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2 years ago

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2.10-kg box is moving to the right with speed 8.50 m>s on a horizontal, frictionless surface. At t = 0 a horizontal force is

applied to the box. The force is directed to the left and has magnitude F1t2 = 16.00 N>s22t2. (a) What distance does the box move from its position at t = 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at t = 3.00 s?

2 answers:

larisa86 [58]2 years ago

8 0

Explanation:

Below is an attachment containing the solution.

vodka [1.7K]2 years ago

4 0

Answer:

a) the distance moved by the object is 14m

b) the speed of the box at 3.00s is - 18.5m/s

The negative sign in the answer indicates that the speed of the box is in opposite direction. Therefore, the direction of the box is towards left.

Explanation:

see the attached file

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A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 22.6 s how long is the slope? As

Nina [5.8K]

Answer:

648 m

Explanation:

Let the length of slope be 'L' m.

Given:

Initial velocity of the child (u) = 0

Angle of inclination of the slope (x) = 15.0°

Time taken to reach the bottom (t) = 22.6 s

Acceleration due to gravity (g) = 9.8 m/s²

No frictional forces act.

Now, acceleration acting on the child along the slope can be determined by resolving the acceleration due to gravity along the slope.

So, acceleration along the slope is given as:

$a_{slope}=g\sin(x)=9.8\sin(15)=2.536\ m/s^2$

Now, use to equation of motion along the slope to find the length of slope.

Therefore,

$L=ut+\frac{1}{2}a_{slope}t^2\\\\L=0+\frac{1}{2}(2.536)(22.6)^2\\\\L=1.268\times 510.76=647.6\approx 648\ m$

Therefore, the slope is 648 m long.

Option (1) is correct.

4 0

2 years ago

Determine the length of a copper wire that has a resistance of 0.172 ? and cross-sectional area of 7.85 × 10-5 m2. The resistivi

KonstantinChe [14]

Answer:

Length of copper wire, l = 785 meters

Explanation:

Given that,

Resistance of the copper wire, R = 0.172 ohms

Area of cross section, $A=7.85\times 10^{-5}\ m^2$

Resistivity of copper, $\rho=1.72\times 10^{-8}\ \Omega-m$

The resistance of a wire is given by :

$R=\rho\dfrac{l}{A}$

$l=\dfrac{RA}{\rho}$

$l=\dfrac{0.172\ \Omega\times 7.85\times 10^{-5}\ m^2}{1.72\times 10^{-8}\ \Omega-m}$

l = 785 meters

So, the length of the copper wire is 785 meters. Hence, this is the required solution.

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2 years ago

Think about how geothermal energy is captured and used. Explain how geothermal energy shows the flow of thermal energy from hot

kumpel [21]

Answer:

People can capture geothermal energy through: Geothermal power plants, which use heat from deep inside the Earth to generate steam to make electricity. Geothermal heat pumps, which tap into heat close to the Earth's surface to heat water or provide heat for buildings

When the weather is cold, the water or refrigerant heats up as it travels through the part of the loop that's buried underground. Once it gets back above ground, the warmed water or refrigerant transfers heat into the building. The water or refrigerant cools down after its heat is transferred.

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2 years ago

You throw a tennis ball (mass 0.0570 kg) vertically upward. It leaves your hand moving at 15.0 m/s. Air resistance cannot be neg

Deffense [45]

Answer:195 J

Explanation:

Given

mass of ball $m=0.0570\ kg$

ball leaves the hand with $u=15\ m/s$

maximum height reached by ball $h=8\ m$

Initial Mechanical energy when ball just leaves the hand

$M.E._1=(P.E.+K.E.)_1$

$M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1$

considering hand to be datum so h_1=0[/tex]

so Potential energy at ground is zero

$M.E._1=\frac{1}{2}\times m\times (15)^2$

$M.E._1=6.41\ J$

Mechanical Energy at highest point

$(M.E.)_2=(P.E.+K.E.)_2$

at highest Point velocity is zero

$(M.E.)_2=mgh_2+0$

$(M.E.)_2=0.0570\times 9.8\times 8$

$(M.E.)_2=4.46\ J$

Decrease in Mechanical energy

$(M.E.)_1-(M.E.)_2=6.41-4.46$

$(M.E.)_1-(M.E.)_2=1.95\ J$

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2 years ago

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

BartSMP [9]

Answer:

35%

Explanation:

Assuming no external torques present during the collision between the record and the turntable, total angular momentum must be conserved.

For a rotating body with some angular velocity and moment of inertia, the angular momentum can be expressed as follows;

L = I* ω

So, as initial angular momentum and final angular momentum must be the same, we have:

Li = Lf

⇒ I₁ * ω₁ = I₂ * ω₂ (1)

where I₁ is the rotational inertia of the turntable, and I₂, is the combined rotational inertia of the turntable and the record:

I₂ = I₁ + 0.54 I₁ = 1.54 I₁

We can solve (1) for the new common angular speed, as follows:

ω₂ = ω₁ / 1.54 (2)

The initial rotational kinetic energy is given by definition for the following equation:

Kroti = 1/2 * I₁ * ω₁² (3)

The final rotational kinetic energy takes into account the new rotational inertia and the common final angular speed:

Krotf = 1/2* I₂ * ω₂² = 1/2* 1.54 I₁* (ω₁/1.54)² (4)

Dividing both sides in (3) and (4), we get:

Krotf/Kroti = 1/1.54 = 0.65

This means that the final rotational kinetic energy, has reduced to 0.65 of the initial value, or that has lost 35% of the initial kinetic energy.

8 0

2 years ago