A boat with a weight of 547 newtons is floating in a harbor. What is the buoyant force on the boat?

How many times could Haley fly bewteen the two flowers in 1 minute (60 seconds)

dmitriy555 [2]

30 seconds is the answer

7 0

2 years ago

If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it

Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

$\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

Explanation:

From the question we are told that

The mass of the rocket is m = 6300 kg

The velocity at gas is being ejected is u = 2000 m/s

The initial acceleration desired is $a = 27.0 \ m/s$

The time taken for the gas to be ejected is t = 1 s

Generally this desired acceleration is mathematically represented as

$a = \frac{u * \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}$

Here $\frac{\Delta m}{\Delta t }$ is the rate at which gas is being ejected with respect to time

Substituting values

$27 = \frac{2000 * \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}$

=> $170100 -27* \frac{\Delta m}{\Delta t} = 2000 * \frac{\Delta m}{\Delta t}$

=> $170100 = 2027 * \frac{\Delta m}{\Delta t}$

=> $\frac{\Delta m}{\Delta t} = \frac{170100}{2027}$

=> $\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

3 0

2 years ago

For nitrogen feel like with its temperature must be within 12.78 Fahrenheit of -333.22 Fahrenheit which equation can be used to

photoshop1234 [79]

Answer:

The following equation can be used.

(32°F − 32) × 5/9=C

7 0

2 years ago

Read 2 more answers

|| Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and r

Otrada [13]

To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.

Newton's second law is defined as

$F = ma$

Where,

m = mass

a = acceleration

From this equation we can figure the acceleration out, then

$a = \frac{F}{m}$

$a = \frac{11*10^3}{80}$

$a = 137.5m/s$

From the cinematic equations of motion we know that

$v_f^2-v_i^2 = 2ax$

Where,

$v_f =$ Final velocity

$v_i =$ Initial velocity

a = acceleration

x = displacement

There is not Final velocity and the acceleration is equal to the gravity, then

$v_f^2-v_i^2 = 2ax$

$0-v_i^2 = 2(-g)x$

$v_i =\sqrt{2gx}$

$v_i = \sqrt{2*9.8*4.8}$

$v_i = 9.69m/s$

From the equation of motion where acceleration is equal to the velocity in function of time we have

$a = \frac{v_i}{t}$

$t = \frac{v_i}{a}$

$t =\frac{9.69}{137.5}$

$t = 0.0705s$

Therefore the time required is 0.0705s

4 0

2 years ago

Read 2 more answers

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

2 years ago