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3 years ago

What is the weight of a ring tailed lemur that has a mass of 10kg

1 answer:

3 0

None of these. A lemur weighs about 2.2kg. Therefore this question doesn't add up correctly with there being more that 10kg worth of a lemur. You would have to split it up evenly between more than one lemur.

I'm just kidding. It's 98oz

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When a force of 20.0 N is applied to a spring, it elongates 0.20 m. Determine the period of oscillation of a 4.0-kg object suspe

mash [69]

Answer:

1.26 secs.

Explanation:

The following data were obtained from the question:

Force (F) = 20 N

Extention (e) = 0.2 m

Mass (m) = 4 Kg

Period (T) =.?

Next, we shall determine the spring constant, K for spring.

The spring constant, K can be obtained as follow:

Force (F) = 20 N

Extention (e) = 0.2 m

Spring constant (K) =..?

F = Ke

20 = K x 0.2

Divide both side by 0.2

K = 20/0.2

K = 100 N/m

Finally, we shall determine the period of oscillation of the 4 kg object suspended on the spring. This can be achieved as follow:

Mass (m) = 4 Kg

Spring constant (K) = 100 N/m

Period (T) =..?

T = 2π√(m/K)

T = 2π√(4/100)

T = 2π x √(0.04)

T = 2π x 0.2

T = 1.26 secs.

Therefore, the period of oscillation of the 4 kg object suspended on the spring is 1.26 secs.

6 0

3 years ago

A cannon is fired from a castle wall at some unknown height above the ground. The cannonball leaves the cannon with speed 30.0m/

Sauron [17]

Answer:

Part a)

$t = 3.85 s$

Part b)

$h = 72.67 m$

Part C)

$v_x = 25.98 m/s$

$v_y = 0$

Part d)

In horizontal direction velocity will remain constant

$v_x = 30 cos30 = 25.98 m/s$

in vertical direction we have

$v_y = -22.77 m/s$

Explanation:

Part a)

Horizontal speed of the cannon

$v = 30.0 m/s$

angle of projection

$\theta = 30^o$

now we have

horizontal speed = $v_x = vcos30 = 30 cos30 =25.98 m/s$

vertical speed = $v_y = vsin30 = 30 sin30 = 15 m/s$

now the time taken by it to cover the distance 100 m from the wall

$x = v_x t$

$100 = 25.98 t$

$t = 3.85 s$

Part b)

Since it hits the ground in the same time

so the height of the castle is given as

$h = \frac{1}{2}gt^2$

$h = \frac{1}{2}(9.81)(3.85^2)$

$h = 72.67 m$

Part C)

At highest point of the projection

the vertical component of the velocity will become zero

so we will have

$v_x = 25.98 m/s$

$v_y = 0$

Part d)

In horizontal direction velocity will remain constant

so we have

$v_x = 30 cos30 = 25.98 m/s$

in vertical direction we have

$v_y = v_i + at$

$v_y = 15 - 9.81(3.85)$

$v_y = -22.77 m/s$

Part e)

8 0

3 years ago

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Electromagnetic waves are sometimes demonstrated by rippling waves in water. However, electromagnetic waves are different from w

algol [13]

The answer is D: all electromagnetic waves can travel in a vacuum

3 0

3 years ago

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Which of the following steps to solve a problem is completed last?

Vesna [10]

Answer:

that would be analyzing data I believe

7 0

3 years ago

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A thin-walled vessel of volume V contains N particles which slowly leak out of a small hole of area A. No particles enter the vo

LuckyWell [14K]

Answer:

$\frac{V}{2av}$

Explanation:

From the question we are told that

Volume V

Contains N particles

Leaks from a small hole of area A

Generally the equation for Flow rate is given as

Volume Flow Rate $V_r = A * v$

Mathematically we find the time taken to flow half way which is given by

$\frac{(V/2)}{A*v}$

Therefore the time taken is

$\frac{V}{2av}$

7 0

3 years ago