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Ksenya-84 [330]

2 years ago

5

Taylor Swift weighing 794 N gets on an elevator. The elevator uses 313 W of power to lift the person 22.0 m. How much time did t

he elevator take to lift the Taylor?

1 answer:

mezya [45]2 years ago

3 0

Answer:

55.80s

Explanation:

Power is calculated using the expression

Power = Work done/Time

Workdone= Force ×distance

Workdone = 794×22

Work done = 17468Joules

From the power formula

Time = Workdone/Power

Time = 17468/313

Time = 55.80seconds

The elevator takes 55.80seconds to life the Taylor

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A sphere with a charge q is fixed at the bottom left corner of the right triangle shown in the figure. Points P and R are at the

Alexus [3.1K]

Answer:

the final potential energy of this system is 3U0/10

Explanation:

We are given

charge at left end and another test charge at point p

Potential energy is given by = $\frac{k*Q1*Q2 }{R}$

where k is electrostatics constant = $9 *10^9$

Q1 = first charge , Q2= test charge

R= distance between charges

potential at point p

U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1

now the test charge moves to point R

using Pytahgoreou theorem

R(distance) = $\sqrt{8^2 + 6^2}$ = 10

New Potential energy

U1 = kq1*q2 / 10

substituting kq1q2 = 3U0 from 1

U1 = 3U0/10

So this is the final potential energy of this system.

5 0

2 years ago

Problem 9: Suppose you wanted to charge an initially uncharged 85 pF capacitor through a 75 MΩ resistor to 90.0% of its final vo

Bingel [31]

Answer:

$t=14.678\times 10^{-3}s$

Explanation:

Given:

Capacitance, C = 85 pF = 85 × 10⁻¹² F

Resistance, R = 75 MΩ = 75×10⁶Ω

Charge in capacitor at any time 't' is given as:

$Q=Q_o(1-e^{-\frac{t}{RC}})$

where,

Q₀ = Maximum charge = CE

E = Initial voltage

t = time

also, Q = CV

V= Final voltage = 90% of E = 0.9E

thus, we have

$C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})$

or

$0.9=1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}$

or

$e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}=1-0.9=0.1$

taking log both sides, we get

$ln(e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})=ln(0.1)=ln(\frac{1}{10})$

or

$-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)$

or

$t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}$

or

$t=14.678\times 10^{-3}s$

3 0

3 years ago

In Fig. 4-41, a ball is thrown up onto a roof, landing 4.00 s later at height h ???? 20.0 m above the release level. The ball’s

Yanka [14]

"Fig is attacted with answer"

Answer:

a) d = 33.72 m

b) $v_{i}$ = 26 m/s

c) β = 71.08°

Explanation:

a)

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

time = t = 4.00 s

Height = h = 20 m

Angle = θ = 60°

Horizontal distance = d = ?

Using 2nd equation of motion

$h = v_{y_{f}}t + \frac{1}{2}gt^{2}$

-20 = $v_{y_{f}}$ (4) + 0.5(-9.8)(4)²

$v_{y_{f}}$ (4) = 58.4

$v_{y_{f}}$ = 14.6 m/s

This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use

$v_{f}$ = $v_{y_{f}}$ / sinθ

$v_{f}$ = 14.6 / sin 60

$v_{f}$ = 16.86 m/s

$v_{x_{f}}$ = $v_{f}$ cosθ

$v_{x_{f}}$ = 16.86 cos 60

$v_{x_{f}}$ = 8.43 m/s

To calculate the horizontal distance

d = $v_{y_{f}}$ t

d = (8.43)(4)

d = 33.72 m

b)

We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

So,

$v_{x_{f}}$ = $v_{x_{i}}$

but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.

So,

g = ( $v_{y_{f}}$ - $v_{y_{i}}$ ) / t

9.8 = 14.6 - $v_{y_{i}}$ ) / 4

$v_{y_{i}}$ = 24.6 m/s

$v_{i}$ = $\sqrt{v_{x_{i}}^{2}+v_{y_{i}}^{2} }$

$v_{i}$ = $\sqrt{8.43^{2}+24.6^{2}}$

$v_{i}$ = 26 m/s

c)

cos β = $v_{x_{i}}$ / $v_{i}$

β = cos⁻¹ (8.43 / 26)

β = 71.08°

3 0

2 years ago

How can you predict the action of the element

patriot [66]

Stark contrast to paths on energy surfaces or even mechanistic reactions, rule-based and inductive computational approaches to reaction prediction mostly consider only overall transformations. Overall transformations are general molecular graph rearrangements reflecting only the net change of several successive mechanistic reactions. For example, Figure 1 shows the overall transformation of an alkene interacting with hydrobromic acid to yield the alkyl bromide along with the two elementary reactions which compose the transformation.

8 0

2 years ago

Which vector should be negative?

pentagon [3]

Answer:

The velocity of a falling object

Explanation:

The positive X axis is towards right and positive Y axis is towards up, so North direction is positive

A vector with less than 1 magnitude is not negative, because its magnitude may be in between 0 and 1 which is positive vector.

Any vector whose magnitude is greater than 1 is never be a negative vector.

The velocity of a falling object is towards bottom, that is towards negative Y axis. So that vector is negative.

7 0

2 years ago