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Triss [41]
3 years ago
7

It is easier to climb up a slanted slope than a vertical slope

Physics
2 answers:
deff fn [24]3 years ago
4 0

The correct answer is it is easier to climb a slanted slope rather then a vertical slope. Vertical is straight up versus a nice slant.

V125BC [204]3 years ago
3 0

IT IS EASIER TO CLIMB A SLANTED SLOPE

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A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo
Serggg [28]

Before the engines fail (0\le t\le3.00\,\rm s), the rocket's horizontal and vertical position in the air are

x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2

y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2

and its velocity vector has components

v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t

v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t

After t=3.00\,\rm s, its position is

x=273\,\rm m

y=362\,\rm m

and the rocket's velocity vector has horizontal and vertical components

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}

After the engine failure (t>3.00\,\rm s), the rocket is in freefall and its position is given by

x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t

y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2

and its velocity vector's components are

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}-gt

where we take g=9.80\,\frac{\rm m}{\mathrm s^2}.

a. The maximum altitude occurs at the point during which v_y=0:

159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s

At this point, the rocket has an altitude of

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve y=0 for t, then add 3 seconds to this time:

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s

So the rocket stays in the air for a total of 37.6\,\rm s.

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute x for this time t:

273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m

5 0
3 years ago
Is how high or low sound is
JulsSmile [24]

Answer:

How high the sound or how low the sound is depending on the pitch of the sound which in this case the frequencies of the sound. The higher of the frequencies , the higher of the pitch but it has the shortest length of wave (λ). That's why AM radio have a longer range but bad audio quality than FM radio that have better audio quality with shorter range.

5 0
3 years ago
A 2.24x10^3 kg car slows down uniformly from 20.0 m/s to a stop (assume constant acceleration) via a braking force of 8.41x10^3
melisa1 [442]

Answer:

Explanation: 139 m

4 0
2 years ago
If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re
valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J

So, the required work done is 16 J.

7 0
3 years ago
Why electric forces are assential to forming compunds?
lidiya [134]
Ions were once atoms with the same number of electrons and protons. Since they have opposite charges atoms are neutral. When they become ions the lose or gain electrons and become unbalanced.These different charges are attracted to each other via electric forces<span>.</span>
8 0
3 years ago
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