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scoray [572]
4 years ago
5

A scientist wants to publish a report on a general feeding habits of a moose in Canada. He should

Physics
2 answers:
saul85 [17]4 years ago
6 0
Observe as many moose as he can in as many locations as possible.
Rom4ik [11]4 years ago
3 0

THE ANSWER IS: <u><em>observe as many moose as he can in as many locations as possible.</em></u>

In order to write a report about general feeding habits of moose over such a large area as Canada, the scientist needs to observe as many moose as he can in as many locations as possible. Studying too few moose or in just one location will not provide enough data to apply his results to all moose in Canada. The scientist can also use studies by other scientists to help in his report.

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A new planet has been discovered and given the name Planet X . The mass of Planet X is estimated to be one-half that of Earth, a
harina [27]

Answer:

    vₐ = v_c  ( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )

Explanation:

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

          Em₀ = K + U = ½ m v_c² - G Mm / R

final point. At a very distant point

         Em_f = U = - G Mm / R₂

energy is conserved

           Em₀ = Em_f

           ½ m v_c² - G Mm / R = - G Mm / R₂

           v_c² = 2 G M (1 /R -  1 /R₂)

if we consider the speed so that it reaches an infinite position R₂ = ∞

           v_c = \sqrt{\frac{2GM}{R} }

now indicates that the mass and radius of the planet changes slightly

            M ’= M + ΔM = M ( 1+ \frac{\Delta M}{M} )

            R ’= R + ΔR = R ( 1 + \frac{\Delta R}{R} )

we substitute

           vₐ = \sqrt{\frac{2GM}{R} } \  \frac{\sqrt{1+ \frac{\Delta M}{M} } }{ \sqrt{1+ \frac{ \Delta R}{R} } }

         

let's use a serial expansion

           √(1 ±x) = 1 ± ½ x +…

we substitute

         vₐ = v_ c ( (1 + \frac{1}{2}  \frac{\Delta M}{M} )  \ ( 1 - \frac{1}{2}  \frac{\Delta R}{R} ))

we make the product and keep the terms linear

        vₐ = v_c  ( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )

5 0
3 years ago
Vector → A points due north and vector → B points due west. Both vectors have the same magnitude. Which of the following vectors
-BARSIC- [3]

Answer:

B - A

Explanation:

For the combination of 2 vector to due southwest, 1 vector must due south and the other vector due west. Since vector B is already due west, vector A should due south. As vector A is already due north, vector -A would due south. So the combination of B + (-A) or B - A should points southwest

5 0
3 years ago
What is a stable and unstable equilibrium
Citrus2011 [14]

Answer:

here

Explanation:

Equilibrium is a state of a system which does not change. ... An equilibrium is considered stable (for simplicity we will consider asymptotic stability only) if the system always returns to it after small disturbances. If the system moves away from the equilibrium after small disturbances, then the equilibrium is unstable.

4 0
3 years ago
Iron man wears an awesome ironsuit.He is flying over high current carrying wire. Will he be affected?
frozen [14]

Answer:

According to super hero logic , nothing will happen to him.

But according to science , yes he will get current shock but good news is that he wouldn't get elected until he is in contact with the wires.

He may / may not be affected but his suit will be damaged for sure as it is made of metal.

HOPE THIS HLEP AND PLSSSSS MARK AS BRAINLIEST AND THNXX :)

8 0
4 years ago
A particle starts from the origin at t = 0 with an initial velocity of 4.8 m/s along the positive x axis.If the acceleration is
tia_tia [17]

Answer:

Part a)

Velocity = 6.9 m/s

Part b)

Position = (3.6 m, 5.175 m)

Explanation:

Initial position of the particle is ORIGIN

also it initial speed is along +X direction given as

v_x = 4.8 m/s

now the acceleration is given as

\vec a = -3.2 \hat i + 4.6 \hat j

when particle reaches to its maximum x coordinate then its velocity in x direction will become zero

so we will have

v_f = v_i + at

0 = 4.8 - 3.2 t

t = 1.5 s

Part a)

the velocity of the particle at this moment in Y direction is given as

v_f_y = v_i + at

v_f_y = 0 + 4.6(1.5)

v_f_y = 6.9 m/s

Part b)

X coordinate of the particle at this time

x = v_x t + \frac{1}{2}a_x t^2

x = 4.8(1.5) - \frac{1}{2}(3.2)(1.5^2)

x = 3.6 m

Y coordinate of the particle at this time

y = v_y t + \frac{1}{2}a_y t^2

y = 0(1.5) + \frac{1}{2}(4.6)(1.5^2)

y = +5.175 m

so position is given as (3.6 m, 5.175 m)

5 0
4 years ago
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