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gtnhenbr [62]
3 years ago
5

A gull is flying horizontally 10.80 m above the ground at 6.00 m/s. The bird is carrying a clam in its beak and plans to crack t

he clamshell by dropping it on some rocks below. (Ignore air resistance.) With what speed relative to the gull does the clam smash into the rocks?
Physics
1 answer:
UkoKoshka [18]3 years ago
8 0

Answer:

v_y = 14.55 m/s

Explanation:

given,

height at which gull is flying = 10.80 m

speed of the gull = 6 m/s

acceleration due to gravity = 9.8 m/s²

Relative to the seagull, the x-speed is 0,

because the seagull has the same x-speed.

Only the y-speed counts:

v_y^2 = u^2 + 2 g h

v_y^2 = 0^2 + 2 g h

v_y = \sqrt{2gh}

v_y = \sqrt(2\times 9.8 \times 10.8)  

v_y = \sqrt(211.68)

v_y = 14.55 m/s

hence, the speed at which the clam smash the rock is v_y = 14.55 m/s

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Answer:

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Explanation:

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Therefore, the correct option is:

<u>D. The tea loses heat to the spoon causing the spoon to become warmer</u>

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Read 2 more answers
She sights two sailboats going due east from the tower. The angles of depression to the two boats are 42o and 29o. If the observ
Reika [66]

Answer:

The boats are  934.65 feet apart

Explanation:

Given:

The angles of depression to the two boats are 42 degrees and 29 degrees

Height of the observation deck i =  1,353 feet

To Find:

How far apart are the boats (y )= ?

Solution:

<em><u>Step 1 : Finding the value of x(Refer the figure attached)</u></em>

We can use the tangent ratio to find the x value

tan(42^{\circ}) = \frac{1353}{x}

x = \frac{1353}{tan(42^{\circ}) }

x = 590.47 feet

<em><u>Step 2 : Finding the value of  z (Refer the figure attached)</u></em>

tan(29^{\circ}) = \frac{1353}{z }

z  = \frac{1353}{tan(29^{\circ})}

z = 1525.12  feet

<em><u>Step 3 : Finding the value of  y (Refer the figure attached</u></em>)

y =  z -x

y = 1525.12 - 590.47

y = 934.65 feet

Thus the two boats are 934.65 feet apart

3 0
4 years ago
As a train enters the station it slows down from 40 m/s to a 10 m/s in 5 seconds
lora16 [44]

Answer:

See the answers below.

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}=v_{o}-a*t

where:

Vf = final velocity = 10 [m/s]

Vo = initial velocity = 40 [m/s]

t = time = 5 [s]

a = acceleration [m/s²]

Now replacing:

10=40-a*5\\40-10=a*5\\30=5*a\\a=6[m/s^{2}]

Note: The negative sign in the above equation means that the velecity is decreasing.

2)

To solve this second part we must use the following equation of kinematics.

v_{f}^{2} =v_{o}^{2} -2*a*x\\

where:

x = distance [m]

(10)^{2} =(40)^{2} -2*6*x\\100=1600-12*x\\12*x=1600-100\\12*x=1500\\x=125[m]

7 0
3 years ago
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