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Yuliya22 [10]
3 years ago
13

An escalator and an elevator can transport a person from one floor to the next. the escalator does it in 15 s and the elevator t

akes 10 s. Which does more work on the person? Which has the greater power output?
Physics
1 answer:
ad-work [718]3 years ago
5 0

Answer:

The escalator does more work and the elevator has a greater power output.

Explanation:

The escalator does more work on the person because it is carrying them up. It also gives the person a chance to walk up the escalator making it a little more challenging than using just an elevator. The elevator has a greater power output because it can transport a person from one floor to another in just 10 seconds where as the escalator can do it in 15.

Hope this helps!

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Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two
slava [35]

Q: Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two charges q1'=2q1 and q2'=3q2 that that are distance d apart?

Answer:

The force = 38.4 N

Explanation:

From coulombs law,

F = kq₁q₂/r² ............................ Equation 1

Where F = Force of attraction or repulsion between the charges, q₁ and q₂ = first and second charge respectively, r = distance between the charges, k = constant of proportionality.

When, F = 6.4 N, r = d m.

6.4 = kq₁q₂/d²......................... Equation 1

When q₁' = 2q₁, q₂' = 3q₂, r = d cm

F = k(2q₁)(3q₂)/d²

F = 6kq₁q₂/d².......................... Equation 2

Dividing Equation 1 by equation 2

6.4/F = kq₁q₂/d²/(6kq₁q₂/d²)

6.4/F = 1/6

F = 6.4×6

F = 38.4 N.

Thus the force = 38.4 N

6 0
3 years ago
Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.00 s, it rotates 14.8 rad. Du
11Alexandr11 [23.1K]

Answer:

Explanation:

Given:

Initial θ = 0 rad (from rest)

Final θ = 14.3 rad

Time, t = 5 s

B.

Angular velocity, w = dθ / dt

= (14.3 - 0)/5

= 2.86 rad/s

A.

Acceleration, ao = dw/dt

Initial angular velocity, wi = 0 rad/s (from rest)

Final angular velocity, wf = 2.86 rad/s

a = (2.86 - 0)/5

= 0.572 rad/s^2

3 0
3 years ago
Read 2 more answers
Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.
Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge q_{1}= 3.8\times10^{-6}\ C

Second charge q_{2}=3.2\times10^{-6}\ C

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

E_{1}=E_{2}

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}

\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}

\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}

x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}

Put the value into the formula

x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}

x=1.69\ m

Hence, The distance is 1.69 m.

5 0
3 years ago
g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k
Salsk061 [2.6K]

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s

Now, we will apply the law of conservation of momentum:

m_1v_1 = m_2v_2

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s

Now, we again use the third equation of motion for the upward motion of the ball:

2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\

<u>h = 3.5 m</u>

6 0
2 years ago
If the mass of one of two objects is increased, the force of attraction between them will
Sonja [21]
It will increase..........................
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3 years ago
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