum emitted by excited atoms of a hot gas with dark spaces in between due to scattered light of specific wavelengths is termed as an atomic spectrum. A hot gas has excited electrons and produces an emission spectrum; the scattered light forming dark bands are called spectral lines.

Fraunhofer closely observed sunlight by expanding the spectrum and a huge number of dark spectral lines were seen.  "Robert Bunsen and Gustav Kirchhoff" discovered that when certain chemicals were burnt using a Bunsen burner, atomic spectra with spectral lines were seen. Atomic spectral pattern is thus a unique characteristic of any gas and can be used to independently identify presence of elements.

The spectrum change does not depend greatly on increasing temperatures and hence no significant change is observed in the emitted spectrum with moderate increase in temperature.

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How do longshore currents shape the land?
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Longshore drift has a very powerful influence on the shape and composition of the coastline. It changes the slopes of beaches and creates long, narrow shoals of land called spits, that extend out from shore. Longshore drift may also create or destroy entire “barrier islands” along a shoreline.
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Read 2 more answers
A 5 kg ball moving to the right at a speed of 6 m/s strikes another 4 kg
Dahasolnce [82]

Answer:

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(5 kg) (6 m/s) + (4 kg) (-5 m/s) = (5 kg) v₁ + (4 kg) v₂

10 m/s = 5 v₁ + 4 v₂

Assuming an elastic collision, kinetic energy is conserved.

½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²

m₁ u₁² + m₂ u₂² = m₁ v₁² + m₂ v₂²

(5 kg) (6 m/s)² + (4 kg) (-5 m/s)² = (5 kg) v₁² + (4 kg) v₂²

280 m²/s² = 5 v₁² + 4 v₂²

Substituting:

v₂ = (10 − 5 v₁) / 4

280 = 5 v₁² + 4 [(10 − 5 v₁) / 4]²

280 = 5 v₁² + (10 − 5 v₁)² / 4

1120 = 20 v₁² + (10 − 5 v₁)²

1120 = 20 v₁² + 100 − 100 v₁ + 25 v₁²

0 = 45 v₁² − 100 v₁ − 1020

0 = 9 v₁² − 20 v₁ − 204

0 = (9 v₁ + 34) (v₁ − 6)

v₁ = -3.78 m/s or 6 m/s

u₁ = 6 m/s, so v₁ = -3.78 m/s.  Solving for v₂:

v₂ = (10 − 5 v₁) / 4

v₂ = 7.22 m/s

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

6 0
3 years ago
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