1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
____ [38]
3 years ago
14

Suppose a fast-pitch softball player does a windmill pitch, moving her hand through a circular arc with her arm straight. She re

leases the ball at a speed of 25.5 m/s (about 57.0 mph ). Just before the ball leaves her hand, the ball's radial acceleration is 1060 m/s2 . What is the length of her arm from the pivot point at her shoulder
Physics
1 answer:
RideAnS [48]3 years ago
4 0

Answer:

61.3 cm

Explanation:

Radial acceleration of the object in circular motion is given by formula

a = \frac{v^2}{R}\\

Given:

a = 1060 m/s^2\\v = 25.5 m/s

 

Plugging in the values in the formula

1060 = \frac{25.5^2}{R}\\R = 0.613 m

so length of his arm is 61.3 cm

You might be interested in
Two pans of a balance are 46.3 cm apart. The fulcrum of the balance has been shifted 0.633 away from the center by a dishonest s
soldier1979 [14.2K]

distance of each pan from the center or fulcrum is given as

r = 23.15 cm

now if dishonest shopkeeper shifted it by 0.633 cm from center

so distance on each side is given as

d_1 = 23.15 - 0.633 = 22.52 cm

d_2 = 23.15 + 0.633 = 23.78 cm

now the weight is balance as

W_1d_1 = W_2d_2

W(22.52) = W_2(23.78)

now we will have

W_2 = 0.95W

now we can find the percentage change as

percentage = \frac{W - W_2}{W} \times 100

percentage = 5%

5 0
4 years ago
Help. Please! I really need help. It’s timed, and I’m loosing points.
Archy [21]

Answer:

I think balanced

Explanation:

because there is a 2 on each arrow

5 0
3 years ago
A rock, with a density of 3.55 g/cm^3 and a volume of 470 cm^3, is thrown in a lake. a) What is the weight of the rock out of th
SIZIF [17.4K]

Answer:

a) Weight of the rock out of the water = 16.37 N

b) Buoyancy force = 4.61 N

c) Mass of the water displaced = 0.47 kg

d) Weight of rock under water = 11.76 N

Explanation:

a) Mass of the rock out of the water = Volume x Density

   Volume = 470 cm³

   Density = 3.55 g/cm³

   Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg

   Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N

b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.

   Volume = 470 cm³

   Density of liquid = 1 g/cm³

   \texttt{Buoyancy force}= \frac{470\times 1\times 9.81}{1000} = 4.61 N

c) Mass of the water displaced = Volume of body x Density of liquid

   Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg

d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force

   Weight of rock under water = 16.37 - 4.61  =11.76 N

3 0
3 years ago
One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i
kati45 [8]

Answer:

The force exerted on an electron is 7.2\times10^{-18}\ N

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance  = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}

E_{1}=1.0183\times10^{3}\ N/C

Using formula of electric field again

E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}

E_{2}=-1.064\times10^{3}\ N/C

We need to calculate the resultant electric field

Using formula of electric field

E=E_{1}+E_{2}

Put the value into the formula

E=1.0183\times10^{3}-1.064\times10^{3}

E=-0.045\times10^{3}\ N/C

We need to calculate the force exerted on an electron

Using formula of electric field

E = \dfrac{F}{q}

F=E\times q

Put the value into the formula

F=-0.045\times10^{3}\times(-1.6\times10^{-19})

F=7.2\times10^{-18}\ N

Hence, The force exerted on an electron is 7.2\times10^{-18}\ N

8 0
3 years ago
Flapping flight is very energy intensive. A wind tunnel test
steposvetlana [31]

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight=\frac{P}{m}

Using the formula

Metabolic power for starting flight=\frac{12}{0.089}

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

4 0
3 years ago
Other questions:
  • Which of the following is NOT an example of a network configuration?
    13·1 answer
  • The contents of the cylinder is then entirely transfered to a partially filled holding tank of volume 10,000 L originally at pre
    5·1 answer
  • A glider of mass 0.240 kg is on a frictionless, horizontal track, attached to a horizontal spring of force constant 6.00 N/m. In
    14·1 answer
  • Green light has a wavelength of 5.20 x 10^-7m. the speed of light is 3.00 * 10^8 m/s. what is the frequency of green light waves
    7·1 answer
  • What is the energy from the sun that is converted to electrical or thermal energy
    13·2 answers
  • An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte
    10·1 answer
  • In diving to a depth of 308 m, an elephant seal also moves 579 m due east of his starting point. What is the magnitude of the se
    8·1 answer
  • Ship A is located 3.90 km north and 2.50 km east of ship B. Ship A has a velocity of 21.0 km/h toward the south and ship B has a
    8·1 answer
  • A basketball is resting on the ground.
    13·1 answer
  • PLEASE HELP!!!!!!
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!