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anzhelika [568]
3 years ago
11

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.
Upon reaching the water, how does the kinetic energy of the smaller child compare with that of the larger child?
Physics
1 answer:
FinnZ [79.3K]3 years ago
7 0

Answer:

THE KINETIC ENERGY OF THE SMALLER CHILD IS LESS THAN THAT OF THE BIGGER CHILD

Explanation: Kinetic energy is the energy that is exerted on a body that is in motion, kinetic energy is affected by both the mass of the object and the velocity of the object.

Mathematically,Kinetic energy is represented as follows;

K.E=1/2MV^{2}

Where M represents the mass of the object in kilograms and V represents velocity of the moving object measured in meters per seconds.

The higher the weight of the object the higher the kinetic energy of the object which means the bigger child will have a higher kinetic energy than the smaller child.

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A CD has a mass of 17 g and a radius of 6.0 cm. When inserted into a player, the CD starts from rest and accelerates to an angul
Stells [14]

Answer:

τ =9.41 * 10⁻⁴ N*m

Explanation:

Kinematics of the CD

The CD rotates with constant angular acceleration and its angular acceleration is calculated as follows:

\alpha = \frac{\omega_{f}- \omega_{i}}{t}   Formula (1)

Where:

α : angular acceleration. (rad/s²)

ωf: final angular velocity  (rad/s)

ωi : initial angular velocity  (rad/s)

t = time interval (s)

Data

ωf= 20 rad/s

ωi =0

t = 0.65 s.

Calculating of the angular acceleration of the CD

We replace data in the formula (1)

\alpha = \frac{20 -0}{0.65}

α  = 30.77 rad/s²

Newton's second law  in rotation:

F = ma has the equivalent for rotation:

τ = I * α   Formula  (2)

where:

τ : It is the net torque applied to the body.  (N*m)

I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Calculating of the moment of inertia  of the CD

The moment of inertia of a disk with respect to an axis perpendicular to the plane  and passing through its center is calculated by the following formula:

I = (1/2) M*R² Formula (3)

Data

M= 17 g = 17/1000 kg = 0.017 kg  : CD mass

R= 6.0 cm= 6/100 m = 0.06 m : CD  radius

We replace data in the formula (3) :

I = (1/2) ( 0.017 kg)*(0.06 m)² = 3.06 * 10⁻⁵ kg*m²

Calculating of the  net torque acting on the CD

Data

α  = 30.77 rad/s²

I =  3.06 * 10⁻⁵ kg*m²

We replace data in the formula (2) :

τ = I * α

τ =( 3.06 * 10⁻⁵ kg*m²) * (30.77 rad/s²)

τ =9.41 * 10⁻⁴ N*m

3 0
3 years ago
Driving 30.7 m/s in your car you see a dog up ahead and slam on your brakes. You stop just before hitting the dog after skidding
Nitella [24]

Answer: 2.74

Explanation:

We can solve this problem using the stopping distance formula:

d=\frac{(V_{o})^{2}}{2 \mu g}

Where:

d=132.1 m is the distance traveled by the car before it stops

V_{o}=30.7 m/s is the car's initial velocity

\mu is the coefficient of friction between the road and the tires

g=9.8 m/s^{2} is the acceleration due gravity

Isolating \mu:

\mu=\frac{2dg}{(V_{o})^{2}}

Solving:

\mu=\frac{2(132.1 m)(9.8 m/s^{2})}{(30.7 m/s)^{2}}

\mu=2.74 This is the coefficient of friction

7 0
3 years ago
A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of
tekilochka [14]

The largest mass of cargo the balloon can lift is 791.06 kg

First, we need to calculate the mass of helium.

Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.

The volume of the balloon also equals the volume of helium present.

Now, the mass of helium m = density of helium, ρ × volume of helium, V

m = ρV

Since ρ = 0.179 kg/m³

m = ρV

m = ρ4πr³/3.

m = 0.179 kg/m³ × 4π(7.15 m)³/3

m = 0.179 kg/m³ × 4π(365.525875 m³)/3

m = 0.179 kg/m³ × 1462.1035π m³/3

m = 261.7165265π/3 kg

m = 822.207/3 kg

m = 274.07 kg

Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.

The weight of this mass W = Mg where g = acceleration due to gravity.

The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.

W' = m'g

Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'

We know that the volume of air displaced, V' = volume of balloon, V

So, V' = V = 4πr³/3.

Since the density of air, ρ' = 1.29 kg/m³,

m' = ρ'V

m = 1.29 kg/m³ × 4π(7.15 m)³/3

m = 1.29 kg/m³ × 4π(365.525875 m³)/3

m = 1.29 kg/m³ × 1462.1035π m³/3

m = 1886.113515π/3 kg

m = 5925.4/3 kg

m = 1975.13 kg

So, the net weight W" that the balloon can lift is W" = W' - W = m'g - Mg = (m' - M )g = (1975.13 kg - 1184.07 kg)g = 791.06g.

So, the net mass m" = W"/g = 791.06g/g = 791.06 kg

This net mass is the largest mass of cargo that the balloon can lift.

Thus, the largest mass of cargo the balloon can lift is 791.06 kg

Learn more about balloons here:

brainly.com/question/21890581

8 0
3 years ago
A closed, rigid container holding 0.2 moles of a monatomic ideal gas is placed over a Bunsen burner and heated slowly, starting
Georgia [21]

Answer:

a) 2250 J

b) 0 J

c) 2250 J

Explanation:

a) Since, the process is isochoric

the change in internal energy

\Delta U = n C_v(T_f-T_i)

Here, n = 0.2 moles

Cv = 12.5 J/mole.K

We have to find T_f so we can use gas equation as

\frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2    [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K

So,  \Delta U= 0.2\times12.5(1200-300)\\=2250 J

b) Since, the process is isochoric no work shall be done.

c) By first law of thermodynamics we have

\Delta U = Q-W\\Since, W = 0\\\Delta U = Q\\Therefore, Q = 2250 J

Since, Q is positive 2250 J of heat will flow into the system.

6 0
3 years ago
A car goes around a curve. The road is so so icy, so the centripetal force ( decreases, increases, stays the same), so the veloc
Katen [24]
The velocity increases because car goes vroom vroom French fries
8 0
3 years ago
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