Ask question

Ask question

All categories

2 years ago

6

a professional baseball player can pitch a baseball with a velocity of 44.7m/s towards home plate. If a baseball weighs 1.4 N, h

ow much momentum does it have when it hits the catcher units

2 answers:

topjm [15]2 years ago

8 0

Answer:

p = 6.25 kg-m/s

Explanation:

It is given that,

Velocity of the baseball, v = 44.7 m/s

Weight of the baseball, W = 1.4 N

Firstly we need to find the mass of the baseball. It can be calculated as :

$m=\dfrac{W}{g}$

$m=\dfrac{1.4\ N}{9.8\ m/s^2}$

m = 0.14 kg

The momentum of an object is equal to the product of mass and velocity. It is given by :

$p=m\times v$

$p=0.14\ kg\times 44.7\ m/s$

p = 6.25 kg-m/s

So, the momentum it have when it hits the catcher is 6.25 kg-m/s. Hence, this is the required solution.

kumpel [21]2 years ago

3 0

1.4 N is a weight so calculating it's mass
1.4/9.8 = 0.1428 kg
momentum will be 0.1428*44.7 = 6.38 kgm/s

You might be interested in

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

kati45 [8]

Answer:

The coefficient of kinetic friction $\mu_k = 0.724$

Explanation:

From the question we are told that

The length of the lane is $l = 36.0 \ m$

The speed of the truck is $v = 22.6\ m/s$

Generally from the work-energy theorem we have that

$\Delta KE = N * \mu_k * l$

Here N is the normal force acting on the truck which is mathematically represented as

$\Delta KE$ is the change in kinetic energy which is mathematically represented as

$\Delta KE = \frac{1}{2} * m * v^2$

=> $\Delta KE = 0.5 * m * 22.6^2$

=> $\Delta KE = 255.38m$

$255.38m = m * 9.8 * \mu_k * 36.0$

=> $255.38 = 352.8 * \mu_k$

=> $\mu_k = 0.724$

6 0

2 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

2 years ago

Which one of the following statements is true concerning an object executing simple harmonic motion?

timurjin [86]

Answer:

D) The objects velocity is zero when its acceleration is a maximum

Explanation:

In a simple harmonic motion, the total energy is constant (if we neglect air resistance and friction), and it is equal to the sum of the elastic potential energy U and the kinetic energy K:

$E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2$ (1)

where

m is the mass

v is the velocity

k is the spring constant

x is the displacement

As a consequence, since E must remain constant, when K increases U decreases, and vice-versa.

Also, in a simple harmonic motion the acceleration of the system is proportional to the negative of the displacement:

$a\propto - x$ (2)

So, combining (1) with (2), we have the following situations:

- When the displacement is zero (x=0), the acceleration is also zero (a=0), and so the velocity is maximum, because the kinetic energy is maximum

- When the displacement is maximum (x=max), the acceleration is also maximum, while the velocity is zero because the kinetic energy is zero

So, the correct statement is

D) The objects velocity is zero when its acceleration is a maximum

4 0

2 years ago

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec

bixtya [17]

Answer:

1340.2MW

Explanation:

Hi!

To solve this problem follow the steps below!

1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid

W=αhQ

α=specific weight for water =9.81KN/m^3

h=height=220m

Q=flow=690m^3/s

W=(690)(220)(9.81)=1489158Kw=1489.16MW

2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator

Wr=(0.9)(1489.16MW)=1340.2MW

the maximum possible electric power output is 1340.2MW

3 0

2 years ago

A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

Ulleksa [173]

(a) 34 V

The average emf induced in the loop is given by Faraday-Newmann-Lenz law:

$\epsilon = -\frac{\Delta \Phi_B}{\Delta t}$ (1)

where

$\Delta \Phi_B$ is the variation of magnetic flux through the coil

$\Delta t = 2.0 ms = 0.002 s$ is the time interval

We need to find the magnetic flux before and after. The magnetic flux is given by:

$\Phi_B = BA$

where

B is the magnetic field intensity

A is the area of the coil

The radius of the coil is r = 12.0 cm = 0.12 m, so its area is

$A=\pi r^2 = \pi (0.12 m)^2 = 0.045 m^2$

At the beginning, the magnetic field is

$B_i = 1.5 T$

so the flux is

$\Phi_i = B_i A = (1.5 T)(0.045 m^2)=0.068 Wb$

while after the removal of the coil, the magnetic field is zero, so the flux is also zero:

$\Phi_f = 0$

so the variation of magnetic flux is

$\Delta \Phi = 0-0.068 Wb=-0.068 Wb$

And substituting into (1) we find the average emf in the coil

$\epsilon=-\frac{-0.068 Wb}{0.002 s}=34 V$

(b) Counterclockwise

In order to understand the direction of the induced current, we have to keep in mind the negative sign in Lenz's law (1), which tells that the direction of the induced current must be such that the magnetic field produced by this current opposes the variation of magnetic flux in the coil.

In this situation, the magnetic flux through the coil is decreasing, since the coil is removed from the field. So, the induced current must be such that it produces a magnetic field whose direction is the same as the direction of the external magnetic field, which is upward along the positive z-direction.

Looking down from above and using the right-hand rule on the loop (thumb: direction of the current, other fingers wrapped: direction of magnetic field), we see that in order to produce at the center of the coil a magnetic field which is along positive z-direction, the induced current must be counterclockwise.

4 0

2 years ago