Ask question

Ask question

All categories

4 years ago

5

A baseball is hit with a bat. The direction of the ball is completely reversed and its speed is doubled. If the actual contact w

ith the bat lasts what is the ratio of the acceleration to the original velocity?

2 answers:

alexdok [17]4 years ago

8 0

Before the impact, let the velocity of the baseball was v m/s.

After being hit by the bat its velocity is -2v
So, change in velocity, Deltav=v-(-2v)=3v
Acceleration is defined as the rate of change in velocity, i.e. actual change in velocity divided by the time taken to change it. Time taken to change velocity is the time of actual contact of the bat and ball, i.e. 0.31 s.

a=(Deltav)/(Deltat)
=(3v)/0.37
Therefore, a/v=3/0.31=9.7 s^-1
So, the ratio of acceleration of the baseball to its original velocity is 9.7.

Lubov Fominskaja [6]4 years ago

6 0

Answer:

ratio = 9.67

Explanation:

given,

initial velocity = v

final velocity = - 2 v

change in velocity = v-(-2v)

= 3 v

assuming the time be t = 0.31 s

acceleration = $\dfrac{change\ in\ velocity}{time}$

= $\dfrac{3v}{0.31}$

= 9.67 v

ratio = $\dfrac{a}{v} = \dfrac{9.67 v}{v}$

ratio = 9.67

You might be interested in

What is one of Kepler's laws of planetary motion?

iragen [17]

Answer:

a. Planets move on elliptical orbits with the Sun at one focus.

Explanation:

Johannes Kepler was an astronomer who discovered that planets had elliptical orbits in the early 1600s (between 1609 and 1619).

The three (3) laws published by Kepler include;

I. The first law of planetary motion by Kepler states that, all the planets move in elliptical orbits around the Sun at a focus.

II. According to Kepler's second law of planetary motion, the speed of a planet is greatest when it is closest to the Sun.

Thus, the nearer (closer) a planet is to the Sun, the stronger would be the gravitational pull of the sun on the planet and consequently, the faster is the speed of the planet in terms motion.

III. The square of any planetary body's orbital period (P) is directly proportional to the cube of its orbit's semi-major axis.

Hence, one of Kepler's laws of planetary motion states that planets move on elliptical orbits with the Sun at one focus. This is his first law of planetary motion.

4 0

3 years ago

In an experiment, students measure the position x of a cart as a function of time t for a cart that starts at rest and moves wit

Minchanka [31]

Given :

Initial velocity , u = 0 m/s² .

To Find :

The acceleration of the cart.

Solution :

Since, acceleration is constant.

Using equation of motion :

$x = ut + \dfrac{at^2}{2}\\\\x = \dfrac{at^2}{2}$

Putting, t = 1 s and x = 4 m in above equation, we get :

$4 = \dfrac{a(1)^2}{2}\\\\a = 8 \ m/s^2$

Therefore, the acceleration of the cart is 8 m/s².

5 0

3 years ago

What happens when magma rises towards the surface?

zzz [600]

It slowly cools an hardens,eventually turning into igneous rock<span />

5 0

3 years ago

Read 2 more answers

One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at

Zolol [24]

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature $(T_i)=300 K$

$P_i=10 atm$

$P_f=2 atm$

Work done in iso-thermal process $=P_iV_iln\frac{P_i}{P_f}$

$P_i$ =initial pressure

$P_f$ =Final Pressure

$W=10\times 2.463\times ln\frac{10}{2}=39.64 J$

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is $=P\Delta V$

$V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L$

$V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L$

$\Delta W=1\times (12.315-2.463)=9.852 J$

$\Delta q=\Delta W=9.852 J$

$\Delta U=0$

8 0

4 years ago

An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10

STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

$P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}$

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi $(\frac{6894.76\ Pa}{1\ psi})$ = 379212 Pa

$\rho$ = density of water = 1000 kg/m³

Therefore,

$v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}$

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

$\frac{V}{t} = Av\\\\t =\frac{V}{Av}$

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = $\frac{\pi (0.015875\ m)^2}{4}$ = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = $[\frac{\pi (9.144\ m)^2}{4}][1.524\ m]$ = 100.1 m³

Therefore,

$t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\$

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

7 0

3 years ago