From her bedroom window a girl drops a water-filled balloon to the ground, 4.77 m below. If the balloon is released from rest, how long is it in the air?

Answers

Answer 1

Answer: We need to use the equation x = vt + (1/2)at^2. We know x = 4.77, v = 0, and a = 9.81m/s^2. Plug in the values. 4.77 = (0)t + (1/2)(9.81)t^2 Solve for t. 4.77 = (4.905)t^2 0.972 = t^2 t = (sq.rt)_/0.972 t = 0.985 So it's in the air 0.985 seconds.

Related Questions

Could you please solve it with shiwing the full work1-A high powered projectile is fired horizontally from the top of a cliff at a speed of 638.6 m/s. Determine the magnitude of the velocity (in m/s) after 5 seconds.Take gravitational acceleration to be 9.81 m/s2.2-A man throws a ball with a velocity of 20.9 m/s upwards at 33.2° to the horizontal. At what vertical distance above the release height (in metres) will the ball strike a wall 13.0 m away ?Take gravitational acceleration to be 9.81 m/s2.3-A particle is moving along a straight path and its position is defined by the equation s = (1t3 + -5t2 + 3) m, where t is measured in seconds. Determine the average velocity (in m/s) of the particle when t = 5 seconds.4-A particle has an initial speed of 26 m/s. The particle undergoes a deceleration of a = (-9t) m/s2, where t is measured in seconds. Determine the distance (in metres) the particle travels before it stops. When t = 0, s = 0.
A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete swing?
Calculate the ionization potential for C+5 ( 5 electrons removed for the C atom) and in addition compute the wavelength of the transition from n=3 to n= 2.
A 2.0 cm thick brass plate (k_r = 105 W/K-m) is sealed to a glass sheet (kg = 0.80 W/K m), and both have the same area. The exposed face of the brass plate is at 80°C, while the exposed face of the glass is at 20 °C. How thick is the glass if the glass brass interface is at 65 C? Ans. 0.46 mm
Please someone help, I’m very confused and it’s due soon, thanks

A piano tuner hears a beat every 2.20 s when listening to a 266.0 Hz tuning fork and a single piano string. What are the two possible frequencies (in Hz) of the string? (Give your answers to at least one decimal place.)

Answers

Answer:

The lower frequency is $f_1 = 265.55 \ Hz$

The higher frequency is $f_2 = 266.4546 \ Hz$

Explanation:

From the question we are told that

The period is $T = 2.20 \ s$

The frequency of the tuning fork is $f = 266.0 \ Hz$

Generally the beat frequency is mathematically represented as

$f_b = (1)/(T)$

substituting values

$f_b = (1)/(2.20)$

$f_b = 0.4546 \ Hz$

Since the beat frequency is gotten from the beat produced by the tuning fork and and the string then

The possible frequency of the string ranges from

$f_1 = f- f _b$

$f_2 = f + f_b$

Now substituting values

$f_1 = 266.0 - 0.4546$

$f_1 = 265.55 \ Hz$

For $f_2$

$f_2 = 266 + 0.4546$

$f_2 = 266.4546 \ Hz$

Beginning 156 miles directly east of the city of Uniontown, a truck travels due south. If the truck is travelling at a speed of 31 miles per hour, determine the rate of change of the distance between Uniontown and the truck when the truck has been travelling for 81 miles. (Do not include units in your answer, and round to the nearest tenth.)

Answers

Answer:

14.3

Explanation:

The distance s as a function of time can be written as:

$s(t) = \sqrt{156^(2) + (31t)^2}$

The rate of change is the derivative of d with respect to time:

$(ds)/(dt) =\frac{961t}{\sqrt{156^(2)+(31t)^(2)}}$

The time t when the track has been traveling for 81 miles is given by:

$81 = 31t\n t = (81)/(31)$

Using t in the previous equation gives:

$(ds)/(dt)((81)/(31) ) =\frac{2511}{\sqrt{156^(2)+81^(2)}}=14.3$

In the study of sound, one version of the law of tensions is:f1= f2 √ (F1/F2)

If f1= 300, F2= 60, and f2=260, find f1 to the nearest unit.

Answers

Answer:

F1 = 80

Explanation:

f1= f2 √ (F1/F2)

Where f1 = 300, f2 = 260 and F2 = 60

Putting in the above formula

300 = 260√(F1/60)

Dividing both sides by 260

=> 1.15 = √(F1/60)

Squaring both sides

=> 1.33 = F1/60

Multiplying both sides by 60

=> F1 = 80

Use the worked example above to help you solve this problem. A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 15.0 m/s. The point of release is h = 46.0 m above the ground.(a) How long does it take for the ball to hit the ground?
(b) Find the ball's speed at impact.
(c) Find the horizontal range of the ball.

Answers

Answer:

Explanation:

An electron is moving in the presence of an electric field of 400 N/C. What force does the electron experience?

Answers

Answer:

Explanation:

Given

Electric Field strength $E=400\ N/C$

charge of electron $q=1.6* 10^(-19)\ C$

Force on the charge particle is given by

$F=qE$

$F=1.6* 10^(-19)* 400$

$F=640* 10^(-19)\ N$

but this force will be acting in the direction opposite to the direction of Electric field because electron is negatively charged

Based on the measured force between objects that are 10 meters apart, how can you find the force between objects that are any distance apart ?

Answers

The force between objects that are any distance apart is expressed as $P'=(100P)/(r^2)$

According to the gravitational law, the force acting on an object is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. Mathematically,

$P=(GMm)/(r^2)$

M and m are the masses

r is the distance between the masses

If the force between objects that are 10 meters apart, hence;

$P=(GMm)/(10^2)\nP=(GMm)/(100)\nGMm = 100P$

To find the force between objects that are any distance apart, we will use the same formula above to have;

$P'=(GMm)/(r^2)\n$

Substitute the result above into the expression to have:

$P'=(100P)/(r^2)$

Hence the force between objects that are any distance apart is expressed as $P'=(100P)/(r^2)$

Learn more on gravitational law here: brainly.com/question/11760568

Answer:

F' = 100 F/r²

Explanation:

The gravitational force of attraction between two objects is given by the Newton's Gravitational Formula. The Newton's Gravitational Formula is as follows:

F = Gm₁m₂/r²

where,

F = Force between objects

G = Universal Gravitational Constant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects = 10 m

Therefore,

F = Gm₁m₂/10²

Gm₁m₂ = 100F --------------------- equation (1)

Now, we consider these objects at any distance r apart. So, the force becomes:

F' = Gm₁m₂/r²