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scoundrel [369]
2 years ago
10

compared to an object that does not roll, but instead slides without friction, should a rolling object be released from the same

,a greater, or a lesser height in order just barely to complete the loop the loop?
Physics
1 answer:
Zinaida [17]2 years ago
6 0

Answer:

from a lesser hight

Explanation:

because you need less force

hope its right if it is mark brainlyest ;)

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Help<br><br><br> What is Gravitational Potenial Energy
frosja888 [35]
Gravitational potential energy is an object that possesses because of its position in a gravitational field
5 0
3 years ago
Write the equation expressing the relationship "y varies directly as x." use k as the constant of proportionality.
Alexeev081 [22]

The equation expressing the statement "y varies directly as x"  is y=kx.

Explanation

The statement that y varies directly as x is analogous to saying that  the ratio of y to x is constant. In other words, when x increases, y likewise increases and that when x decreases, y decreases proportionally.

Mathematically, we express the relationship that the ration of y is to x is constant is expressed as; \frac{y}{x} =k where k is the constant of proportionality.

We can then solve the relationship for y to determine the correct form of the relationship as shown below,

\frac{y}{x} =k\\\rightarrow y=kx


6 0
3 years ago
Read 2 more answers
a heavy jar sits on top of a 3.4 m shelf with a gravitational potential energy of 180 j. What is the mass of the jar?
sp2606 [1]

Answer:

5.2941176471 kg or 5294.1 grams

Explanation:

g.p.e= mgh

g.p.e/gh=m

180j/10×3.4= m

180/34= m

5.2941 kg= m

6 0
3 years ago
How many types of quarks are there?<br> a. 2<br> b. 4<br> c. 6<br> d. 8
Nostrana [21]
There are 6 types of quarks.

Answer: C. 6
8 0
3 years ago
Read 2 more answers
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
3 years ago
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