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3 years ago

How far can i hit a golf ball?

2 answers:

3 0

It depends on the trajectory path of the golf ball. You must consider air friction, surface friction, and if there going to be any impediments in front of the ball to prevent it from further movement. Also, your potential energy that is turned into kinetic energy and the force of excretion will count in this process.<span />

neonofarm [45]3 years ago

3 0

Only you can answer this because ur the one hitting the golfball

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How much physical activity do children and trens need?

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When is a white dwarf formed?

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3 years ago

Read 2 more answers

A circuit consists of a 6 ohm resistor, a 0.2 farad capacitor, and an AC voltage source supplying V(t) = 120 cos(20 t) volts. Wr

Zinaida [17]

Answer:

q = 24 cos (20t) (1- e (-t / 1.2))

Explanation:

To work this circuit we use the mesh equation (Kirchoff) that states that the voltage in a closed circuit is zero

Vg + Vr + Vc = 0

Where Vg is the generator voltage, Vr the resistance voltage and Vc the capacitor voltage

Vg = 120 cos 20t = V

Vr = i R

Vc = q / C

We see that in one term we have the current (i) and in another the load (q), but there is a relationship between the two

i = dq / dt

Let's replace in the initial equation

V + R dq / dt + q / C = 0

Reorder the terms

Rdq / dt + q / C - V = 0

dq / dt + q / rC - V / R = 0

dq / dt = V / r -q / RC

dq / (V / R -q / RC) = dt

we integrate

∫I dq / (V / R - 1 / RC q) = ∫ dt

We change variables

u = (V / R - 1 / RC q)

du = -1 / RC dq

∫ dq / (V / R - 1 / RC q) = -RC ∫ du / u

-RC ln u = -RC ln (V / R - 1 / RC q)

We evaluate between the limits of integration, the lower t = 0 q (0) = 0

-RC [ln (V / R - 1 / RC q) - ln (V / R)] = t

[ln (V / R - 1 / RC q) - ln (V / R)] = -t / RC

Ln (V / R - 1 / RC q) / (V / R)] = -t / RC

Ln (1 - 1 / VC q)) = (-t / RC)

Ln (VC- q) = ln (VC) (-t / RC)

VC-q = VC e (-t / RC)

q = VC - Vc e (-t / RC)

q = VC (1- e (-t / RC)

We substitute the values they give us

q = (120 cos (20t) 0.2) (1- e (-t / 6 0.2))

q = 24 cos (20t) (1- e (-t / 1.2))

7 0

4 years ago

During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a vel

Nata [24]

Answer:

133 kg

Explanation:

Parameters given:

Mass of tackler, m = 133 kg

Initial velocity of tackler, u = 3.4 m/s

Final velocity of tackler and receiver, v = 1.7 m/s

Since momentum is conserved, we apply the principle of conservation of momentum:

Total initial momentum = Total final momentum

mu + MU = (m + M)v

Where U = initial velocity of receiver = 0 m/s

M = mass of receiver

Therefore:

(133 * 3.4) + (M * 0) = (133 + M) * 1.7

452.2 = (133 + M) * 1.7

(133 + M) = 452.2/1.7

133 + M = 266

=> M = 266 - 133

M = 133 kg

The mass of the receiver is 133 kg.

4 0

4 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Juliette [100K]

(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

$U=-\frac{GMm}{R+h}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

$U_A=-\frac{GMm}{R+h_A}$

while potential energy of satellite B is

$U_B=-\frac{GMm}{R+h_B}$

Therefore the ratio of the potential energy of satellite B to that of satellite A is

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}$

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

$\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473$

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

$K=\frac{GMm}{2(R+h)}$

So the kinetic energy of satellite A is

$K_A=\frac{GMm}{2(R+h_A)}$

while kinetic energy of satellite B is

$K_B=\frac{GMm}{2(R+h_B)}$

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}$

which is identical to before, so it gives

$\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473$

(c) Satellite B

The total energy of a satellite in orbit is given by

$E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}$

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) $1.03\cdot 10^7 J$