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maria [59]
3 years ago
11

- The school zone in front of your school has a posted speed limit of 25 mi/h, which is about 11 m/s. Let's

Physics
1 answer:
miv72 [106K]3 years ago
5 0

Answer:

s = 38.7 m

Explanation:

First we calculate the distance covered during uniform motion of reaction time.

s₁ = vt

where,

s₁ = distance covered during uniform motion = ?

v = uniform speed = 11 m/s

t = time = 2.3 s

Therefore,

s₁ = (11 m/s)(2.3 s)

s₁ = 25.3 m

Now, we calculate the distance covered during decelerated motion:

2as₂ = Vf² - Vi²

where,

a = deceleration = -4.5 m/s²

s₂ = distance covered during decelerated motion = ?

Vf = Final Velocity = 0 m/s

Vi = Initial Velocity = 11 m/s

Therefore,

2(-4.5 m/s²)s₂ = (0 m/s)² - (11 m/s)²

s₂ = (-121 m²/s²)/(-9 m/s²)

s₂ = 13.4 m

the total distance will be:

s = s₁ + s₂

s = 25.3 m + 13.4 m

<u>s = 38.7 m</u>

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Answer:

A fuse and circuit breaker both serve to protect an overloaded electrical circuit by interrupting the continuity, or the flow of electricity. ... Fuses tend to be quicker to interrupt the flow of power, but must be replaced after they melt, while circuit breakers can usually simply be reset.

6 0
3 years ago
The fictional rocket ship Adventure is measured to be 65 m long by the ship's captain inside the rocket.When the rocket moves pa
kipiarov [429]

Answer:

1.04 s

Explanation:

The computation is shown below:

As we know that

t = t' × 1 ÷ (√(1 - (v/c)^2)

here

v = 0.5c

t = 1.20 -s

So,

1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)

1.20 = t' × 1 ÷ (√(1 - (0.5)^2)

1.20 = t' ÷ √0.75

1.20 = t' ÷ 0.866

t' = 0.866 × 1.20

= 1.04 s

The above formula should be applied

8 0
3 years ago
The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an
Mrrafil [7]

Answer:

v_1  = 3.5 \ m/s

Explanation:

Given that :

mass of the SUV is  = 2140 kg

moment of inertia about G , i.e I_G = 875 kg.m²

We know from the conservation of angular momentum that:

H_1= H_2

mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2

2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2

1637.1 v_1 = 3841.575 \omega_2

\omega_2 = \frac{1637.1 v_1}{3841.575}

\omega _2 = 0.4626 \ v_1

From the conservation of energy as well;we have :

T_2 +V_{2  \to 3} = T_3 \\ \\ \\  \frac{1}{2} I_A \omega_2^2 - mgh =0

[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0

706.93 \ v_1^2 - 8657.49 =0

706.93 \ v_1^2  = 8657.49

v_1^2  =  \frac{8657.49}{706.93 }

v_1 ^2 =  12.25

v_1  = \sqrt{ 12.25

v_1  = 3.5 \ m/s

6 0
3 years ago
What is the relationship between thickness of lens and focal length?​
Ymorist [56]

Thick lens will have shorter and consequently thin lens will have greater focal length. Because, For a thick lens, the optical path length of the light is more, than for a thin lens, thus, the bending of light will be more in case of a thicker lens. Consequently, it has a shorter focal length.

7 0
3 years ago
large truck exerts 7000 n of force on a piston with an area of 0.4m squared and the is ton can only support 16,000 pa of pressur
sasho [114]
7000 N / 0,4m^2  = 17 500 Pa.
6 0
3 years ago
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