The acceleration is given as:

a = g sin(30°) where g is the gravitational acceleration

For g = 10 m/s^2, we get

a = 10 sin(30°) = 10 * 1/2 = 5 m/s^2

7 0

2 years ago

A rock of mass 0.340 kg is spun horizontally at the end of a wire that has a diameter of 1.00 mm. If the wire gets stretched by

borishaifa [10]

Answer:

Y = 78.13 x 10⁹ Pa = 78.13 GPa

Explanation:

First we will find the centripetal force acting on the wire as follows:

F = mv²/r

where,

F = Force = ?

m = mass of rock = 0.34 kg

v = speed = 19 m/s

r = length of wire

Therefore,

F = (0.34)(19)²/r

F = 122.74/r

now, we find cross-sectional area of wire:

A = πd²/4

where,

A = Area = ?

d = diameter of wire = 1 mm = 0.001 m

Therefore,

A = π(0.001)²/4

A = 7.85 x 10⁻⁷ m²

Now, we calculate the stress on wire:

Stress = F/A

Stress = (122.74/r)/(7.85 x 10⁻⁷)

Stress = 1.56 x 10⁸/r

Now, we calculate strain:

Strain = Δr/r

where,

Δr = stretch in length = 2 mm = 0.002 m

Therefore,

Strain = 0.002/r

now, for Young's modulus (Y):

Y = Stress/Strain

Y = (1.56 x 10⁸/r)/(0.002/r)

8 0

2 years ago

An object whose weight is 100 lbf experiences a decrease in kinetic energy of 500 ft lbf and an increase in potential energy of

Aleonysh [2.5K]

Answer:

a) $v_2=35.60ft/sec$

b) $h_2=45ft$

Explanation:

From the question we are told that:

Weight $W=100lbf$

Decrease in kinetic energy $dK.E= 500 ft lbf$

Increase in potential energy $dP.E =1500 ft lbf.$

Velocity $V_1=40$

Elevation $h=30ft$

$g=32.2 ft/s2$

Generally the equation for Change in Kinetic Energy is mathematically given by

$dK.E=\frac{1}{2}m(v_1^2-v_2^2)$

$500=\frac{1}{2}*\frac{100}{32.2}(v_1^2-v_2^2)$

$v_2=35.60ft/sec$

Generally the equation for Change in Potential Energy is mathematically given by

$dP.E=mg(h_2-h_1)$

$1500=mg(h_2-h_1)$

$h_2=45ft$

4 0

2 years ago