All categories

2 years ago

Function do lappet which faced vultures serve in a temperate grassland

Physics

2 answers:

vfiekz [6]2 years ago

7 0

Um this doesn't make since to me since you did not clearly state your awnser

Blababa [14]2 years ago

3 0

Answer:

free

Explanation:

You might be interested in

A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is require

g100num [7]

Answer:

a. 145 N/m b. 1.29 Hz c. 1.62 m/s d. 0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²

Explanation:

a. The force constant of the spring

The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m

k = 29.0 N/0.200 m = 145 N/m

b. The frequency of oscillations, f

f = 1/2π√(k/m) m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at 0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m = 145 × 0.2/(2.2 × 3)= 4.39 m/s²

8 0

3 years ago

A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?

dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0

2 years ago

A block slides down a rough ramp with a 30-degree incline as shown.

In-s [12.5K]

Answer:

Image 4 ?

Explanation:

5 0

3 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

3 years ago

Sultan throws a ball horizontally from his window, 12 m above the garden. It reaches the ground after

Kay [80]

Answer:

Explanation:

<h2><em>Hope it help mark as Brainlist</em></h2>

8 0

3 years ago