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Snezhnost [94]
3 years ago
5

A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th

e time the clam was dropped?
Physics
1 answer:
ankoles [38]3 years ago
4 0
<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 3 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 3 + 0.5 x 9.81 x 3²

                      s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

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A 2290 kg car traveling at 10.5 m/s collides with a 2780 kg car that is initially at rest at the stoplight. The cars stick toget
avanturin [10]

Answer:

0.41

Explanation:

given,

mass of the car, m = 2290 Kg

initial speed = 10.5 m/s

mass of another car, M = 2780 Kg

distance moved = 2.80 m

coefficient of friction = ?

conservation of energy

m u = (M + m) V

2290 x 10.5 = (2290 + 2780) V

V = 4.74 m/s

using equation of motion

v² = u² + 2 a s

4.74² = 2 x a x 2.8

a = 4.02 m/s²

now using equation

a = μ g

4.02 = μ x 9.8

μ = 0.41

7 0
2 years ago
Which of the following would be the magnitude of the vector given a horizontal component of 30 and a
Scorpion4ik [409]

Answer:

50

Explanation:

Use the Pythagorean theorem to find the length of the diagonal, or the hypotenuse of an imaginary triangle. 30^2 + 40^2 = 2500, which is 50^2. So, the magnitude is 50.

Brainliest, please :)

7 0
1 year ago
How would you change the distance between two charged particles to increase the electric force between them by a factor of 16
Naily [24]

The electrostatic force between two charges is inversely
proportional to the square of the distance between them.

So if you want to multiply the force by, say, ' Q ',
you need to multiply the distance by  ( 1 / √Q ) .

We want to multiply the force by 16, so we need to
multiply the distance by    ( 1 / √16 )  =  ( 1 / 4 ) .

The distance should be changed to  1/4  of what it is now.

4 0
3 years ago
Read 2 more answers
A BMX bicycle rider takes off from a ramp at a point 2.4 m above the ground. The ramp is angled at 40 degrees from the horizonta
adoni [48]

Answer:

The BMX lands 5.4 m from the end of the ramp.

Explanation:

Hi there!

The position of the BMX is given by the position vector "r":

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t

x0 = initial horizontal position

v0 = initial velocity

α = jumping angle

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

Please, see the attached graphic for a better understanding of the situation. At final time, when the bicycle reaches the ground, the vector position will be "r final" (see figure). The y-component of the vector "r final" is - 2.4 m (placing the origin of the frame of reference at the jumping point). With that information, we can use the equation of the y-component of the vector "r" (see above) to calculate the time of flight. With that time, we can then obtain the x-component (rx in the figure) of the vector "r final". Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

-2.4 m = 0 m + 5.9 m/s · t · sin 40° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 5.9 m/s · t · sin 40° + 2.4 m

Solving the quadratic equation:

t = 1.2 s

Now, we can calculate the x-component of the vector "r final" that is the horizontal distance traveled by the bicycle:

x = x0 + v0 · t · cos α

x = 0 m + 5.9 m/s · 1.2 s · cos 40°

x = 5.4 m

The BMX lands 5.4 m from the end of the ramp.

Have a nice day!

8 0
2 years ago
Si el periodo de oscilación de resorte es de 0,44 segundos cuando oscila atado a una masa de 2 Kg. ¿Cuál será el valor de la con
boyakko [2]

Answer:

i d k h b u lol I wish I knew it sorry

7 0
2 years ago
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