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3 years ago

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A heavy dart and a light dart are launched vertically by identical ideal springs. Both springs were initially compressed by the

same amount. There is no significant air resistance. Which of the following statements about these darts are correct? (There could be more than one correct choice.)a. Both darts reach the same maximum height. b. The heavy dart goes higher than the light dart. c. Both darts began moving upward with the same initial speed. d. The light dart goes higher than the heavy dart At the maximum height, both darts have the same gravitational potential energy.

1 answer:

sveticcg [70]3 years ago

8 0

Answer:

D

Explanation:

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The aqueous humor in a person's eye is exerting a force of 0.242 N on the 1.21 cm2 area of the cornea. What pressure is this in

Virty [35]

Answer:

<h2><em>15.00124mmHg</em></h2>

Explanation:

Pressure is defined as the ratio of force applied to an object to its area.

Pressure = Force/Area

Given parameters

Force = 0.242N

Area = 1.21cm²

Required parameters

Pressure = 0.242/1.21

Pressure = 0.2N/cm²

Using the conversion to convert the pressure to mmHg

1N/cm² = 75.0062mmHg

0.2N/cm² = y

y = 0.2 * 75.0062

y = 15.00124mmHg

<em>Hence the pressure in mmHg is 15.00124mmHg</em>

4 0

3 years ago

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

3 years ago

car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastw

sergejj [24]

Answer:

The force of car 3 on car 2 ≈ 1810.82 N

Explanation:

The equation for the change in momentum of the two cars are;

Conservation of linear momentum

150( 2.2 - v) = 265(1.5-v)

150 × 2.2 - 265×1.5 = (150+265)v

150 × 2.2 - 265×1.5 = -67.5 = 415×v

∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East

The impulse of the net force is the amount of momentum change experienced given by the equation;

Impulse force = $m \times v_f$ - $m \times v_0$

Where;

$v_f$ = The final velocity

$v_0$ = The initial velocity

For the the 265 kg mass, we have;

$v_f$ = 0.1627 m/s

$v_0$ = 1.5 m/s

Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s

The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J

Whereby the distance moved in one second is 0.1627 m, we have;

Work done = Force × Distance = Force × 0.1627 = 294.62

Force = 294.62/0.1627 = 1810.82 N.

8 0

3 years ago

It is rate for any motion to

Gnoma [55]

Answer:

a. stay the same for very long

Explanation:

It is rare for any motion to stay the same for a very long time. The force applied on a body causes changes in the magnitude of motion.

For motion to remain constant, there must not be a net force acting on the body
All the forces on the body must be balanced.
This is very hard to come by.
Motion changes very frequently.

3 0

3 years ago

An automobile engine can produce 153 N · m of torque. Calculate the angular acceleration (in rad/s^2) produced if 85.2% of this

galina1969 [7]

Answer:

46.2 rad/s2

Explanation:

Angular acceleration works very similar to linear acceleration, it follows this equation:

$\gamma = \frac{Mt}{J}$

Where:

γ: angular acceleration

Mt: torque

J: moment of inertia of the load from its turning axis

Since we have the torque we just need the moment of inertia. We have to add together the moments of the drive shaft, tires, wheel walls and wheels.

The wheels act like disks. For disks the moment of inertia is:

$J = \frac{1}{2} * m * r^2$

$Jwheel = \frac{1}{2} = 15 * 0.18^2 = 0.243 kg*m^2$

The wheel walls act like annular rings, for these the moment of inertia is:

$J = \frac{1}{2} * m * (re^2 - ri^2)$

$Jwall = \frac{1}{2} * 2 * (0.32^2 - 0.18^2) = 0.07 kg * m^2$

The tread acts like a hoop, as in mass concentrated into a circunference, for these:

$J = m * r^2$

$Jtread = 10 * 0.33^2 = 1.09 kg*m^2$

The axle acts like a rod, which is the same as the disk:

$Jaxle = \frac{1}{2} * 14.1 * 0.02^2 = 0.0028 kg*m^2$

The drive shaft acts like a rod too:

$Jshaft = \frac{1}{2} * 31.7 * 0.032^2 = 0.016 kg*m^2$

SO, the total moment of inertia is:

J = 2*Jwheel + 2*Jwall + 2*Jtread + Jaxle + Jshaft

J = 2*0.243 + 2*0.07 + 2*1.09 + 0.0028 + 0.016 = 2.82 kg*m2

Finally the angular acceleration is:

$\gamma = \frac{0.852 * 153}{2.82} = 46.2 \frac{rad}{s^2}$

4 0

3 years ago