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4 years ago

8

Explain why it is easier to climba mountain on a zigzag path rather than one straight up the side. Is your increase in gravitati

onal potential energy the same in both cases? Is your energy consumption the same in both?

1 answer:

DerKrebs [107]4 years ago

3 0

Answer:

it is easier in zig zag because youll fell less tired and feel more energetic and it will will feel as if it were shorter to go zig zag

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How does energy and temperature affect the density of fluids (liquids and gases)?

Alex Ar [27]

Answer: When a liquid or gas is heated, the molecules move faster, bump into each other, and spread apart. Because the molecules are spread apart, they take up more space. ... The molecules move more slowly and take up less space. Therefore temperature can affect density.

Explanation:

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3 years ago

High friction materials

Harrizon [31]

Answer:

For the dry and static friction materials, some of these are Rubber, Aluminum, Gold, Platinum,

Explanation:

High friction materials has higher coefficient of friction (COF).

7 0

3 years ago

An electric vehicle starts from rest and accelerates at a rate of 2.4 m/s2 in a straight line until it reaches a speed of 27 m/s

DENIUS [597]

A) use v=u+at for both

First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.

b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.

First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.

6 0

3 years ago

Professional Application: A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 11

Nataliya [291]

Answer:

0.182 m/s

Explanation:

m1 = 30,000 kg, m2 = 110,000 kg, u1 = 0.85 m/s

let the velocity of loaded freight car is v

Use the conservation of momentum

m1 x u1 + m2 x 0 = (m1 + m2) x v

30,000 x 0.85 = (30,000 + 110,000) x v

v = 0.182 m/s

5 0

3 years ago

A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio

Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

$f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}$

$f=2.85 Hz$

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2$

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

$kA^2 =mV^2 +ky^2$

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

$1700*A^2=5.3*1.7^2 +1700*(0.045)^2$

Solving for A,

$A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}$

$A ^2 = 0.011035$

$A=0.105 m \approx 10.5 cm$

PART C) Finally, the total mechanical energy is given by the equation

$E = \frac{1}{2}kA^2$

$E=\frac{1}{2}1700*(0.105)^2$

$E= 9.3712 J$

3 0

3 years ago