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3 years ago

How much force is necessary to stretch a spring 0.5 m when the spring constant is 190N/m

Physics

1 answer:

Sauron [17]3 years ago

4 0

Answer:

190×0.5=95N

Explanation:

The spring constant is 190 N/m . Therefore the force is necessary to stretch a spring 0.5 m when the spring constant is 190 N/m is 95 N

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•• Your roommate is working on his bicycle and has the bike upside down. He spins the 60-cm-diameter wheel, and you notice that

Y_Kistochka [10]

Answer:

$v=0.57\frac{m}{s}$

$a_c=10.83\frac{m}{s^2}$

Explanation:

We have an uniform circular motion, therefore, the pebble’s speed is given by the distance traveled in a revolution $(2\pi r)$ and the period (T), since this is the time pebble’s takes to complete a revolution:

$v=\frac{2\pi r}{T}$

The period is inversely proportional to the frequency:

$T=\frac{1}{f}$

So, we have:

$v=\frac{2\pi r}{\frac{1}{f}}\\v=2\pi rf\\$

Recall that the radius is the half of the diameter and one revolution per is equal to one Hz:

$v=2\pi (30*10^{-2}m)(3Hz)\\v=0.57\frac{m}{s}$

The centripetal acceleration is defined as:

$a_c=\frac{v^2}{r}\\a_c=\frac{(0.57\frac{m}{s})^2}{30*10^{-2}m}\\\\a_c=10.83\frac{m}{s^2}$

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3 years ago

A 20 kg sled stars at the top of a hill which is 10 m above the bottom and slides a distance of 50 m, ending at the bottom of th

katrin [286]

Answer:

80 Joules.

Explanation:

Given that a 20 kg sled stars at the top of a hill which is 10 m above the bottom and slides a distance of 50 m, ending at the bottom of the hill with a speed of 8 m/s. During the slide, the work done by gravity is about ?

The parameters given are:

Mass M = 20 kg

Height h = 10 m

Velocity V = 8 m/s

The workdone by gravity will be equal to the total energy

At the bottom of the hill, the total energy will be equal to the maximum kinetic energy.

Maximum K.E = 1/2mv^2

K.E = 1/2 × 20 × 8

K.E = 80J

Therefore, the workdone by the gravity is equal to 80 Joules.

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3 years ago

A model rocket blasts off and moves upward with an acceleration of 12 m/s2 until it reaches a height of 26 m. at that height, it

Diano4ka-milaya [45]

initial acceleration of rocket is given as

a = 12 m/s^2

h = 26 m

now we can use kinematics to find its speed

$v_f^2 - v_i^2 = 2 a h$

$v_f^2 - 0 = 2 * 12*26$

$v_f = 24.98 m/s$

now after this it will be under free fall

so now again using kinematics

$v_f = 0$

at maximum height

$v_f^2 - v_i^2 = 2 a s$

$0 - 24.98^2 = 2 * (-9.8)* h$

$h = 31.8 m$

total height from the ground = 31.8 + 26 = 57.8 m

Part b)

now after reaching highest height it will fall to ground

So in order to find the speed we can use kinematics again

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2*9.8*57.8$

$v_f = 33.67 m/s$

Part c)

first rocket accelerate to reach height 26 meter and speed becomes 24.98 m/s

now we have

$v_f - v_i = a t$

$24.98 - 0 = 12*t_1$

$t_1 = 2.1 s$

after this it will reach to highest point and final speed becomes zero

$v_f - v_i = at$

$0 - 24.98 = -9.8 * t$

$t_2 = 2.55 s$

now from this it will fall back to ground and reach to final speed 33.67 m/s

now we have

$v_f - v_i = at$

$33.67 - 0 = 9.8 * t$

$t_3 = 3.44 s$

so total time is given as

<em>t = 3.44 + 2.55 + 2.1 = 8.1 s</em>

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3 years ago

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Answer:

e is correct

Explanation:

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