Ask question

Ask question

All categories

Amiraneli [1.4K]

3 years ago

7

A major-league pitcher throws a baseball toward home plate. The ball rotates at 1560 rpm, and it travels the 18.5 meters to the

plate at an average translational speed of 40.2 m/s. How many revolutions does the ball make during this trip?

1 answer:

DedPeter [7]3 years ago

4 0

Answer:

no of revolutions = 56.62 rev

Explanation:

given data

rotates = 1560 rpm

travels distance = 18.5 meters

speed = 40.2 m/s

solution

first we convert here rev/min to rev/sec

rotates = 1560 rpm = 1560 × 0.0167 = 26.052 rev/sec

we get here no of revolutions that is

no of revolutions = spin rotate ÷ time ..........1

here time = distance ÷ speed

time = 18.5 ÷ 40.2

time = 0.4601 s

put value in equation 1 we get

no of revolutions = $\frac{26.052}{0.4601}$

no of revolutions = 56.62 rev

You might be interested in

What is the relationship between the center of gravity and the support base for an object that is in stable equilibrium?

Norma-Jean [14]

Answer:

when the center of gravity is within the washing area, the torque returns in the body to its initial position and is in a stable equilibrium

Explanation:

The concept of center of gravity is equivalent to the concept of center of mass, in this place all external forces applied can be considered.

When we analyze the balance of a body that is the torque it is the one that defines the balance

τ = F xd

If the torque tends to restore the body to the initial position the balance is stable, but if the torque has to increase the body's rotation the balance is unstable

. When the body tends to rotate the torque with respect to the pivot point at the base it decreases because the distance from the center of gravity to the end of the base decreases in value, but it has to return it to the initial position, the balance is stable. The critical point of this process is when the center of gravity is at the limit of the body base area in this case the torque is zero; If the body rotates a little more the center of gravity is outside the base, the torque changes sign and has to increase the turn, going to an unstable balance.

In summary, when the center of gravity is within the washing area, the torque returns in the body to its initial position and is in a stable equilibrium.

8 0

3 years ago

Which numbers on the ph scale indicate an acid

KiRa [710]

Answer:

0 - 6.9 --> Acidic

Explanation:

4 0

3 years ago

Gennadij [26K]

Answer:

3.675 m

Explanation:

$a_{x} =0$ $v_{xo}=100$ $a_{y} =-g$ $v_{yo}=0$

X-direction | Y-direction

$R=x_{o}+ v_{xo} t$ | $y=y_{o}+v_{yo}t+\frac{1}{2}a_{y}t^2$

$75=100t$ | $y=0+0+\frac{1}{2} (9.8)(0.75)$

$\frac{75}{100} =t$ | $y=3.675 m$

$0.75s=t$

Hope it helps

3 0

3 years ago

A student stretches a spring, attaches a 1.20 kg mass to it, and releases the mass from rest on a frictionless surface. The resu

Sphinxa [80]

Answer:

the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s

Explanation:

Given that;

mass of the object m = 1.20 kg

period of oscillation = 0.750 s

Amplitude ( A/x) = 15.0 cm = 0.15 m

now;

a) Determine the oscillation frequency;

oscillation frequency f = 1/T

we substitute

f = 1 / 0.750 s

f = 1.33 Hz

Therefore, the oscillation frequency is 1.33 Hz

b) Determine the spring constant;

we solve for spring constant from the following expression;

T = 2π√(m/k)

k = 4π²m / T²

so we substitute

k = (4π² × 1.20) / (0.750)²

k = 47.3741 / 0.5625

k = 84.22 N/m

Therefore, the spring constant is 84.22 N/m

c) determine the speed of the mass when it is halfway to the equilibrium position

So, at equilibrium, the energy is equal to K.E

such that;

1/2mv² = 1/2kx²

mv² = kx²

v² = kx² / m

v = √( kx²/m)

we substitute

v = √( 84.22×(0.15 m)²/ 1.2 )

v = √( 1.89495 / 1.2 )

v = √ 1.579125

v = 1.26 m/s

Therefore, the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s

3 0

3 years ago

Pun zapato de golf tiene 10 tacos cada uno con un área de 0.01 pulgadas en contacto con el piso suponga que caminar hay un insta

Mars2501 [29]

Answer:

Las presión ejercida por los tacos sobre el suelo es 900 libras por pulgada cuadrada.

Explanation:

Según la Física, comprendemos que la presión es igual a la fuerza dividida por área. En este caso, el peso de la persona dividida por área total de los tacos de los zapatos de golf suponiendo una distribución uniforme de la fuerza. Es decir:

$\sigma = \frac{W}{n\cdot A_{T}}$ (Eq. 1)

Donde:

$\sigma$ - Presión, medida en libras por pulgada cuadrada.

$W$ - Peso de la persona, medida en libras.

$n$ - Cantidad de tacos en los zapatos, adimensional.

$A_{T}$ - Área del taco, medida en pulgadas cuadradas.

Si conocemos que $W = 180\,lb$ , $n = 20$ y $A_{T} = 0.01\,in^{2}$ , la presión es:

$\sigma = \frac{180\,lb}{(20)\cdot (0.01\,in^{2})}$

$\sigma = 900\,psi$

Las presión ejercida por los tacos sobre el suelo es 900 libras por pulgada cuadrada.

3 0

3 years ago