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3 years ago

an eagle carries a fish up 50 m into the sky using 90 n of force. how much work did the eagle do on the fish

Physics

2 answers:

Colt1911 [192]3 years ago

6 0

Answer:

The answer is 4,500

Explanation:

W = FD

W = 90 x 50

W = 4,500

Mashutka [201]3 years ago

4 0

The answer would be 1.8 acceleration

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A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightl

USPshnik [31]

Given Information:

Wavelength = λ = 39.1 cm = 0.391 m

speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

tension = T = 160 N

Required Information:

Length of the vibrating string = L = ?

Answer:

Length of the vibrating string = 0.28 m

Explanation:

The frequency of beautiful note is

f = v/λ

f = 344/0.391

f = 879.79 Hz

As we know, the speed of the wave is

v = √T/μ

v = √160/0.00066

v = 492.36 m/s

The wavelength of the string is

λ = v/f

λ = 492.36/879.79

λ = 0.5596 m

and finally the length of the vibrating string is

λ = 2L

L = λ/2

L = 0.5596/2

L = 0.28 m

Therefore, the vibrating section of the violin string is 0.28 m long.

3 0

3 years ago

8. An object accelerates 12.0 m/s2 when a force of 6.0 newtons is applied to it. What is the mass of the object? _______________

Sholpan [36]

Answer:

Mass of object is 0.5kg

Explanation:

Given the following data;

Force = 6N

Acceleration = 12m/s²

Mass =?

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

$F = ma$

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

Making mass (m) the subject, we have;

$Mass (m) = \frac{F}{a}$

Substituting into the equation;

$Mass (m) = \frac{6}{12}$

Mass, m = 0.5kg.

Therefore, the mass of the object is 0.5kg

5 0

3 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

3 years ago

Which set of terms represent mechanical waves?

Helga [31]

Energy that transfers through the medium

4 0

3 years ago

A block oscillating on a spring has period T = 2.8 s . (Note: You do not know values for either m or k. Do not assume any partic

olasank [31]

Answer:

Part a)

T = 3.96 s

Part b)

T = 1.98 s

Part c)

T = 2.8 s

Explanation:

As we know that time period of spring block system is given as

$T = 2\pi\sqrt{\frac{m}{k}}$

T = 2.8 s

Part a)

If the mass of the block attached is doubled

then we will have

$T' = 2\pi\sqrt{\frac{2m}{k}}$

$T' = \sqrt2 T$

$T' = 3.96 s$

Part b)

If the spring constant is doubled

then we have

$T' = 2\pi\sqrt{\frac{m}{2k}}$

$T' = \frac{T}{\sqrt2}$

$T' = 1.98 s$

Part c)

If the amplitude is halved but mass and spring constant will remain the same

so here we know that time period does not depends on Amplitude

so we will have

$T = 2\pi\sqrt{\frac{m}{k}}$

T = 2.8 s

7 0

3 years ago