When preparing an emergency supply kit for a family, which is the minimun number of days the supplies shoulds last

A dynamite blast propels a heavy rock straight up with a launch velocity of 160ft/sec (about 109 mph). Write a position for the

Flura [38]

A) Using:
2as = v² - u², where v will be 0 at max height
s = -(160)² / 2 x -32.174
s = 397.8 ft

b) Using:
s = ut + 1/2 at²
256 = 160t - 16.1t²
solving for t,
t = 2.0, t = 7.9
Now, v = u + at, for both times:
v(2) = 160 - 32.174(2)
v(2) = 95.7 ft/sec on the way up

v(7.9) = 160 - 32.174(7.9)
v(7.9) = -94.3 ft/sec; 94.3 ft/sec on the way down

c) -32.174 ft/s², which is the acceleration due to gravity.

d) s = 0
0 = 160t - 1/2 x 32.174t²
t = 9.94 seconds

3 0

4 years ago

A mass is attached to an ideal spring. At time t = 0 the spring is at its natural length and the mass is given an initial veloci

JulijaS [17]

Answer:

t = T/4

Explanation:

The power delivered to the mass by the spring is work done by the spring per second.

$P = \frac{dW}{dt}$

The work done by the spring is equal to the elastic potential energy stored in the spring.

$U = \frac{1}{2}kx^2$

The maximum energy stored in the spring is at the amplitude of the oscillation.

$U_{max} =\frac{1}{2}kA^2$

So the first time the mass reaches to its amplitude can be found by the following equation of motion:

$x = A\cos(\omega t + \phi)\\\phi = \pi/2 ~because ~at ~t= 0, ~ x = 0\\0 = A\cos(0 + \pi/2)\\x = A\cos(\omega t + \pi/2)$

When the mass reaches the amplitude:

$A = A\cos(\omega t + \pi/2)\\1 = \cos(\omega t + \pi/2)\\\omega t + \pi/2 = \pi$

because cos(π) = 1.

$\omega t = \pi/2$

Using ω = 2π/T,

$\omega t = \pi/2\\\frac{2\pi}{T}t = \pi/2\\t = \frac{T}{4}$

4 0

4 years ago

This diagram shows that there are high tides on the side of Earth closest to the moon and on the side opposite the moon, and low

hram777 [196]

Answer:

Explanation:

The water moving into the high tide areas has moved out of the low tide areas. Water is incompressible.

5 0

3 years ago

a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi

zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

v = u + at

a = (v - u) / t

Substituting the given values in the above equation

a = (10 - 0) / 3

= 3.33 m/s²

Using the second equations of motion

s = ut + 1/2 at²

= 0 + 0.5 x 3.33 x 3²

= 15 m

The force exerted by the car

F = m x a

= 1500 Kg x 3.33 m/s²

= 5000 N

The work done by the car,

W = F x S

= 5000 N x 15 m

= 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

P = W / t

= 75,000 J / 3 s

= 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0

4 years ago

If a vehicle accelerating at 2.7 m/s2what is it's velocity at 20 meters 12.63m/s,1.03m/s,10.39m/s,6.39m/s

netineya [11]

Answer:

The final velocity of the vehicle is 10.39 m/s.

Explanation:

Given;

acceleration of the vehicle, a = 2.7 m/s²

distance moved by the vehicle, d = 20 m

The final velocity of the vehicle is calculated using the following kinematic equation;

v² = u² + 2ah

v² = 0 + 2 x 2.7 x 20

v² = 108

v = √108

v = 10.39 m/s

Therefore, the final velocity of the vehicle is 10.39 m/s.

5 0

3 years ago