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3 years ago

Can I get answers 14 through 18 please? Plz show work number 18!!!

Physics

1 answer:

kicyunya [14]3 years ago

8 0

14). c
15). c
16). a little less than d
17). a

18).
-- The parallel combination of 4Ω, 6Ω, and 10Ω is 1.935Ω .

-- The battery sees 1.935Ω in series with 2Ω. That's 3.935Ω .

-- The voltage across the 3 parallel resistors is (1.935/3.935)
of the total battery voltage. That's about 5.9 volts.

-- I = V/R .
The current in the 10Ω resistor is (5.9v/10Ω) = 0.59 Amp. (choice-a)

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Fish are hung on a spring scale to determine their mass (most fishermen feel no obligation to report the mass truthfully). (a) W

bija089 [108]

Answer:

1411.8 N/m

Explanation:

From Hooke's law;

F= Ke

Where

F= force on the spring

K= force constant

e = extension

But e= 8.50 × 10^-2m

F= weight = 12.0 kg × 10 = 120 N

K = F/e = 120/8.50 × 10^-2

K= 1411.8 N/m

7 0

3 years ago

A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at halfway

$K_f$ is the final kinetic energy, at halfway

The equation can be rewritten as

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 2.7 kg is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0.54$ is the initial height

u = 0 is the initial speed

$h_f = 0.25 m$ is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

$v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s$

The total mechanical energy of the block can be calculated from the initial conditions, and it is

$E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J$

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

$E=E_e = \frac{1}{2}kx^2$

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m$

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$