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3 years ago

While riding in a hot air balloon, which is steadily descending at a speed of 1.01 m/s, you accidentally drop your cell phone?

Physics

1 answer:

mote1985 [20]3 years ago

5 0

While riding in a hot air balloon, which is steadily at a speed of 1.01 m/s, and your phone accidentally falls.

(a) The speed of your phone after 4 s is:

V= u + at

V= 1.01 + (9.8)(4)

V= 40.21 m/s

(b) The balloon is ____ far:

V = u + at

V= 1.01 + (9.8)(1)

V=10.81 –distance at 1 one second

V= u + at

V= 1.01 + (9.8)(2)

V= 20.61-distance at 2 seconds

V= u+ at

V= 30.41- distance at 3 seconds

V= 40.21- distance at 4 seconds

D= 102.04 m

(c) If the balloon is rising steadily at 1.01 m/s:

V= -1.1 m/s

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3 years ago

Define the fundamental difference between kinematics and dynamics. .

Tcecarenko [31]

In kynematics you describe the motion of particles using vectors and their change in time. You define a position vector r for a particle, and then define velocity v and acceleration a as

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3 years ago

What must the diver's minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, wh

Dafna1 [17]

Answer:

1.08 m/s

Explanation:

This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.

Time taken to fall 9.5 m

vertical acceleration = a = 9.8 m/s^2.

vertical velocity = 0, (since there is only horizontal component for velocity, )

distance traveled s = 9.5 m.

Substituting these values in the equation

$s= u \timest+0.5at^{2}$

$t= \sqrt{\frac{2s}{g} }$

$t=\sqrt{\frac{2\times9.5}{9.8} }$

⇒ t= 1.392 sec

Velocity needed

We know the time taken (1.392 s) to travel 1.5 m,

So velocity = 1.5 m / 1.392 s = 1.08 m/s

hence velocity of the diver must be at least 1.08 m/s

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3 years ago

The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

Gennadij [26K]

Answer:

Part a)

$k = 6.06 \times 10^4 N/m$

Part b)

$b = 1795.4 kg/s$

Explanation:

Part a)

as the mass of the suspension system is given as

$m = 2100 kg$

also we have

$x = 8.5 cm$

so now for force balance we have

$mg = kx$

$(525)(9.81) = k(0.085)$

$k = 6.06 \times 10^4 N/m$

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

$A = 0.37 A_o$

also we know

$A = A_o e^{-bt/2m}$

$0.37 A_o = A_o e^{-bt/2m}$

$\frac{bt}{2m} = 1$

$b = \frac{2m}{t}$

here t = time period of one oscillation

so it is

$t = 2\pi\sqrt{\frac{m}{k}}$

$t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}$

$t = 0.58 s$

now damping constant is

$b = \frac{2(525)}{0.58}$

$b = 1795.4 kg/s$

7 0

3 years ago

"The wave" is a particular type of pulse that can

scoray [572]

Answer:

t = 36π seconds

Explanation:

For resolving this problem, we are going to consider a representative stadium of the United States. The Mercedes-Benz Stadium located in Atlanta, Georgia has an average radius of 90 m.

Then, its circumference measures:

L = 2πr

L = 2π(90)

L = 180π m

First, we estimate the wave's velocity: the average width of an person is 0.5 m, then the velocity is:

v = x/t

Where x: person's width

t: time

v = 0.5/0.1 = 5 m/s

The time required for the pulse to make one circuit around the stadium is:

t = x/v = 180π/5 = 36π seconds

4 0

3 years ago