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MariettaO [177]
3 years ago
7

What does increasing the speed of an object do to its potential energy?

Physics
2 answers:
JulijaS [17]3 years ago
7 0
It does not affect the objects potential energy.
olchik [2.2K]3 years ago
7 0
The ____ energy of an object increases with its height. The kinetic energy of an object increases as its ____ increases. Increasing the speed of an object ____ its potential energy. ... Energy may change from one form to another, but the total of energy never changes.
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If x2 = 60, what is the value of x? plus or minus square root 30 plus or minus square root 60 ±30 ±120
Zielflug [23.3K]
X2 = 60
/ 2 / 2
x = 30
Plus or minus square root 60
4 0
3 years ago
Which statement best describes why a machine is useful?
sdas [7]

There are no options here. Could you provide the different answers?

7 0
3 years ago
Read 2 more answers
A skier starts from rest at the top of a 20 degree incline and skis in a straight line to the bottom of the slope, a distance d
SSSSS [86.1K]

Answer:

Explanation:

Loss of potential energy = mgh.

h = d sin 20

= 400 sin20 = 136.8 m

Loss of potential energy = m x 9.8 x  136.8

= 1340.64 m

negative work done by friction = μ mg cosθ x d

= .2 x m x 9.8 x cos 20 x 400

= 736.72 m

Net loss of potential energy = 1340.64 m - 736.72 m

= 603.92 m

= gain of kinetic energy = 1/2 m v²

1/2 m v² = 603.92 m

v² = 1207.84

v = 34.75 m /s .

6 0
3 years ago
A power plant produces 1000 MW to supply a city 40 km away. Current flows from the power plant on a single wire of resistance 0.
nikitadnepr [17]

Answer:

Current = 8696 A

Fraction of power lost = \dfrac{80}{529} = 0.151

Explanation:

Electric power is given by

P=IV

where I is the current and V is the voltage.

I=\dfrac{P}{V}

Using values from the question,

I=\dfrac{1000\times10^6 \text{ W}}{115\times10^3\text{ V}} = 8696 \text{ A}

The power loss is given by

P_\text{loss} = I^2R

where R is the resistance of the wire. From the question, the wire has a resistance of 0.050\Omega per km. Since resistance is proportional to length, the resistance of the wire is

R = 0.050\times40 = 2\Omega

Hence,

P_\text{loss} = \left(\dfrac{200000}{23}\right)^2\times2

The fraction lost = \dfrac{P_\text{loss}}{P}=\left(\dfrac{200000}{23}\right)^2\times2\div (1000\times10^6)=\dfrac{80}{529}=0.151

3 0
3 years ago
Points P and Q are located at (0, 2, 4) and (-3, 1,5). Calculate
Akimi4 [234]

Answer:

a) The position vector of P is \vec P =(0, 2,4).

b) The distance vector from P to Q is \overrightarrow{PQ} = (-3,-1,1).

c) The distance between P and Q is \|\overrightarrow{PQ}\|=\sqrt{11}.

d) A vector parallel to PQ with magnitude of 10 is \vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right).

Explanation:

a) The position vector of a point is the vector displacement from the origin to the location of the point. That is:

\vec P = (0,2,4)-(0,0,0)

\vec P = (0-0, 2-0, 4-0)

\vec P =(0, 2,4)

The position vector of P is \vec P =(0, 2,4).

b) First, we calculate the position vector of point Q:

\vec Q = (-3,1,5)-(0,0,0)

\vec Q = (-3-0,1-0,5-0)

\vec Q =(-3,1,5)

The distance vector from P to Q is define by the following vectorial expression:

\overrightarrow{PQ} = \vec Q - \vec P (1)

\overrightarrow{PQ} = (-3,1,5)-(0,2,4)

\overrightarrow{PQ} =(-3-0,1-2,5-4)

\overrightarrow{PQ} = (-3,-1,1)

The distance vector from P to Q is \overrightarrow{PQ} = (-3,-1,1).

c) There are two approaches to calculate the distance between P and Q:

First Method - Pythagorean Theorem:

\|\overrightarrow{PQ}\| = \sqrt{(-3)^{2}+(-1)^{2}+1^{2}}

\|\overrightarrow{PQ}\|=\sqrt{11}

Second Method - Dot Product:

\|\overrightarrow{PQ}\| = \sqrt{\overrightarrow{PQ}\,\bullet\,\overrightarrow{PQ}} (2)

\|\overrightarrow{PQ}\| = \sqrt{(-3,-1,1)\,\bullet (-3,-1,1)}

\|\overrightarrow{PQ}\|=\sqrt{11}

The distance between P and Q is \|\overrightarrow{PQ}\|=\sqrt{11}.

d) To determine a vector parallel to PQ with a given magnitude is determined by the following expression:

\vec v = \frac{k}{\|\overrightarrow{PQ}\|} \cdot \overrightarrow{PQ} (3)

Where k is the scale factor.

If we know that \overrightarrow{PQ} = (-3,-1,1), \|\overrightarrow{PQ}\|=\sqrt{11} and k = 10, then the vector is:

\vec v = \frac{10}{\sqrt{11}}\cdot (-3,-1,1)

\vec v = \left(-\frac{30}{\sqrt{11}},-\frac{10}{\sqrt{11}},\frac{10}{\sqrt{11}}\right)

\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)

A vector parallel to PQ with magnitude of 10 is \vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right).

8 0
3 years ago
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