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3 years ago

What is the velocity of a quarter drooped from a towards after 10secounds

1 answer:

6 0

U=0
t=10 
a=9.8m/s/s 
v is velocity (the tower must be very high to be able to fall for 10 seconds!!!) 

you work out the result now

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What are the four planets that orbit closets to the sun

Elden [556K]

Answer:

Explanation:The four closest to the sun — Mercury, Venus, Earth and Mars — are the terrestrial planets. They have rocky surfaces enclosed by relatively shallow atmospheres.

8 0

3 years ago

Two inductors, L1 and L2, are in parallel. L1 has a value of 25 mH and L2 a value of 50 mH. The parallel combination is in serie

Bond [772]

Answer:

Explanation:

For parallel inductors ,

$\frac{1}{L_R} = \frac{1}{L_1} +\frac{1}{L_2}$

$\frac{1}{L_R} =\frac{1}{25} +\frac{1}{50}$

$L_R=16.67 mH.$

For series combination

Total inductance

= 16.67 + 20

= 36.67 mH .

reactance of total inductance at 300 kHz

= ω $L_{total}$ where ω is angular frequency

= 2πf $L_{total}$

= 2 x 3.14 x 300 x 10³ x 36.67 x 10⁻³

= 69.1 x 10³ ohm

Total rms current = Vrms / reactance

= 60 / 69.1 x 10³ A

= .87 x 10⁻³ A

= .87 mA

7 0

2 years ago

A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary

Ann [662]

The distance traveled by the wood after the bullet emerges is 0.16 m.

The given parameters;

mass of the bullet, m = 23 g = 0.023 g
speed of the bullet, u = 230 m/s
mass of the wood, m = 2 kg
final speed of the bullet, v = 170 m/s
coefficient of friction, μ = 0.15

The final velocity of the wood after the bullet hits is calculated as follows;

$m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s$

The acceleration of the wood is calculated as follows;

$\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2$

The distance traveled by the wood after the bullet emerges is calculated as follows;

$v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m$

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.

Learn more here:brainly.com/question/15244782

7 0

3 years ago

If it takes 200 joules of energy to lift a bucket of water 3 meters in 2 seconds, how much power would be required to do the sam

Harlamova29_29 [7]

200 joules of work energy are involved. That's all we need to know to answer the question. Once we know that 200 joules of work energy are involved, we don't care what was lifted, or how far, or how long it took, or how many people worked on it, or how much they were paid, or what was the distribution of their gender identities, or the ethnic diversity among the team. or what day each of them celebrates as their sabbath. Any other information besides the 200 joules is only there to distract us, and see whether we're paying attention.

Power = (work or energy) / (time to do the work or move the energy)

Power = (200 joules) / (5 seconds)

Power = 40 watts

3 0

3 years ago

Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the

balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:

$a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)$

Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:

$v_{f} ^{2} -v_{o} ^{2} = 2* a* \Delta x (2)$

Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

$\Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)$

We can find the time needed to come to an stop, applying the definition of acceleration, as follows:

$v_{f} = v_{o} + a*\Delta t (4)$

Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
Replacing a and v₀ as we did in (3), we can solve for Δt as follows:

$\Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s (5)$

7 0

2 years ago