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3 years ago

15

Lucifer is sitting in a tree that is 15 m above the ground. She jumps straight off the tree in the horizontal direction and land

s 7 m away from the base of the tree. How fast did she jump off the tree?

1 answer:

-BARSIC- [3]3 years ago

4 0

Answer:

huppp huppp huppp huppp huppp huppp

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A pendulum clock gives correct time at 20°c how many sec/day will it gain or loose when the temperature fall to 5°c? cofficient

rewona [7]

Answer:

129.6 seconds

Explanation:

Given that :

α = 0.0002°c-1

θ1 = 20°C

θ2 = 5°C

Time t = one day ; Converting to seconds ; number of seconds in a day ; (24 * 60 * 60) = 86400 seconds

Let dT= change in time

Using the relation :

dT = 0.5* α * dθ * t

dθ = (20 - 5) = 15°C

dT = 0.5 * 0.0002 * 15 * 86400

dT = 129.6 seconds

5 0

2 years ago

Standard weight and scales in the market should be checked in every two year why?

disa [49]

Answer:

to check for errors.

Explanation:

since some of the components of the scale are metallic, they are prone to rusting which would cause error in reading of the scale so there is organized checking if the scale to correct that.

hope it helps .

7 0

3 years ago

Read 2 more answers

You are skateboarding in your friend's driveway. You hit a rock and your

nekit [7.7K]

Answer:

the rock hit the board you go flying because you board stops but you have energy still going so there for you go flying

Explanation:

3 0

2 years ago

Suppose that you are standing on a train accelerating at 0.39g. What minimum coefficient of static friction must exist between y

timurjin [86]

Answer:

0.39

Explanation:

In order not to slide, you must have exactly the same acceleration of the train:

$a=0.39 g$

where

g = 9.81 m/s^2

There is only one force acting on you: the static frictional force that "pulls" you forward, and it is given by

$F_s = \mu_s mg$

According to Newton's second law, the net force acting on you (so, the frictional force) must be equal to your mass times the acceleration, so we have

$F= ma = \mu_s mg$

from which we find

$\mu_s = \frac{a}{g}=\frac{0.39 g}{g}=0.39$

so, the minimum coefficient of static friction must be 0.39.

5 0

3 years ago

A Pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3300

adell [148]

Answer:

v = 306.76 Km/h

Explanation:

given,

height of the aircraft = 3000 m

differential pressure reading = 3300 N/m²

density of air = 0.909 Kg/m³

speed of aircraft = ?

Assuming the air flowing above air craft is in-compressible, irrotational and steady so, we can use Bernoulli's equation to solve the problem.

using Bernoulli's equation

$\dfrac{v^2}{2} = \dfrac{\Delta P}{\rho}$

where ρ is the density of the air at 3000 m

$v= \sqrt{\dfrac{2 \times \Delta P}{\rho}}$

$v= \sqrt{\dfrac{2 \times 3300}{0.909}}$

$v = \sqrt{7260.726}$

v = 85.21 m/s

$v= 85.21 \times \dfrac{3600}{1000}$

v = 306.76 Km/h

4 0

3 years ago