NEED HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Gulls are often observed dropping clams and other shellfish from a height to the rocks below, as a means of opening the shells.

kumpel [21]

Answer:

v = 17.71 m / s

Explanation:

We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity

v² = v₀² - 2 g (y -y₀)

v² = 0 - 2g (y -y₀)

when it hits the stone the height is zero and part of the height of the seagull I

v² = 2g y₀

v = Ra (2g i)

let's calculate

v =√ (2 9.8 16)

v = 17.71 m / s

8 0

3 years ago

The earth orbits a start called the ____?

Aleks [24]

Answer:

the earth orbits a star called the sun

6 0

3 years ago

A trampoline has a spring constant of 3430 N/m how far will the trampoline sink when a 70 kg person steps on it

Dominik [7]

F = ma = -kx

a = 9.81 m/s²
k = 3430 N/m
m = 70 kg

x = - ma/ k = 0.2m

3 0

3 years ago

Read 2 more answers

An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a p

kotykmax [81]

Answer:

a) $v=1.77\times 10^6\ m/s$

b) $v=3.872\times 10^6\ m/s$

c) $v=5.5\times 10^6\ m/s$

d) $v=6.7\times 10^6\ m/s$

e) $v=7.7\times 10^6\ m/s$

Explanation:

Given that

d = 2 cm

V = 200 V

$u=4\times 10^5\ m/s$

We know that

F = E q

F = m a

E = V/d

So

m a = q .V/d b

$a=\dfrac{q.V}{m.d}$ ---------1

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The charge on electron

$q=1.6\times 10^{-19}\ C$

Now by putting the all values in equation 1

$a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2$

$a=1.5\times 10^{15}\ m/s^2$

We know that

$v^2=u^2+2as$

a)

s = 0.1 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}$

$v=\sqrt{3.16\times 10^{12}}\ m/s$

$v=1.77\times 10^6\ m/s$

b)

s = 0.5 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}$

$v=\sqrt{1.5\times 10^{13}}\ m/s$

$v=3.872\times 10^6\ m/s$

c)

s = 1 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1\times 10^{-2}$

$v=\sqrt{3.06\times 10^{13}}\ m/s$

$v=5.5\times 10^6\ m/s$

d)

s = 1.5 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1.5\times 10^{-2}$

$v=\sqrt{4.5\times 10^{13}}\ m/s$

$v=6.7\times 10^6\ m/s$

e)

s = 2 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 2\times 10^{-2}$

$v=\sqrt{6.06\times 10^{13}}\ m/s$

$v=7.7\times 10^6\ m/s$

5 0

3 years ago

Read 2 more answers

Efficincy of simple machine is 60% what does it mean?<br>

umka2103 [35]

Explanation:

By the statement, Efficiency of a machine is 60% it is meant that 'The work output is 60% of the work input'. If you apply 100J of energy for a machine, it will be able to life load of 60 J only

7 0

2 years ago