Ask question

Ask question

All categories

Delicious77 [7]

4 years ago

8

hot-air balloon is ascending at the rate of 14 m/s and is 84 m above the ground when a package is dropped over the side. (a) How

long does the package take to reach the ground? (b) With what speed does it hit the ground?

2 answers:

timurjin [86]4 years ago

4 0

Answer:

a) t = 4.14 s

b) Speed with which it hits the ground = 40.58 m/s

Explanation:

Using the equations of motion,

g = 9.8 m/s², y = H = 84 m,

Initial velocity, u = 0 m/s,

final velocity, v = ?

Total Time of fall, t = ?

a) y = ut + gt²/2

84 = 0 + 9.8t²/2

4.9t² = 84

t² = 84/4.9

t = 4.14 s

b) v = u + gt

v = 0 + (9.8 × 4.14)

v = 40.58 m/s

Korolek [52]4 years ago

3 0

Answer:

A. t = 4.14 s.

B. 40.62 m/s.

Explanation:

Using the equations of motion,

Given:

g = 9.8 m/s²

S = 84 m,

S = vo*t + (a*t²)/2

vo = initial velocity

= 0 m/s

vi = final velocity

t = Time of motion

84 = 0 + (9.8*t²)/2

t² = 84/4.9

= sqrt(17.143)

t = 4.14 s.

B.

vi = vo + a*t

= 0 + (9.81 * 4.14)

= 40.62 m/s.

You might be interested in

If a diode at 300°K with a constant bias current of 100μA has a forward voltage of 700mV across it, what will the voltage drop a

MAVERICK [17]

Answer:

the voltage drop across this same diode will be 760 mV

Explanation:

Given that:

Temperature T = 300°K

current $I_1$ = 100 μA

current $I_2$ = 1 mA

forward voltage $V_r$ = 700 mV = 0.7 V

To objective is to find the voltage drop across this same diode if the bias current is increased to 1mA.

Using the formula:

$I = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}$

$I_1 = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}$

where;

$V_r$ = 0.7

$I_1 = I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix}$

$I_2 = I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix}$

$\dfrac{I_1}{I_2} = \dfrac{ I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }$

$\dfrac{100 \ \mu A}{1 \ mA} = \dfrac{ \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }$

Suppose n = 1

$V_T = \dfrac{T}{11600} \\ \\ V_T = \dfrac{300}{11600} \\ \\ V_T = 25. 86 \ mV$

Then;

$e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { nV_T} -1} \end {pmatrix}$

$e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { 25.86} -1} \end {pmatrix}$

$e^{\dfrac{V_r'}{nv_T}-1} = 5.699 \times 10^{12}$

${e^\dfrac{V_r'}{nv_T}} = 5.7 \times 10^{12}$

${\dfrac{V_r'}{nv_T}} =log_{e ^{5.7 \times 10^{12}}}$

${\dfrac{V_r'}{nv_T}} =29.37$

$V_r'=29.37 \times nV_T$

$V_r'=29.37 \times 25.86$

$V_r'=759.5 \ mV$

$Vr' \simeq$ 760 mV

Thus, the voltage drop across this same diode will be 760 mV

3 0

3 years ago

A 94 g particle undergoes SHM with an amplitude of 8.3 mm, a maximum acceleration of magnitude 7.8 x 103 m/s2, and an unknown ph

Lelechka [254]

Answer:

a) T = 6.49*10^-3 s

b) v = 8 m/s

c) E = 3 J

d) F = 733 N

e) F = 366.5 J

Explanation:

Given

Mass of particle, m = 94 g = 0.094 kg

Amplitude of the particle, A = 8.3 mm = 8.3*10^-3 m

Maximum acceleration of particle, a = 7.8*10^3 m/s²

the equation describing Simple Harmonic Motion is given as

x = A cos (wt +φ)

To fond the acceleration of this relationship, we would have to integrate. Twice, the first would be a Velocity, and the second acceleration that we need.

Velocity = dx/dt = -Aw sin(wt + φ)

Acceleration = d²x/dt = -Aw² cos(wt + φ)

From the question, we were given, magnitude of acceleration to be 7.8*10^3 m/s²

Aw² = 7.8*10^3

w² = 7.8*10^3 / A

w² = 7.8*10^3 / 8.3*10^-3

w² = 939759

w = √939759

w = 969

Recall, T = 2π/w, so that

T = (2 * 3.142) / 969

T = 6.49*10^-3 s

Maximum speed = Aw

Maximum speed = 8.3*10^-3 * 969

Maximum speed = 8.0 m/s

Total mechanical energy oscillator =

mgx + 1/2mx² =

1/2mv(max)² =

1/2 * 0.094 * 8² =

3 J

Maximum displacement

x = A cos(wt + φ)

For x to be maximum here, then cos(wt + φ) Must be equal to 1

Acceleration = d²x/dt² = -Aw²

And force = mass * acceleration

Force = 0.094 * 7.8*10^3

Force = 733 N

x = A cos(wt + φ), where cos(wt + φ) = 1/2

d²x/dt² = -Aw² * 1/2

d²x/dt² = 733 * 0.5

= 366.5 N

7 0

3 years ago

How can being physically fit help you emotionally

Svetlanka [38]

It can help you feel better about your self and body

7 0

4 years ago

Read 2 more answers

The average velocity for a trip has a positive value. Isit

Luba_88 [7]

Answer:

Yes, it's possible.

Explanation:

The average velocity is a mean value:

$Vavg=\frac{displacement}{time taken}$ .

during that displacement, it may occur that the acceleration would negative at any time so at that moment if the velocity goes in the same direction with the acceleration, the velocity will be negative, it may take just a few moments and then go positive again. The velocity can also take negative values if for a moment the object was going backward (opposite direction). so the average velocity only means that the major of the velocity was positive.

5 0

4 years ago

A boy of mass 30.0 kg is sledding down a 70.0-m slope starting from rest. The slope is angled at 15.0° below the horizontal. Aft

Studentka2010 [4]

The speed of the boy and his friend at the bottom of the slope is 16.52 m/s.

<h3>Their speed at the bottom</h3>

Apply the principle of conservation of energy,

E(up) - E(friction) = E(bottom)

mg sin(15) + ¹/₂(M + m)u² - μ(M + m)cos 15 = ¹/₂(M + m)v²

$v = \sqrt{2[\frac{mgd \ sin15 \ + \frac{1}{2}(M + m)u^2 \ -\mu (M + m)g cos\ 15 }{M + m}] }$

where;

u is the speed of the after 28 m

u = √2gh

u = √(2gL sin15)

u = √(2 x 9.8 x 28 x sin 15)

u = 11.92 m/s

$v = \sqrt{2[\frac{(30)(9.8)(70) \ sin15 \ + \frac{1}{2}(30 + 50)(11.92)^2 \ - 0.12 (30 + 50)9.8 cos\ 15 }{30 + 50}] }\\\\v = 16.52 \ m/s$

Thus, the speed of the boy and his friend at the bottom of the slope is 16.52 m/s.

Learn more about speed here: brainly.com/question/6504879

#SPJ1

8 0

2 years ago