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3 years ago

What is a change in the speed of an object called?

Physics

2 answers:

mina [271]3 years ago

5 0

Acceleration........................................

Tcecarenko [31]3 years ago

3 0

Acceleration ..........

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A rope exerts a 280 N force while pulling an 80 Kg skier upward along a hill inclined at 12o. The rope pulls parallel to the hil

user100 [1]

Answer:

The speed of the skier after moving 100 m up the slope are of V= 25.23 m/s.

Explanation:

F= 280 N

m= 80 kg

α= 12º

μ= 0.15

d= 100m

g= 9,8 m/s²

N= m*g*sin(α)

N= 163 Newtons

Fr= μ * N

Fr= 24.45 Newtons

∑F= m*a

a= (280N - 24.5N) / 80kg

a= 3.19 m/s²

d= a * t² / 2

t=√(2*d/a)

t= 7.91 sec

V= a* t

V= 3.19 m/s² * 7.91 s

V= 25.23 m/s

4 0

2 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago

How many hours does it take for the sun's light to reach the edge of the solar system?

Georgia [21]

32 hours is what I got

7 0

3 years ago

Read 2 more answers

Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years i

Naddik [55]

Answer:

r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

F = ma

G m M / r² = m a

The centripetal acceleration is given by

a = v² / r

For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

v = d / t

The distance traveled Esla orbits, in a circle the distance is

d = 2 π r

Time in time to complete the orbit, called period

v = 2π r / T

Let's replace

G m M / r² = m a

G M / r² = (2π r / T)² / r

G M / r² = 4π² r / T²

G M T² = 4π² r3

r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

T = 1.03 10⁸ s

Let's calculate

r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

r = ∛ (21.44 10³⁵ / 39.478)

r = ∛(0.0543087 10 36)

r = 0.3787 10¹² m

r = 3.787 10¹¹ m

7 0

3 years ago

It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar had a constant accelerati

Soloha48 [4]

Answer: 47.6 m/s

Explanation: Please see attached for the calculation and formula.

5 0

3 years ago