Ask question

Ask question

All categories

3 years ago

11

A 10 kg ball is swung in a horizontal circle at the end of a 2 - m rope over a persons head. The ball makes 30 revolutions per m

inute. How much time does it take for the ball to go around once and what is the speed of the ball?

1 answer:

steposvetlana [31]3 years ago

8 0

Answer : T = 2 s and Speed = 6.28 m/s.

Explanation :

It is given that,

Mass of the ball, m = 10 kg

Radius of the circle, r = 2 m

The ball makes 30 revolutions per minute. So, frequency is given by :

$f=\dfrac{30}{60\ s}$

$f=\dfrac{1}{2}\ Hz$

We know that time period is defined as the reciprocal of frequency i.e.

$T=\dfrac{1}{f}$

$T=2\ s$

So, the time period for the ball to go around once is 2 s.

Now, speed is defined as distance travelled per unit time. Distance travelled by ball in circular path is equal to the circumference of the circle.

$Speed =\dfrac{distance}{time}$

$S=\dfrac{2\pi r}{T}$

$S=\dfrac{2\times 3.14\times 2\ m}{2\ s}$

$S=6.28\ m/s$

Speed of the ball is 6.28 m/s.

You might be interested in

What happens to the boiling point of water when the pressure is increased?

Artemon [7]

You can drink it nothing will happen you caN BOIL ME SOME EGGS IF YOU WANT TOO

4 0

3 years ago

A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° no

Rzqust [24]

Answer:

a) R = 2.5 mi b) To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

Cos 20 = Aₓ / A

sin 20 = $A_{y}$ / A

Aₓ = A cos 160 = 2 cos 160

$A_{y}$ = A sin160 = 2 sin160

Aₓ = -1,879 mi

$A_{y}$ = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

cos 2600 = Bₓ / B

sin 260 = $B_{y}$ / B

Bₓ = B cos 260

$B_{y}$ = B sin 260

Bₓ = 4 cos 260

$B_{y}$ = 4 sin 260

Bₓ = -0.6946mi

$B_{y}$ = - 3,939 mi

Third vector C = 3 mi to 15 north east

cos 15 = Cₓ / C

sin15 = $C_{y}$ / C

Cₓ = C cos 15

$C_{y}$ = C sin15

Cₓ = 3 cos 15

$C_{y}$ = 3 sin 15

Cₓ = 2,898 mi

$C_{y}$ = 0.7765 mi

Now we can find the final position of the person

X = Aₓ + Bₓ + Cₓ

X = -1.879 -0.6949 + 2.898

X = 0.3241 mi

Y = $A_{y}$ + $B_{y}$ + $C_{y}$

Y = 0.684 - 3.939 +0.7765

Y = -2.4785 mi

a) We use Pythagoras' theorem

R = √ (x2 + y2)

R = √ (0.3241 2 + (-2.4785) 2)

R = 2.4996 mi

R = 2.5 mi

b) let's use trigonometry

Tan θ = y / x

Tanθ = -2.4785 / 0.3241

θ = tan⁻¹ (-7,647)

θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

3 0

3 years ago

Calculate the force of attraction between

vagabundo [1.1K]

Answer:

So the force of attraction between the two objects is 3.3365*10^-6

Explanation:

m1=10kg

m2=50kg

d=10cm=0.1m

G=6.673*10^-11Nm^2kg^2

We have to find the force of attraction between them

F=Gm1m2/d^2

F=6.673*10^-11*10*50/0.1^2

F=3.3365*10^-8/0.01

F=3.3365*10^-6

4 0

3 years ago

Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1

lesantik [10]

Answer:

a) $\lambda=6.63\times10^{-31}m$

b) $\lambda=6.63\times10^{-39}m$

c) $\lambda=9.97\times10^{-11}m$

d) $\lambda=4.03\times10^{-36}$ m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

$\lambda=\frac{h}{P}=\frac{h}{mv}$

with h the Plank's constant ( $6.63\times10^{-34}\frac{m^{2}kg}{s}$ ) and P the momentum of the object that is mass (m) times velocity (v).

a) $\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}$

$\lambda=6.63\times10^{-31}m$

b) $\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}$

$\lambda=6.63\times10^{-39}m$

c) $\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}$

$\lambda=9.97\times10^{-11}m$

d) $\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}$

$\lambda=4.03\times10^{-36}$ m

e) $\lambda=\frac{6.63\times10^{-34}}{(74*0)}$

λ=∞

6 0

3 years ago

A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a dia

Bingel [31]

Answer

given,

diameter of the pipe is = (14 ft)4.27 m

minimum speed of the skater must have at very top = ?

At the topmost point of the pipe the normal force will be equal to zero.

F = mg

centripetal force acting on the skateboard

$F = \dfrac{mv^2}{r}$

equating both the force equation

$mg = \dfrac{mv^2}{r}$

$v = \sqrt{gr}$

r = d/2 = 14/ 2 = 7 ft

or

r = 4.27/2 = 2.135 m

g = 32 ft/s² or g = 9.8 m/s²

$v = \sqrt{32 \times 7}$

v = 14.96 ft/s

or

$v = \sqrt{9.8 \times 2.135}$

v = 4.57 m/s

5 0

3 years ago