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2 years ago

URGENT!!!!

Physics

1 answer:

Nastasia [14]2 years ago

3 0

Answer:

8 cm

Ladder is safer than to walk on it

Explanation:

Streched = Final length - initial length

= 66 - 58

= 8 cm

When your are walking along frozen ice pond your weight is not distributed .It's acting on directly on the ice pond because of contact Area is too small between pond and you. There for According to pressure equation ,

P = F/A

When area is less pressure is less.

If use ladder instead of going by foot. Contact area will be high between pond and ladder. There for pressure is reduced .

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Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight. Use this co

natima [27]

Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight is given below.

Explanation:

Measure unstretched length of spring, L. E.g. L = 0.60m.

Set mass to a convenient value (e.g. m = 0.5kg).

Hang mass.

Measure new spring length, L'. E.g. L' = 0.70m.

Calculate extension: e = L' - L = 0.70 – 0.60 = 0.10m

Use mg = ke (in equilibrium weight = tension)

k = mg/e

Don't know what value you are using for example. Suppose it is 10N/kg (same thing as 10m/s²).

k = 0.5*10/0.10 = 50 N/m

Repeat for a few different masses. (L always stays the same.)

Take the average of your k values.

5 0

1 year ago

Read 2 more answers

A machine produces photo detectors in pairs. Tests show that the first photo detector is acceptable with probability 3/5. When t

klasskru [66]

Answer:

a. $a. \ \frac{7}{25}$

$b.\ \ \ P(D_1D_2)=\frac{6}{25}$

Explanation:

a. Find the probability that exactly one photo detector of a pair is acceptable:

Let $A_i=i^{th}$ photo is accepted and the probability $D_i=i^{th}$ is defected.

Therefore:

$P(A_i)=3/5,\ P(A_2|A_1)=4/5,\ \ P(A_2|D_1)=2/5\\\\\\=P(A_1D_2)+P(D_1A_2)\\\\=\frac{3}{5}\times\frac{1}{5}+\frac{2}{5}\times\frac{2}{5}\\\\=\frac{7}{25}$

#The probability of exactly one photo detector of a pair is accepted is 7/25

b.Find the probability that both photo detectors in a pair are defective,P(D1D2):

$P(D_1D_2)=\frac{2}{5}\times \frac{3}{5}\\\\=\frac{6}{25}$

Hence, from out tree diagram,the probability that both photo detectors in a pair are defective is 6/25

4 0

2 years ago

Before leaving the house in the morning, you plop some stew in your slow cooker and turn it on Low. The slow cooker has a 160 Oh

guajiro [1.7K]

Answer:

Total charge flow through the cooker is 21600 C

Explanation:

As we know that the current flow through the cooker is given by Ohm's law

here it is given as

$V = i R$

$i = \frac{V}{R}$

$i = \frac{120}{160}$

$i = \frac{3}{4} A$

now the charge flow through it is given as

$Q = i t$

total time is t = 8 hours

$Q = \frac{3}{4}(8 \times 60 \times 60)$

$Q = 21600 C$

7 0

2 years ago

A motorcycle of mass 100 kilograms travels around a flat, circular track of radius 10 meters with a constant speed of 20 meters

JulijaS [17]

Answer:

100/10 = 10 , 10 × 10 = 100÷20 = 5

I'm pretty sure its wrong

8 0

2 years ago

Read 2 more answers

Official (Closed) - Non Sensitive

Pavlova-9 [17]

Answer:

The minimum running time is 319.47 s.

Explanation:

First we find the distance covered and time taken by the train to reach its maximum speed:

We have:

Initial Speed = Vi = 0 m/s (Since, train is initially at rest)

Final Speed = Vf = 29.17 m/s

Acceleration = a = 0.25 m/s²

Distance Covered to reach maximum speed = s₁

Time taken to reach maximum speed = t₁

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)

t₁ = 116.68 s

Using 2nd equation of motion:

s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²

s₁ = 1701.78 m = 1.7 km

Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

We have:

Final Speed = Vf = 0 m/s (Since, train is finally stops)

Initial Speed = Vi = 29.17 m/s (The train must maintain max. speed for min time)

Deceleration = a = - 0.7 m/s²

Distance Covered to stop = s₂

Time taken to stop = t₂

Using 1st equation of motion:

Vf = Vi + at₂

t₂ = (Vf - Vi)/a

t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)

t₂ = 41.67 s

Using 2nd equation of motion:

s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²

s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²

s₂ = 607.78 m = 0.6 km

Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

The remaining distance is:

s₃ = 7 km - s₂ - s₁

s₃ = 7 km - 0.6 km - 1.7 km

s₃ = 4.7 km

Now, for uniform speed we use the relation:

s₃ = vt₃

t₃ = s₃/v

t₃ = (4700 m)/(29.17 m/s)

t₃ = 161.12 s

So, the minimum running time will be:

t = t₁ + t₂ + t₃

t = 116.68 s + 41.67 s + 161.12 s

5 0

2 years ago