PHYSICS COLLEGE

In college softball, the distance from the pitcher's mound to the batter is 43 feet. If the ball leaves the bat at 110 mph , how much time elapses between the hit and the ball reaching the pitcher?

Answers

Answer 1

Answer:

The time taken should be 0.000074 hours or 0.2664 seconds.

Calculation of the time taken:

Here we assume the time be t

And, The distance from the pitcher's mound to the batter is 43 feet, d = 43 feet = 0.00814 miles

So, the following formula should be used.

$= 0.00814 / 110$

= 0.000074 hours or 0.2664 seconds.

Learn more about the time here: brainly.com/question/15094745

Answer 2

Answer:

Explanation:

It is given that,

The distance from the pitcher's mound to the batter is 43 feet, d = 43 feet = 0.00814 miles

Speed with which ball leaves the ball, v = 110 mph

Let t is the time elapses between the hit and the ball reaching the pitcher. It is given by :

$t=(d)/(v)$

$t=(0.00814)/(110)$

t = 0.000074 hours

t = 0.2664 seconds

So, the time between the hit and the ball reaching the pitcher is 0.2664 seconds. Hence, this is the required solution.

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List Five examples from daily life in which you see periodic motion caused by a pendulum(Marking Brainliest)

Answers

Answer:

by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave.

Explanation:

If I am given a total capacitence of two capacitors, their capacitence togather is 22 F. What capacitence would the individual capacitors have if they are connected in parallel or connected in series.

Answers

Answer:

In parallel combination, the capacity of each capacitor is 11 F.

In series combination, the capacity of each capacitor is 44 F.

Explanation:

Let there are two capacitors each of capacitance C.

When they are connected in parallel:

In parallel combination, the effective capacitance is Cp.

Cp = C1 + C2 = C + C

22 = 2 C

C = 11 F

When they are connected in series:

In parallel combination, the effective capacitance is Cs.

1 / Cs = 1 / C1 + 1 / C2 = 1 / C + 1 / C = 2 / C

1 / 22 = 2 / C

C = 44 F

A rocket exhausts fuel with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with fuel comprising 80 per cent of the total mass. When all the fuel has been exhausted its speed is:________

Answers

The speed when all the fuel has been exhausted is 2415m/s

According to this question, the following information was given:

Exhaust velocity of fuel, V(e)= 1500 m/s
Initial speed of rocket, V₁ = 0 m/s
Final speed of rocket, V₂ = ?
Fuel weight = 80% of total weight

Using Tsiolkovsky rocket equation as follows:

∆V = V(e) ln(m1/m2)

m1 = initial mass
m2 = final mass without repellant

m2 = m1 - 80%m1

m2 = m1 - 0.8m1

m2 = 0.2m1

∆V = V2 - V1

Hence;

V2 - 0 = 1500 × ln (m1/0.2m1)

V2 = 1500 ln(1/0.2)

V2 = 1500 × 1.609

V2 = 2415m/s.

Therefore, the speed when all the fuel has been exhausted is 2415m/s.

Learn more at: brainly.com/question/19531823?referrer=searchResults

Answer:

$v_2 =2414\ m/s$

Explanation:

given,

exhaust velocity of fuel(v_e) = 1500 m/s

initial speed of rocket,v₁ = 0 m/s

final speed of rocket, v₂ = ?

fuel weigh = 80 % of total weight

using Tsiolkovsky rocket equation

$\Delta v = v_e ln((m_1)/(m_2))$

Δ v = v₂ - v₁

v_e is the exhaust speed

m₁ is the initial total mass.

m₂ is the is the final total mass without propellant.

m₂ = m₁ - 0.8 m₁

m₂ = 0.2 m₁

$v_2-v_1 = 1500* ln((m_1)/(0.2 m_1))$

$v_2 = 1500* ln((m_1)/(0.2 m_1))$

$v_2 =2414\ m/s$

When all the fuel is exhausted speed of the fuel is equal to $v_2 =2414\ m/s$

A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change the period to 2.07 s?

Answers

Answer:

The add mass = 5.465 kg

Explanation:

Note: Since the spring is the same, the length and Tension are constant.

f ∝ √(1/m)........................ Equation 1 (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)

Therefore,

T ∝ √(m)

Therefore,

T₁/√m₁ = k

where k = Constant of proportionality.

T₁/√m₁ = T₂/√m₂ ........................ Equation 3

making m₂ the subject of the equation

m₂ = T₂²(m₁)/T₁²........................... Equation 4

Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.

Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.

Substituting into equation 4

m₂ = (2.07)²(0.5)/(1.18)²

m₂ = 4.285(1.392)

m₂ = 5.965 kg.

Added mass = m₂ - m₁

Added mass = 5.965 - 0.5

Added mass = 5.465 kg.

Thus the add mass = 5.465 kg

"At time t = 0 a 2330-kg rocket in outer space fires an engine that exerts" an increasing force on it in the +x-direction. This force obeys the equation Fx=At2, where t is time, and has a magnitude of 781.25 N when t = 1.27 s .What impulse does the engine exert on the rocket during the 1.50- s interval starting 2.00 s after the engine is fired?

Answers

Answer:

$Imp = 5626.488\,(kg\cdot m)/(s)$

Explanation:

First, it is required to model the function that models the increasing force in the +x direction:

$a =(781..25\,N)/((1.27\,s)^(2))$

$a = 484 (N)/(s^(2))$

The equation is:

$F_(x) = 484\,(N)/(s^(2))\cdot t^(2)$

The impulse done by the engine is given by the following integral:

$Imp=484\,(N)/(s^(2)) \int\limits^(3.50\,s)_(2\,s) {t^(2)} \, dt$

$Imp = 161.333\,(N)/(s^(2))\cdot [(3.50\,s)^(3)-(2\,s)^(3)]$

$Imp = 5626.488\,(kg\cdot m)/(s)$

A car is initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at a constant acceleration.(a) how large was the acceleration in m/s ^2
(b)how large was the acceleration, in units go g= 9.80 m/s ^2

Answers

acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
= 37.5 m/s
velocity initial = 35 km/hr x 1hr /3600 s x 1000 m/1 km
= 9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units = (2.646m/s^2)/(9.8m/s^2)
= 0.27 units

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