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scoray [572]
3 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]3 years ago
6 0
Observe as many moose as he can in as many locations as possible.
Rom4ik [11]3 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 78.5-kg man is standing on a frictionless ice surface when he throws a 2.40-kg book horizontally at a speed of 11.3 m/s. With
kirill [66]

Answer:

The man moves across the ice with a speed of 0.345m/s.

Explanation:

From the conservation of linear momentum, we have that the total linear momentum before the book throw is equal to the total linear momentum just after it. Since the initial velocity of the system is zero (so the initial momentum is zero), we have that:

m_mv_m+m_bv_b=0\\\\v_m=-\frac{m_b}{m_m} v_b

Where m_m is the mass of the man, m_b is the mass of the book, and v_m and v_b are their velocities. Plugging in the given values, we can compute the speed of the man (ignoring the negative sign, because we care about the magnitude, not the direction):

v_m=\frac{2.40kg}{78.5kg}(11.3m/s)=0.345m/s

In words, the resulting speed of the man is 0.345m/s.

8 0
3 years ago
Read 2 more answers
A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It ta
My name is Ann [436]

Answer:

a).K=528.92 \frac{N}{m}

b).m=4.84kg

Explanation:

a).

The work of the spring is find by the formula:

w_s =\frac{1}{2}*k*x^2

So knowing the work can find the constant K'

3.2J =\frac{1}{2}*k*(0.11m)^2

Solve for K'

K=\frac{2*W_s}{x^2}=\frac{2*3.2J}{0.11m^2}

K=528.92 \frac{N}{m}

b).

The force of the spring realice a motion so using the force and knowing the accelerations can find the mass

F=m*a

m=\frac{F}{a}=\frac{K*x}{a}

m=\frac{528.9*0.11m}{12m/s^2}

m=4.84kg

8 0
2 years ago
A car is initially moving at 20 m/s east and a little while later it is moving at 10 m/s north. Which of the following best desc
Nina [5.8K]

Answer:d

Explanation:

Given

First car is moving  towards east with velocity 20 m/s

\vec{v_1}=20\hat{i}

then it turns towards north then velocity is  

\vec{v_2}=10\hat{j}

suppose car takes t sec to change its path so average acceleration is given by

a=\frac{v_2-v_1}{t}

a=\frac{1}{t}(10\hat{j}-20\hat{i})

So average  acceleration is towards North of west.

5 0
3 years ago
Drag the correct labels to the images. Each label can be used more than once.
coldgirl [10]

Answer:

plato answer.

Explanation:

4 0
3 years ago
A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C d
Alex73 [517]

Answer

given,

Side of copper plate, L = 55 cm

Electric field, E = 82 kN/C

a) Charge density,σ = ?

  using expression of charge density

 σ = E x ε₀

ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

now,

 σ = 82 x 10³ x 8.85 x 10⁻¹²

 σ = 725.7 x 10⁻⁹ C/m²

 σ = 725.7 nC/m²

change density on the plates are 725.7 nC/m² and -725.7 nC/m²

b) Total change on each faces

   Q = σ  A

   Q = 725.7 x 10⁻⁹ x 0.55²

   Q = 219.52 nC

Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC

7 0
3 years ago
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