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2 years ago

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A student pushes against a tree with a force of 10 newtons (N). The tree does not move. What is the amount of force exerted by t

he tree on the student?

1 answer:

Andreas93 [3]2 years ago

8 0

10 N according to third law of emotion

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Which of the following ways is usable energy lost?

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A. Friction or all of the above

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A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr

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Answer:

a) m = 3 for λ1 and m= 2 for λ2 will overlap since they have the same values

b) At what angle does this overlap occur = 34.54 degree

Explanation:

The detailed steps and appropriate formula is as shown in the attachment.

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Which statement correctly compares and contrasts the information represented by the chemical formula and model of a compound?

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Models show how the atoms in a compound are connected.

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A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

The average velocity is given by the ratio between the displacement and the time taken:

$v=\frac{\Delta x}{\Delta t}$

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

The position at t = 2.00 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

So the displacement is

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The time interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

And so, the average velocity in this interval is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

While the position at t = 4.00 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

So the average velocity here is

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

While the position at t = 4 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Learn more about average velocity:

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2 years ago

A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy

GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

$v=\sqrt{v_1^2+v_2^2}$

$v=\sqrt{5^2+8^2}=9.43\ m/s$

Let E is the kinetic energy of the plane. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2$

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

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