Ask question

Ask question

All categories

2 years ago

15

a 0.145 kg baseball pitched at 39.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 52.0 m/s. if the c

ontqct time bewteen bat and ball is 3.00x10^-3 calculate the average force between the bat and ball during contact

1 answer:

velikii [3]2 years ago

3 0

consider the velocity towards the pitcher as positive

m = mass of the baseball = 0.145 kg

v₀ = initial velocity of the baseball = - 39 m/s

v = final velocity of the baseball = 52 m/s

t = time of contact = 3 x 10⁻³ sec

F = average force between bat and ball

Using impulse-change in momentum equation

F t = m (v - v₀ )

F (3 x 10⁻³) = (0.145) (52 - (- 39))

F = 4398.33 N

You might be interested in

A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a

sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0

3 years ago

Uma wants to recreate the discovery of superconductivity. Which describes the materials needed?

zaharov [31]

Answer:

B)a tool to drop temperatures, mercury, an electric current, and a tool to measure resistance

7 0

2 years ago

Read 2 more answers

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

4 0

2 years ago

What are the ranks in order on your profile in Brainly

velikii [3]

Answer:

All I know is that first you're Beginner second you're Ambitious and third you're Virtuoso

Explanation:

8 0

2 years ago

Read 2 more answers

The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 s, at which time it

Keith_Richards [23]

Answer:

The no. of revolutions does the tub turn while it is in motion is = 52.51 revolutions

Explanation:

Given data

$\omega_1$ = 0

$\omega_2$ = 5 $\frac{rev}{sec}$

Time taken = 7 sec

(1). The angular acceleration is given by

$\alpha = \frac{d \omega}{dt}$

$\alpha = \frac{5}{7}$

$\alpha = 0.714 \frac{rev}{s^{2} }$

We know that from the equation of motion

$\omega_2^{2} = \omega_1^{2} + 2 \alpha \theta_1$

$5^{2} = 0 + 2 (0.714) \theta_1$

$\theta_1 = 17.5 \ rev$ -------- (1)

(2). The angular acceleration is given by

$\alpha = \frac{d \omega}{dt}$

$\alpha = - \frac{5}{14}$

$\alpha =$ - 0.357 $\frac{rev}{s^{2} }$

We know that from the equation of motion

$\omega_2^{2} = \omega_1^{2} + 2 \alpha \theta_2$

$0^{2} = 5^{2} + 2 (-0.357) \theta_2$

$\theta_2$ = 35.01 rev ------- (2)

Total no of revolution made by the machine is

$\theta = \theta_1 + \theta_2$

$\theta =$ 17.5 + 35.01

$\theta =$ 52.51 rev

Therefore the no. of revolutions does the tub turn while it is in motion is = 52.51 rev

8 0

3 years ago