TheStability of atomic nuclei seems to be related to the ratio of what

The mass of a regulation tennis ball is 57 g. The ball is in contact with the tennis racket for 30 ms. The ball was served at 73

yaroslaw [1]

Answer:

Correct answer: F₂ = 104.5 N

Explanation:

Given:

m = 57 g = 57 · 10⁻³ kg

Δt = 30 ms = 30 · 10 ⁻³ seconds

V₁ = 73.14 m/s service speed

V₂ = 55 m/s returned speed

M = m · V Momentum or Impulse

You forgot to indicate what time the ball contact when returning.

We will assume that the time is the same Δt = 30 ms = 30 10 ⁻³ seconds.

The formula for calculating force is according to Newton's second law is:

F = ΔM / Δt = m · ΔV / Δt

Force during service is:

F₁ = 57 · 10⁻³ · 73.14 / 30 · 10 ⁻³ = 138.97 N

F₁ = 138.97 N

Returned force:

F₂ = 57 · 10⁻³ · 55 / 30 · 10 ⁻³ = 104.5 N

F₂ = 104.5 N

God is with you!!!

8 0

3 years ago

An L-R-C series circuit has L = 0.450 H, C=2.50×10^−5F, and resistance R.

Alex777 [14]

Answer:

298rad/s and 116.96 ohms

Explanation:

Given an L-R-C series circuit where

L = 0.450 H,

C=2.50×10^−5F, and resistance R= 0

In this situation we have a simple LC circuit with angular frequency

Wo = 1√LC

= 1/√(0.450)(2.50×10^-5)

= 1/√0.00001125

= 1/0.003354

= 298rad/s

B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.

Wi/W° = (100-10)/100

Wi/W° = 90/100

Wi/W° = 0.90 ............... 1

Angular frequency of oscillation

The complete aspect of the solution is attached, please check.

6 0

3 years ago

A bus accelerates to 60 m/s to the east in 10 s. What is the buses acceleration? 2. A car traveling at 10.0 m/s to the west acce

professor190 [17]

Answers:

1) $a=6\frac{m}{s^{2}}$

2) $t=8s$

Explanation:

1) Acceleration $a$ is defined as the variation of Velocity $V$ in time $t$ :

$a=\frac{V}{t}$ (1)

A body also has acceleration when it changes its direction.

In this case we have a bus with a velocity of 60m/s to the east, that accelerates in a time 10s. So, we have to find the bus's acceleration:

$a=\frac{60m/s}{10s}$ (2)

$a=6m/s^{2}$ (3) This is the bus's accelerration

2) Now we have a car that accelerates $2m/s^{2}$ to the west in order to reach a speed of $16m/s$ in the same direction, and we have to find the time $t$ it takes to the car to reach that velocity.

Therefore we have to find $t$ from (1):

$t=\frac{V}{a}$ (4)

$t=\frac{16m/s}{2m/s^{2}}$ (5)

Finally:

$t=8s$ (6)

3 0

3 years ago

A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of

SVEN [57.7K]

Answer:

Acceleration due to gravity is 20 $m/sec^2$

So option (E) will be correct answer

Explanation:

We have given length of the pendulum l = 2 m

Time period of the pendulum T = 2 sec

We have to find acceleration due to gravity g

We know that time period of pendulum is given by

$T=2\pi \sqrt{\frac{l}{g}}$

$2=2\times 3.14 \sqrt{\frac{2}{g}}$

$0.3184= \sqrt{\frac{2}{g}}$

Squaring both side

$0.1014= {\frac{2}{g}}$

$g=19.71=20m/sec^2$

So acceleration due to gravity is 20 $m/sec^2$

So option (E) will be correct answer.

8 0

3 years ago

Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reac

Charra [1.4K]

Answer:

Time of flight A is greatest

Explanation:

Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.

So

H = u₁² sin²θ₁ /2g

H = u₂² sin²θ₂ /2g

H = u₃² sin²θ₃ /2g

On the basis of these equation we can write

u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃

For maximum range we can write

D = u₁² sin2θ₁ /g

1.5 D = u₂² sin2θ₂ / g

2 D =u₃² sin2θ₃ / g

1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁

1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )

u₂ cosθ₂ >u₁ cosθ₁

u₂ sinθ₂ < u₁ sinθ₁

2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g

Time of flight B < Time of flight A

Similarly we can prove

Time of flight C < Time of flight B

Hence Time of flight A is greatest .

8 0

3 years ago