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3 years ago

Work is the product of force and an object's

Physics

2 answers:

Roman55 [17]3 years ago

8 0

Answer:

displacement

Explanation:

Neporo4naja [7]3 years ago

7 0

Answer:

C. displacement

Explanation:

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What is the best free energy source? Nuclear Solar Natural Gas

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Solar it is the cheapest and widely used energy source

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2 years ago

NEED HELP ASAP

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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3 years ago

What is true about energy that is added to a closed system?

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The correct answer is B

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3 years ago

Read 2 more answers

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

andrew11 [14]

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}$

For second order

$\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

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2 years ago

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