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Darya [45]
3 years ago
13

With what minimum speed must you toss a 200 g ball straight up to just touch the 15-m-high roof of the gymnasium if you release

the ball 1.6 m above the ground? solve this problem using energy.
Physics
1 answer:
victus00 [196]3 years ago
4 0
Total energy before the toss:
E_1 = \frac{1}{2} mv_1^2 + mgh_1

Total energy at the roof:
 E_2 = \frac{1}{2} mv_2^2 + mgh_2

Energy must be conserved so :
E_1 = E_2 \\ \\ \frac{1}{2} mv_1^2 + mgh_1 = \frac{1}{2} mv_2^2 + mgh_2 \\ \\ \frac{1}{2} m (v_1^2 - v_2^2) = mg(h_2 - h_1) \\ \\ \frac{1}{2} (v_1^2 - v_2^2) = g(h_2 - h_1) \\ \\ v_1^2 - v_2^2 = 2g(h_2 - h_1)

If v₁ is the initial velocity and v₂ is the final velocity, which is zero at the roof:
v_1^2 = 2g(h_2 - h_1) \\  \\ v_1 =  \sqrt{2g(h_2 - h_1)}

v_1 =  \sqrt{(2)(9.81 \frac{m}{s^2} )(15m - 1.6m)} = 16,2  \frac{m}{s}

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A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo
-BARSIC- [3]

Answer:

It would take \tau(\ln 9 - \ln 8) time for the capacitor to discharge from q_0 to \displaystyle \frac{8}{9} \, q_0.

It would take \tau(\ln 9 - \ln 7) time for the capacitor to discharge from q_0 to \displaystyle \frac{7}{9}\, q_0.

Note that \ln 9 = 2\,\ln 3, and that\ln 8 = 3\, \ln 2.

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is \tau, and the initial charge of the capacitor be q_0. Then at time t, the charge stored in the capacitor would be:

\displaystyle q(t) = q_0 \, e^{-t / \tau}.

<h3>a)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{8}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}.

\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9.

t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8).

<h3>b)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{7}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}.

\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9.

t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7).

7 0
3 years ago
With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons
Shtirlitz [24]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

m=\frac{Y-Y_{1}}{X-X_{1}}  (1)

Where:

m=5.7 is the slope of the line

Y_{1}=2390.7pounds is the airplane weight with  51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

X_{1}=51gallons is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

(X_{1},Y_{1})=(51,2390.7)

Rewritting (1):

Y=m(X-X_{1})+Y_{1}  (2)

As Y is a function of X:

Y=f_{(X)}=m(X-X_{1})+Y_{1}  (3)

Substituting the known values:

f_{(X)}=5.7(X-51)+2390.7  (4)

f_{(X)}=5.7X-290.7+2390.7  (5)

f_{(X)}=5.7X+2100  (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

f_{(81)}=5.7(81)+2100  (7)

f_{(81)}=2561.7  (8)   This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

3 0
3 years ago
Describe a situation where you can be traveling at a low speed but have an extremely high velocity
Reil [10]

Answer:

Cruising at 35,000 feet in an airliner, straight toward the east,

at 500 miles per hour

Explanation:

3 0
3 years ago
Read 2 more answers
24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5
il63 [147K]

Answer:

...do

Explanation:

24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5

cm. What is the length of the book in mm?

25. Explain the modifications

3 0
3 years ago
A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2. Compu
boyakko [2]

Answer:

0.42 m/s²

Explanation:

r = radius of the flywheel = 0.300 m

w₀ = initial angular speed = 0 rad/s

w = final angular speed = ?

θ = angular displacement = 60 deg = 1.05 rad

α = angular acceleration = 0.6 rad/s²

Using the equation

w² = w₀² + 2 α θ

w² = 0² + 2 (0.6) (1.05)

w = 1.12 rad/s

Tangential acceleration is given as

a_{t} = r α = (0.300) (0.6) = 0.18 m/s²

Radial acceleration is given as

a_{r} = r w² = (0.300) (1.12)² = 0.38 m/s²

Magnitude of resultant acceleration is given as

a = \sqrt{a_{t}^{2} + a_{r}^{2}}

a = \sqrt{0.18^{2} + 0.38^{2}}

a = 0.42 m/s²

8 0
3 years ago
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