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3 years ago

5

Why do streets and highways have speed limits rather than velocity limits?

1 answer:

andrew11 [14]3 years ago

8 0

They actually DO have velocity limits. There are legal restrictions on both speed and direction.

-- Speeds are limited according to the black numbers on white signs that you see on sign-posts everywhere.

-- Directions are limited by the layout of the pavement and curbs on all the highways, avenues, roads, boulevards and streets, as well as the countless signs that say "One Way", "No Left Turn", "Keep Right", "Keep Left", etc. Violate one of these, and you get nailed as sure as if you had exceeded a posted speed limit.

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If each pull-up requires 300 J and Ben does a pull-up in 2 seconds, what is his power? 150 watts 300 watts 600 watts 750 watts

leonid [27]

Answer:

150 watts

Explanation:

300/2 = 150 watts

6 0

2 years ago

Read 2 more answers

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

olga nikolaevna [1]

Explanation:

It is given that,

Spring constant of the spring, k = 15 N/m

Amplitude of the oscillation, A = 7.5 cm = 0.075 m

Number of oscillations, N = 31

Time, t = 15 s

(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Total number of oscillation per unit time is called frequency of oscillation. Here, $f=\dfrac{31}{15}=2.06\ Hz$

$m=\dfrac{k}{4\pi^2f^2}$

$m=\dfrac{15}{4\pi^2\times 2.06^2}$

m = 0.0895 kg

or

m = 89 g

(b) The maximum speed of the ball that is given by :

$v_{max}=A\times \omega$

$v_{max}=A\times 2\pi f$

$v_{max}=0.075\times 2\pi \times 2.06$

$v_{max}=0.970\ m/s$

$v_{max}=97\ cm/s$

Hence, this is the required solution.

5 0

3 years ago

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.75 m/s. If the roof is pitched at

dalvyx [7]

Answer:

a) t = 0.528 s

b) D = 1.62 m

Explanation:

given,

speed of the baseball = 3.75 m/s

angle made with the horizontal = 35°

height of the roof edge = 2.5 m

using equation of motion

$s = ut +\dfrac{1}{2}gt^2$

$2.5 = vsin\theta \ t +\dfrac{1}{2}gt^2$

$2.5 = 3.75\ sin35^0 \ t +\dfrac{1}{2}\times 9.8 t^2$

4.9 t² + 2.15 t - 2.5 = 0

on solving the above equation

t = 0.528 s

b) D = v cos θ × t

D = 3.75 × cos 35° ×0.528

D = 1.62 m

3 0

3 years ago

A block with mass 0.470 kg sits at rest on a light but not long vertical spring that has spring constant 85.0 N/m and one end on

a_sh-v [17]

Answer: elastic potential energy = 20.27 J

Explanation:

Given that the

Mass M = 0.470 kg

Height h = 4.40 m

Spring constant K = 85 N/m

The maximum elastic potential will be equal to the maximum kinetic energy experienced by the block.

But according to conservative of energy, the maximum kinetic energy is equal to the maximum potential energy experienced by the block of mass M.

That is

K .E = P.E = mgh

Where g = 9.8m/s^2

Substitutes all the parameters into the formula

K.E = 0.470 × 9.8 × 4.4

K.E = 20.27 J

Where K.E = maximum elastic potential energy stored in the spring during the motion of the blocks after the collision which is 20.27J.

4 0

3 years ago

8.6 hours of daylight 8.6 hours of darkness

Gelneren [198K]

???????
Are you okay?

4 0

3 years ago