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marin [14]
3 years ago
8

A rock dropped from a 20 meter bridge falls into the river below. How long did it take to reach the water? What velocity did it

hit the water with.
Physics
1 answer:
Brut [27]3 years ago
6 0

Answer:

2 s, -20 m/s

Explanation:

Given:

y₀ = 20 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t and v

y = y₀ + v₀ t + ½ at²

0 = 20 + 0 + ½ (-9.8) t²

0 = 20 − 4.9 t²

t ≈ 2 s

v² = v₀² + 2a(y − y₀)

v² = 0 + 2(-9.8)(0 − 20)

v ≈ ±20 m/s

Since the rock is falling, v = -20 m/s.

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The potential difference between the plates of a capacitor is 145 V. Midway between the plates, a proton and an electron are rel
aniked [119]

Answer:

= 2.52 x 10^ 6 m/s        

Explanation:

The force that acts on charged particles between capacitor plates =

F = (q) (Δv)  ÷ d

Here,  d = distance between the two plates

          q = charge of the charged particle

         Δv = voltage

Normally, the force that makes both proton and electron released from rest, giving the charge acceleration is F=m X a. where m= mass and a = acceleration

Poting this equation with the first one, we have:

m X a =  (q) (Δv)  ÷ d

So, the acceleration of a proton when moving towards a negatively charged plate is

a = (q) (Δv)  ÷ (d) (m) {proton}

Likewise, the acceleration of an electron when moving towards a positively charged plate is

a = (q) (Δv)  ÷ (d) (m) {electron}

Dividing the proton acceleration formula by the electron acceleration formula we have:

a (proton) / a (electron) = m (proton) / m(electron)

inserting equation of motion to get distance, s

s = ut + 1/2 at^2

recall that electron travel distance, d/2

d/2 = 1/2 at^2

making t the subject of the formula

we have, t =√(d ÷ a(electron))

The distance of proton:

d/2 =  ut + 1/2 at^2 [proton}

put d/2 =  ut + 1/2 at^2 [proton} into t =√(d ÷ a(electron))

Initial speed, ui = √(d ÷ a(electron)) = (d/2) - (1/2) x (d) (a(proton) + a(electron))

since acceleration wasn't given in the question, lets use mass(elect

ron)  ÷ mass(proton) rather than use (a(proton) + a(electron))

Therefore, intial speed= 1/2√((e X Δv) ÷ m(electron)) (1- m(electron)/ m(proton))

   Note, e = 1.60 x 10^-19

           m(electron) = 9.11 X 10^-31

            m(proton) = 1.67  X 10^-27

Input these values into the formula above, initial speed, UI =  

           = 2.52 x 10^ 6 m/s          

7 0
3 years ago
URGENT
Advocard [28]

Answer:

give \\ mass(m) = 1350kg  \\ acceleration(a) = 1ms {}^{ - 2}  \ \\ sln\ \\  from \: our \: formula \\  \: f = ma \\1350kg \times 1ms {}^{ - 2}  \\ f = 1350newton

the force you applied to your car =1350N

5 0
3 years ago
22. Explain why it is better to increase the biodiversity of ecosystem.
Marizza181 [45]

Answer: because increasing biodiversity can influence ecosystem functions such as productivity and variety and even the likelihood that a particular species is discovered by a comminity

Explanation:

7 0
3 years ago
A weight lifter lifts a dumbbell a certain height in 2.0 s, while a competitor does the same workin 1.0 s. Compared to the power
Aloiza [94]

Answer:

a. one-half as great

Explanation:

The power developed by the first lifter is one-half as great as that of the second person.

  Power is defined as the rate at which work is done;

          Power  = \frac{workdone}{time}

Since the two lifters do the same work at different time, let us estimate their power;

       P₁ = \frac{workdone}{2}                     P₂ = \frac{workdone }{1}

   We see that for P₁, power is half of the work done whereas in P₂ power is the same as the work done.

Therefore,

           The power of the first weight lifter is one-half the second lifter.

4 0
2 years ago
A large crane consists of a 20 m, 3,000 kg arm that extends horizontally on top of a vertical tower. The arm extends 15 m toward
Anni [7]

Answer:

m=18000kg

Explanation:

From the question we are told that:

Crane Length l=20m

Crane Mass m_a=3000kg

Arm extension at lifting end l_l=15m

Arm extension at counter weight end l_c=5m

Load M_l=5000kg

Generally the equation for Torque Balance is mathematically given by

T_1 *l_c-(m_a*g) *l_c-(T_2)*l_l=0

mg*5 *-(3000*9.8) *5-(5000*9.8)*15=0

m=18000kg

7 0
3 years ago
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