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ss7ja [257]
3 years ago
6

Use the drop-down menu to answer the question. according to "take a closer look", what type of lens is used to correct nearsight

edness?
Physics
1 answer:
Reil [10]3 years ago
6 0

Answer: concave lens

Explanation:

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What do you notice about where the independent and dependent variables are located on each hypothesis above?!
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Answer:

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4 0
3 years ago
A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately
Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car V_{c} = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

V_{f}  = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

V_{f} =  V_{i} + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁  = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁  will be

d_{c} = V_{c} × t₁

we substitute

d_{c} = 22.352 × 7

d_{c}  = 156.46 m

now distance between polive car and speeding car

Δd =  d_{c} - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( V_{f} - V_{c} )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = V_{f} × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0
3 years ago
A 2.3 kg , 20-cm-diameter turntable rotates at 110 rpm on frictionless bearings. Two 460 g blocks fall from above, hit the turnt
boyakko [2]

Answer:

The correct solution is "64 RPM".

Explanation:

The given values are:

Mass,

M = 2.3 kg

Diameter,

D = 20 cm

i.e.,

   = 0.2 m

Rotates at,

N = 110 rpm

Mass of block,

m = 460 g

i.e.,

   = 0.46 kg

According to angular momentum's conservation,

⇒  I_1\omega_1=I_2\omega_2

then,

⇒  I_1=\frac{1}{2}MR_2

On substituting the values, we get

⇒      =\frac{1}{2}\times 2.3\times (0.1)^2

⇒      =\frac{1}{2}\times 0.023

⇒      =0.0115 \ kg \ m^2

Now,

⇒  I_2=I_1+2mR^2

        =0.0115+2\times 0.46\times (0.1)^2

        =0.0115+0.0092

        =0.02 \ kg \ m^2

then,

⇒  0.0115\times 110=0.02\omega_2

⇒              1.265=0.02\omega_2

⇒                  \omega_2=\frac{1.265}{0.02}

⇒                       =63.25 \ or \ 64 \ RPM

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3 years ago
What unit do we use for pressure
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Answer:

Pascal

Explanation:

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