(e)

A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre

Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

$h=\sqrt{(36m)^2+(227m)^2}=229.83m$ (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

$y=y_o+v_ot+\frac{1}{2}gt^2$ (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

$3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s$

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

$v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}$

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

$\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°$

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0

3 years ago

In three sentences, please describe SIMPLE HARMONIC motion, and give two examples. Thank you! :-)

11Alexandr11 [23.1K]

A system that repeats to and from its mean or rest point. that executes harmonic motion. a few examples I've heard of are since the springtime a mass-spring system,a swing, simple pendulum, one more example is a steel ball rolling in a curved is this what you need or do you need three more sentences dish. to get S.H.M a body just displaced away from the resting position and of course then is released. the human body oscillates due to the reinforce that pulls it back do you need anything else answered on this and I'll answer it

3 0

3 years ago

Solenoid 2 has twice the radius and six times the number of turns per unit length as solenoid 1. The ratio of the magnetic field

Xelga [282]

Answer:

6

Explanation:

The magnetic field inside a solenoid is given by the following formula:

$B = \mu_{0}nI$

where,

B = Magnetic Field Inside Solenoid

μ₀ = permittivity of free space

n = No. of turns per unit length

I = Current Passing through Solenoid

For Solenoid 1:

$B_{1} = \mu_{0}n_{1}I ------------------- equation 1$

For Solenoid 2:

n₂ = 6n₁

Therefore,

$B_{1} = \mu_{0}n_{2}I\\B_{1} = 6\mu_{0}n_{1}I ----------------- equation 2$

Diving equation 1 and equation 2:

$\frac{B_{2}}{B_{1}} = \frac{6\mu_{0}nI}{\mu_{0}nI}\\\\\frac{B_{2}}{B_{1}} = 6$

Hence, the correct option is:

8 0

2 years ago

Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus

charle [14.2K]

Answer:

The total energy of the composite system is 7.8 J.

Explanation:

Given that,

Height = 0.15 m

Radius of circular arc = 0.27 m

Suppose, the entire track is friction less. a bullet with a m₁ = 30 g mass is fired horizontally into a block of wood with m₂ = 5.29 kg mass. the acceleration of gravity is 9.8 m/s.

Calculate the total energy of the composite system at any time after the collision.

We need to calculate the total energy of the composite system

Total energy of the system at any time = Potential energy of the system at the stopping point

$E=mgh+Mgh$