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zimovet [89]
2 years ago
12

Two children stand on a platform at the top of a curving slide next to a backyard swimming pool. At the same moment the smaller

child hops off to jump straight down into the pool, the bigger child releases herself at the top of the frictionless slide.(i) Upon reaching the water, how does the kinetic energy of the smaller child compare with that of the larger child?a. It is greater. It is less.
b. The two energies are equal. (ii) Upon reaching the water, how does the speed of the smaller child compare with that of the larger child is?
a. It is greater. It is less.
b. The two speeds are equal. (iii) During their motions from the platform to the water, how does the average acceleration of the smaller child compare with that of the larger child?
a. It is greater.
b. It is less.
c. The two accelerations are equal.d. The First and third are wrong
Physics
1 answer:
victus00 [196]2 years ago
5 0

1.

Answer:

a) It is less

Explanation:

By energy conservation we can say that initial potential energy of both child must be equal to the final kinetic energy of the two child.

Since initially they are at same height so we will say that initial potential energy will be given as

mgH and MgH

so the child with greater mass has more energy and hence smaller child will reach with smaller kinetic energy

2.

Answer:

b. The two speeds are equal.

Explanation:

As we know by mechanical energy conservation law we have

mgh = \frac{1}{2}mv^2

v = \sqrt{2gh}

since both child starts at same height so here they both will reach the bottom at same speed

3.

Answer:

c. The two accelerations are equal

Explanation:

Since we know that average acceleration of the motion is given as

a = \frac{v_f - v_i}{\Delta t}

since here initial and final speeds are same so they both must have same average acceleration here.

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A cubical shell with edges of length a is positioned so that two adjacent sides of one face are coincident with the +x and +y ax
Bingel [31]

Answer:

Q = ba⁴ * ε₀

Explanation:

From Gauss's Law, we know that

flux Φ = Q / ε₀

where ε₀ = 8.85e-12 C²/N·m²

and also,

Φ = EAcosθ

The field is directed along the x-axis, so that all of the flux passes through the side of the cube at x = a. This means that θ = 0º, and thus

Φ = EAcos0

Φ = EA

E = bx² meanwhile, we are interested in the point where x = a, so we substitute and then

E = ba²

Since A = a² for the cube face, we have

Q / ε₀ = E * A

Q / ε₀ = ba² * a²

so that

Q = ba⁴ * ε₀

5 0
2 years ago
An object on a number line moved from x = 15 cm to x = 165 cm and then
olasank [31]

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

v_{avg}=\frac{x_{all}}{t}

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

3 0
2 years ago
Read 2 more answers
A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

4 0
2 years ago
How much power does it take to lift a 24 kg gift box 6m above the floor in 4 s?
Mrac [35]

Answer:

<h2>5.6kW</h2>

Explanation:

Step one:

given

mass m= 24kg

distance moved= 6m

time taken= 4seconds

Step two:

Required

power

but work done is the force applied at a distance, and the power is the work done time the time taken

Work done= F*D

F=mg

W= mg*D

W=24*9.81*6

W=1412.6J

Power P= work * time

P=1412.6*4

p=5650.5W

P=5.6kW

3 0
1 year ago
A CCD has a greatest possible pixel value of 4095. what is the bit level of this CCD?
Shtirlitz [24]
<span>Bit level for a CCD (Charged coupled device) with a greatest possible pixel value of 4095:The relationship between the bit level and pixel value is given as:pixel value = 2^bit level.Most charged coupled devices (CCDs) have 8-bit, 16-bit, 32-bit levels.Using simple mathematics, we can see that 2^12 = 4096.Since the maximum number of pixels is 4095, the bit level is 12., i.e. the CCD has 12-bit level.</span>
8 0
2 years ago
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