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tekilochka [14]
3 years ago
10

To Mr. H's disgust, a 450-g black crow is raiding the recently-filled bird feeder. As Mr. H runs out the back door with his broo

m in an effort to scare the crow away, it pushes off the 670-gram feeder with a takeoff speed of 1.5 m/s. Determine the speed at which the feeder initially recoils backwards.
Physics
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer:

Actually it's 2.50 m/s, sorry

Explanation:

It is solved by using momentum conservation equation

combined mass of crow and feeder = 450+670=1120 gm

let the recoil speed of feeder be v m/s

Then applying momentum conservation we get;

1120×1.5 = 670×v

v= 2.50 m/s

the speed at which the feeder initially recoils backwards = 2.50 m/s

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The answer is C but man if you have leak or a meltdown good luck to anyone downstream.

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3 years ago
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In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the
Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, u_0= 10 m/s.

Angle of launch with the horizontal, \theta = 53 ^{\circ}

So, the vertical component of the initial velocity,

u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i).

The horizontal component of the initial velocity,

u_0\cos\theta=10 \cos 53 ^{\circ}

Let, t be the time of flight, to the horizontal distance covered

D=10 \cos (53 ^{\circ})t\cdots(ii).

Not, applying the equation of motion in the vertical direction.

s= ut +\frac 1 2 at^2

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, u =10 \sin 53 ^{\circ} (from equation (i), s=0 (as the final height is same as the launch height) and a = -9.81 m/s^2 (negative sign is due to the downward direction).

\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2

\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63 seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m.

Now, for the maximum height, H, applying the equation of motion as

v^2=u^2+2as

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, 0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H

\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}

\Rightarrow H = 3.25 m.

Hence, the maximum height covered is 3.25 m.

8 0
3 years ago
A student measures the mass 8cm block of brown sugar to be 12.9g. what is the density of the brown sugar.
antoniya [11.8K]

Correction

A student measures the mass <em><u>8cm3</u></em> block of brown sugar to be 12.9g. what is the density of the brown sugar

Answer:

1.6\ g/cm^{3}

Explanation:

Density is defined as mass per unit volume of an object expressed as \rho=\frac {m}{v} where \rho is the density, m is the mass of sugar and v is the volume of the sugar. Considering that the volume is given as 8cm3 for sugar then we substitute this for v and mass of 12.9 g we substitute for g then the density will be

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8 0
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Reflection of the images example
NISA [10]

Answer:

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Water in a tank is pressurized by air and pressure measured using a multi-fluid manometer. Determine the gage pressure of air in
sattari [20]

Answer:

The gauge pressure of air is 110 kpa

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Therefore, the gauge pressure of air is 110 kpa

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4 years ago
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