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3 years ago

An object completes 50 cycles in 150 s.

Physics

1 answer:

Lisa [10]3 years ago

3 0

Answer:

a. 3 seconds

b. 0.33 Hz

Explanation:

a. Period is defined by how long, in seconds, it takes for an object to complete one revolution, or cycle. If the object completes 50 cycles in 150 seconds, then it must complete one cycle every 150/50 = 3 seconds. Therefore, the period of the object is $\boxed{3\text{ seconds}}$ .

b. Frequency is given by $f=\frac{1}{T}$ where $T$ is the period, in seconds. Since we just found that the period is 3 seconds, plug in $T=3$ :

$f=\frac{1}{3}\approx \boxed{0.33\text{ Hz}}$

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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

A sphere is charged with electrons to

pishuonlain [190]

A sphere is charged with electrons to −9 × 10−6 C. The value given is the total charge of all the electrons present in the sphere. To calculate the number of electrons in the sphere, we divide the the total charge with the charge of one electron.

N = 9 × 10−6 C / 1.6 × 10−19 C
N = 5.6 x 10^13

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3 years ago

a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature

igomit [66]

Answer:

Explanation:

Given

Volume of fixed chamber $V=1 m^3$