Ask question

Ask question

All categories

3 years ago

12

A fast pitch softball player does a "windmill" pitch moving her hand through a vertical circular arc to pitch a ball at 70mph. T

he 0.19 kg ball is 55 cm from the pivot point at her shoulder.
Just before the ball leaves her hand, what is its centripetal acceleration?

1 answer:

uysha [10]3 years ago

8 0

Answer:

Explanation:

The hand throwing ball moves on a circular path having radius r = .55 m

mass of ball m = .19 kg

velocity v = 70 mph

= 70 x 1600 / 60 x 60

= 31.11 m /s

centripetal force

= mv² / r

= .19 x 31.11² / .55

= 334.34 N

You might be interested in

three condensers are connected in series across a 150 volt supply. The voltages across them are 40,50 and 60 volts respectively,

ioda

Explanation:

Given that,

The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.

(a) Capacitance of capacitor 1,

$C_1=\dfrac{Q}{V_1}\\\\C_1=\dfrac{6\times 10^{-8}}{40}\\\\C_1=1.5\times 10^{-9}\ F\\\\C_1=1.5\ nF$

Capacitance of capacitor 2,

$C_2=\dfrac{Q}{V_2}\\\\C_2=\dfrac{6\times 10^{-8}}{50}\\\\C_2=1.2\times 10^{-9}\ F\\\\C_2=1.2\ nF$

Capacitance of capacitor 3,

$C_3=\dfrac{Q}{V_3}\\\\C_3=\dfrac{6\times 10^{-8}}{60}\\\\C_3=1\times 10^{-9}\ F\\\\C_3=1\ nF$

(b) The equivalent capacitance in series combination is :

$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\\\\\dfrac{1}{C}=\dfrac{1}{1.5}+\dfrac{1}{1.2}+\dfrac{1}{1}\\\\C=0.4\ nF$

Hence, this is the required solution.

5 0

3 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago

6. A Cadillac Escalade has a mass of 2 569.6 kg, if it accelerates at 4.65m/S<br> force on the car?

SVETLANKA909090 [29]

Answer:

F= 2569.6 X 4.65 = 11,948.64

*Multiply the mass and the acceleration to find the force

Explanation:

5 0

4 years ago

Read 2 more answers

Renewable energy sources are ____________________?

Nina [5.8K]

Answer:

Eco-friendly

Explanation:

6 0

4 years ago

A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in t

Margarita [4]

Answer: 996m/s

Explanation:

Formula for calculating velocity of wave in a stretched string is

V = √T/M where;

V is the velocity of wave

T is tension

M is the mass per unit length of the wire(m/L)

Since the second wire is twice as far apart as the first, it will be L2 = 2L1

Let V1 and V2 be the speed of the shorter and longer wire respectively

V1 = √T/M1... 1

V2 = √T/M2... 2

Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1

The equations will now become

249 = √T/(m/L1) ... 3

V2 = √T/(m/2L1)... 4

From 3,

249² = TL1/m...5

From 4,

V2²= 2TL1/m... 6

Dividing equation 5 by 6 we have;

249²/V2² = TL1/m×m/2TL1

{249/V2}² = 1/2

249/V2 = (1/2)²

249/V2 = 1/4

V2 = 249×4

V2 = 996m/s

Therefore the speed of the wave on the longer wire is 996m/s

3 0

3 years ago