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Dahasolnce [82]
2 years ago
4

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

ional potential energy the same in both cases? Is your energy consumption the same in both?
Physics
1 answer:
Nady [450]2 years ago
7 0

Answer:

Explanation:

Work is being done by a force when it displaces the object in the direction of the force. Work done is the same while climbing a mountain on a zigzag path and for one straight up the side as net displacement is the same. but it is easier to climb on the zigzag path because work is done along a longer path and gravitational pull is less at any instant as compared to one straight up. The distance is shortest when climbing the straight path thus gravitational force experience is more and it is difficult to climb.

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Adam observed properties of four different waves and recorded observations about the frequency and volume of each one in his cha
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I believe this answer is B but my believing sucks so
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2 years ago
You start with spring that's already been stretched an unknown amount from equilibrium. After stretching it an additional 2.0 cm
maxonik [38]

Answer: 35*10^3 N/m

Explanation: In order to explain this problem we know that the potential energy for spring is given by:

Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.

We  also know that with additional streching of 2 cm of teh spring,  the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

k=2*7/(0.02)^2=35000 N/m

7 0
2 years ago
The image on the left shows the charges on a balloon after it’s been rubbed with a wool cloth. The image on the right is a piece
taurus [48]

When wool is rubbed with a balloon, the wool is left with a positive charge as electrons have travelled from the wool to the balloon which means the balloon now has a negative charge.

Now that the balloon has a negative charge, you need to know:
The tissue paper originally contains electrons and protons
The fact that the balloon has a negative charge, it will ATTRACT protons because protons are POSITIVE and electrons are NEGATIVE.
So once they are attracted, they will move closer to one another.
3 0
2 years ago
Read 2 more answers
A 1.0 104 kg spacecraft is traveling through space with a speed of 1200 m/s relative to Earth. A thruster fires for 2.0 min, exe
aniked [119]
We are given information:
m=1.0* 10^{4} kg \\ v=1200m/s \\ t=2min=120s \\ F = 25kN = 25000N

If we apply Newton's second law we can calculate acceleration:
F = m * a
a = F / m
a = 25000 / 10000
a = 2.5 m/s^2

Now we can use this information to calculate change of speed.
a = v / t
v = a * t
v = 2.5 * 120
v = 300 m/s

Force is being applied in direction that is opposite to a direction in which space craft is moving. This means that final speed will be reduced.
v = 1200 - 300
v = 900 m/s

Formula for momentum is:
p = m * v
Initial momentum:
p = 10000 * 1200
p = 12 000 000
p = 12 *10^6 kg*m/s
Final momentum:
p = 10000 * 900
p = 9 000 000
p = 9 *10^6 kg*m/s

7 0
2 years ago
In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co
labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m

a)

The final velocity of cart 1 after collision is given as:

v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm

7 0
2 years ago
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