A projectile is launched horizontally at a speed of 45.0 m/s from a

Paying off your debts will most likely

g100num [7]

I think the answer is d

4 0

2 years ago

A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.

Ede4ka [16]

(a) The work required to get the coaster to the top of the first hill is 1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

P = Fd/t

P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

Learn more about energy here: brainly.com/question/13881533

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6 0

2 years ago

He formula p=m x v to answer th<br>2,000 kg car moving at 30 m/s

aivan3 [116]

The momentum of the car is 60,000 kg m/s

Explanation:

The momentum of an object is a vector quantity, whose magnitude is given by

$p=mv$

where

m is the mass of the object

v is its velocity

For the car in this problem, we have

m = 2,000 kg

v = 30 m/s

Therefore, the magnitude of its momentum is

$p=(2000)(30)=60,000 kg m/s$

Learn more about momentum here:

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brainly.com/question/9484203

#LearnwithBrainly

7 0

2 years ago

Define atomic orbital

oksian1 [2.3K]

According to chem.purdue.edu, Atomic Orbital is "... a representation of the three-dimensional volume (i.e., the region in space) in which an electron is most likely to be found..."
Hope this helps!

6 0

2 years ago

A typical jetliner lands at a speed of 146 mi/h and decelerates at the rate of (10.4 mi/h)/s. If the jetliner travels at a const

zzz [600]

jetliner is moving with speed 146 mi/h

now we will have

$v= 146 \frac{1609 m}{3600 s}$

$v = 65.25 m/s$

now if it moves with this speed for 1.5 s

so the displacement for above time

$d_1 = 1.5(65.25) = 97.9 m$

now the deceleration of jet is given as

$a = - 10.4 mi/h/s$

$a = -(10.4)(\frac{1609 m}{3600 s})\frac{1}{s}$

$a = - 4.65 m/s^2$

now by kinematics

$v_f^2 - v_i^2 = 2ad$

$0 - 65.25^2 = 2(-4.65)(d_2)$

$d_2 = 457.8 m$

so total displacement is given as

$d = d_1 + d_2$

$d = 97.9 + 457.8 = 555.7 m$

so displacement will be 555.7 m

7 0

2 years ago

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