A ball thrown in the air will never go as far as physics ideally would predict because?

A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in

navik [9.2K]

Answer:

(A) 1.58 Hz

(B) 0.99 m/s

(C) 0.1 kg

(D) 0.4 m

Explanation:

extension of the spring (x) = 10 cm = 0.1 m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) force = mg = kx

where m = mass

g = acceleration due to gravity

x = extension

mg = kx

substituting the values 0f g and x we have

9.8m = 0.1k

therefore k = 9.8m/0.1

k = 98m

formula for frequency (f) = $\frac{1}{2π}.\sqrt{\frac{k}{m} }$

inserting the value of spring constant (k) as 98m into the equation above

f = $\frac{1}{2π}.\sqrt{\frac{98m}{m} }$

f = $\frac{1}{2π}.\sqrt{98}$

f = 1.58 Hz

(B) find the speed of the object when it is 8 cm below its initial position

from the conservation of energy,

initial potential energy (U) + kinetic energy (K.E) = 0

( $0.5ky^{2} - mgy) + 0.5mv^{2}$

v = $\sqrt{2gy - \frac{k}{m}.y^{2}}$

where y = position of the spring = 8 cm (0.08m) and k = 98m as in (A) above

v = $\sqrt{2 x 9.8 x 0.08 - \frac{98m}{m}.0.08^{2}}$

v = $\sqrt{1.568 - (98 x 0.0064}$

v = 0.99 m/s

(C) find the mass (m) of the object when an object of mass 300 g is attached to the first object, after which the system oscillates with half the original frequency.

after addition of the 300 g mass

new frequency = half the initial frequency

$\frac{1}{2π}.\sqrt{\frac{k}{m + 300} }$ = 0.5 x $\frac{1}{2π}.\sqrt{\frac{k}{m} }$

$\sqrt{\frac{k}{m + 300} }$ = 0.5 x $\sqrt{\frac{k}{m} }$

$(\sqrt{\frac{k}{m + 300} })^{2} =(0.5 x \sqrt{\frac{k}{m} })^{2}$

$\frac{k}{m + 300} =0.25 x \frac{k}{m}$

$\frac{1}{m + 300} =0.25 x \frac{1}{m}$

0.25 (m + 300) = m

m - 0.25m = 0.25 x 300

m- 0.25m = 75

0.75 m = 75

m = 100 g = 0.1 kg

(D) find the new equilibrium position

from mg = kx we can find the new equilibrium position (x)

where m = m + 300 = m + 0.3 (in kg)

(m+0.3)g = kx

x = $\frac{(m+0.3)g}{k}$

recall that k = 98m

x = $\frac{(m+0.3)g}{98m}$

now substituting the values of m and g into the equation

x = $\frac{(0.1 + 0.3) x 9.8}{98 x 0.1}$

x = 0.4 m

6 0

3 years ago

COMPARE: How were the outcomes of Pasteur’s and Redi’s experiments similar? Both experiments proved that life can come from rott

Katen [24]

Both experiments showed that the living things only come from living things

7 0

3 years ago

Read 2 more answers

A single coil of wire, with a radius of 0.13 m is rotated in a uniform magnetic field such that the angle between the field vect

Vedmedyk [2.9K]

The magnitude of emf induced in the single coil of wire rotated in the uniform magnetic field is 0.171 V.

<h3>Induced emf</h3>

The emf induced in the loop is determined by applying Faraday's law of electromagnetic induction.

emf = N(dФ/dt)

where;

N is number of turns of the wire
Ф is magnetic flux

Ф = BA

where;

B is magnetic field strength
A is the area of the loop

emf = NBA/t

A = πr²

A = π x (0.13)²

A = 0.053 m²

emf = NBA/t

emf = (1 x 3.746 x 0.053)/(1.16)

emf = 0.171 V

Thus, the magnitude of emf induced in the single coil of wire is 0.171 V.

Learn more about induced emf here: brainly.com/question/13744192

5 0

2 years ago

Diamonds are a very dense material. Predict what would happen to the light ray if you projected it from air through a diamond

Kipish [7]

Answer:

It [the diamond] would act like a prism, and make a rainbow, or, the light would break up and disappear

Explanation:

that's what I would think at least

4 0

2 years ago

Read 2 more answers

Nicole throws a ball straight up. Chad watches the ball from a window 5.0 m above the point where Nicole released it. The ball p

WARRIOR [948]

Answer:

14.14 m/s

Explanation:

As total mechanical energy is conserved, if the potential energy is the same at Chad's position no matter if the ball is travelling up or down, then its kinetic energy, and speed, is also the same too.

Therefore, the ball would have a speed of 10m/s when it's passing Chad the 1st time. Since we know that Chad is at 5m high from the release point, we can use the following equation of motion:

$v^2 - v_0^2 = 2g\Delta s$

where v = 10 m/s is the velocity of the ball when it passes Chad, v_0 is the initial velocity of the ball when it releases, g = 10 m/s2 is the deceleration of the can, and $\Delta s = 5 m$ is the distance traveled between Nicole and Chad. We can solve for v0:

$10^2 - v_0^2 = 2*(-10)*5$

$100 - v_0^2 = -100$

$v_0^2 = 200$

$v_0 = \sqrt{200} = 14.14 m/s$

4 0

3 years ago