A leaky faucet drips 40 times in 30.0 s. what is the frequency of the dripping?

Dźwig podniósł kontener o masie m = 80 kg na wysokość h = 10 m. Pierwsze 5 m kontener przebył z przy-

Nady [450]

Answer:

a) W_total = 8240 J , b) W₁ / W₂ = 1.1

Explanation:

In this exercise you are asked to calculate the work that is defined by

W = F. dy

As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.

W = F dy = F Δy

let's apply this formula to our case

a) Let's use Newton's second law to calculate the force in the first y = 5 m

F - W = m a

W = mg

F = m (a + g)

F = 80 (1 + 9.8)

F = 864 N

The work of this force we will call it W1

We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)

F₂ - W = 0

F₂ = W

F₂ = 80 9.8

F₂ = 784 N

The work of this fura we will call them W2

The total work is

W_total = W₁ + W₂

W_total = (F + F₂) y

W_total = (864 + 784) 5

W_total = 8240 J

b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use

W₁ / W₂ = F y / F₂ y

W₁ / W₂ = 864/784

W₁ / W₂ = 1.1

7 0

2 years ago

Two roads intersect at right angles, one going north-south, the other east-west. an observer stands on the road 60 meters south

Sloan [31]

observer is standing at distance d = 60 m south from the intersection

cyclist is travelling at speed v = 10 m/s

now after t = 8 s its displacement from intersection is given by

$x = 10*8 = 80 m$

so the position of cyclist makes an angle with the observer

$\theta = tan^{-1}\frac{80}{60} = 53 degree$

now the component of velocity of cyclist along the line joining its position with the observer is given as

$v = v_o cos\phi$

here

$\phi = 90 -\theta$

$\phi = 90 - 53 = 37 degree$

$v = 10 cos37 = 8 m/s$

so at this instant cyclist is moving away with speed 8 m/s

7 0

2 years ago

Read 2 more answers

a student drew the following model: volcano cooling crust motion plates tension which landform should the student put next in th

Andrei [34K]

Answer:

C) rift valley

Explanation:

A rift valley is a lowland region formed by the interaction of Earth's tectonic plates. This small rift valley has a typical formation—long, narrow, and deep. It was formed by the Thingvellir rift, where the North American and Eurasian tectonic plates are tearing, or rifting, apart over a hotspot on the island of Iceland.

3 0

1 year ago

Read 2 more answers

An ambulance moving at 42 m/s sounds its siren whose frequency is 450 hz. a car is moving in the same direction as the ambulance

Korvikt [17]

(a) Since the ambulance and the car are moving one relative to each other, we have to use the general formula of the Doppler effect, which gives us the shift of the frequency of the siren as heard by an observer in the car:
$f'=( \frac{v+v_o}{v+v_s} )f$
where
f' is the apparent frequency as heard by the observer in the car
v is the velocity of the wave
$v_o$ is the velocity of the observer (positive if it is moving towards the source, negative if it is moving away)
$v_s$ is the velocity of the source (positive if the source is moving away from the observer, negative if is is moving towards it)
f is the real frequency of the sound

In the first part of the problem:
$v=343 m/s$ (speed of the sound wave)
$v_o =-25 m/s$ (the car is moving away from the ambulance)
$v_s = -42 m/s$ (the ambulance is moving towards the car)
$f=450 Hz$ (original frequency of the sound)

If we plug the numbers into the formula, we find
$f'=( \frac{343 m/s-25 m/s}{343 m/s-42 m/s} )(450 Hz)=475 Hz$

b) This time, the ambulance passes the car, so the ambulance is now moving away from the car; this means that $v_s$ must be positive:
$v_s=+42 m/s$
Moreover, the car is now moving towards the ambulance, so we should reverse also the sign of $v_o$ :
$v_o=+25 m/s$
All the other data do not change, so if we use the same formula as before, we find
$f'=( \frac{343 m/s+25 m/s}{343 m/s+42 m/s} )(450 Hz)=430 Hz$

8 0

2 years ago

The weights of a large number of miniature poodles are approximately normally distributed with a mean of 99 kilograms and a stan

Karo-lina-s [1.5K]

Answer:

See explanation below

Explanation:

As I say in the comments, the question is incomplete, however, I will try to answer this by using data that I found on another site.

This is the part of the question that is not here:

If measurements are recorded to the nearest

tenth of a kilogram, find the fraction of these poodles

with weights

(a) over 9.5 kilograms;

(b) of at most 8.6 kilograms;

So, assuming a mean of 8 kg, and 0.9 of standard deviation, let X represents the weight of the poodles

The expression to calculate the fraction of poodle needed is:

Z = X - u / d

u: weight of the large number of poodle

d: standard deviation

Replacing data of a) wer have:

Z = 9.5 - 8 / 0.9

Z = 1.67

With this value, we need to take the value of Z, and see the area under the curve of standard deviation (see table attached)

Therefore:

P (X > 9.5) = P(Z > 1.67) = 0.5 - P (Z < 1.67) = 0.5 - 0.4525 = 0.0475

b) In this part, is the same as part a) so:

Z = 8.6 - 8 / 0.9 = 0.67

The value for area in the curve is 0.2486 so:

P = 0.5 + 0.2486 = 0.7486

Hope this helps

8 0

2 years ago