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1 year ago

PLEASE HELP!!!!!!! MT TIME FOR MY TEST IS ALMOST OVER!!!

Physics

1 answer:

pav-90 [236]1 year ago

3 0

Answer:

50 N

Explanation:

Let the natural length of the spring = L

100 = k(40 - L) (1)

200 = k(60 - L) (2)

(2)/(1): 2 = (60 - L)/(40 - L)

60 - L = 2(40 - L)

60 - L = 80 - 2L

2L - L = 80 - 60

L = 20

Sub it into (1):

100 = k(40 - 20) = 20k

k = 100/20 = 5 N/in

Now

X = k(30 - L) = 5(30 - 20) = 50 N

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A certain dielectric with a dielectric constant = 24 can withstand an electric field of 4 107 V/m. Suppose we want to use this d

mr_godi [17]

Answer:

Explanation:

Electric field E = 4 x 10⁷ V / m

Dielectric constant k = 24

capacitance of capacitor

C = kε₀ A / d

d = plate separation

A = plate area

C = .89 x 10⁻⁶

V / d = electric field

for minimum d , electric field will be maximum

V / d = 4 x 10⁷

1930 / d = 4 x 10⁷

d = 1930 / 4 x 10⁷

d = 482.5 x 10⁻⁷ m

= 48.25 x 10⁻⁶ m

C = kε₀ A / d

.89 x 10⁻⁶ = 24 ε₀ A / d

A = .89 x 10⁻⁶ X d / 24 ε₀

A = .89 x 10⁻⁶ X 48.25 x 10⁻⁶ / 24 x 8.85 x 10⁻¹²

= 42.9 / 212.4

= .2019 m²

8 0

3 years ago

Read 2 more answers

2. Kevin works as a janitor, and he is pushing a fully-

dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

weight of the load, w = 557 N
force applied , F = 410 N
angle of force, = 15°
coefficient of kinetic friction = 0.46
distance moved, d = 6.5 m

The net horizontal force on the recycling bin is calculated as follows;

$Fcos\theta - F_k = ma$

where;

m is the mass of the recycling bin
$F_k$ is the frictional force

W = mg

$557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg$

The net horizontal force on the recycling bin is calculated as;

$Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2$

The time taken for him to move the bin 6.5 m is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2} \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s$