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disa [49]
3 years ago
14

You are driving to the grocery store at 14 m/s. You are 115 m from an intersection when the traffic light turns red. Assume that

your reaction time is 0.50 s and that your car brakes with constant acceleration. How far are you from the intersection (in m) when you begin to apply the brakes
Physics
1 answer:
kogti [31]3 years ago
6 0

Answer:

111.5 m

Explanation:

Given that You are driving to the grocery store at 14 m/s. You are 115 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration. 

Use first equation of motion

V = U - at

Since the car is going to rest, V = 0 and a = negative

0 = 14 - a × 0.5

0.5a = 14

a = 14 /0.5

a = 28 m/s^2

Let us use second equation of motion

S = Ut - 1/2at^2

S = 14 × 0.5 - 0.5 × 28 × 0.5^2

S = 7 - 3.5

S = 3.5 m

115 - 3.5 = 111.5

Therefore, you are 111.5 metres from the intersection (in m) when you begin to apply the brakes.

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notka56 [123]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

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5 0
2 years ago
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What happens to the current as we increase the amount of stepping of our transformer? Does this help explain why the primary was
matrenka [14]

Answer:

Current will decrease.

Explanation:

When we increase the number of stepping in transformer, the voltage will increase as its is directly proportional to the number of turn of stepping. Thus as the voltage will increase, current will decrease. As per the equation of ideal transformer,   E1 / E2 = I2 / I1

E1 and E2 are the voltages in primary and secondary winding and I1 and I2 are the current.

As the number of turns will be increased more inevitable losses will be generated that dissipates heat thus warming the primary.

Though the conservation of energy is obeyed but losses occur in this scenario hence step-up transformers cannot be used to create free energy.

7 0
2 years ago
A torsion-bar spring consists of a prismatic bar, usually of round cross section, that is twisted at one end and held fast at th
ra1l [238]

Answer:

d₁ = 0.29 in

d₂ = 0.505 in

Explanation:

Given:

T = 1500 lbf in

L = 10 in

x = 0.5 L = 5 in

T_{1} =\frac{T(L-x)}{L} =\frac{1500*(10-5)}{10} =750lbfin

First case: T = T₁ + T₂

T₂ = T - T₁ = 1500 - 750 = 750 lbf in

If the shafts are in series:

θ = θ₁ + θ₂

θ = ((T₁ * L₁)/GJ) + ((T₂ * L₂)/GJ)

Second case: If d₁ ≠ d₂

θ = ((T₁ * L₁)/GJ₁) + ((T₂ * L₂)/GJ₂) = 0 (eq. 1)

t₁ = t₂

\frac{16T_{1} }{\pi d_{1}^{3}  } =\frac{16T_{2} }{\pi d_{2}^{3}  } (eq. 2)

T₁ + T₂ = 1500 (eq. 3)

θ₁ first case = θ₁ second case

Replacing:

\frac{750*5}{G(\frac{\pi }{32})*0.5^{4}  } =\frac{T_{1}*3.7 }{G(\frac{\pi }{32})*d_{1} ^{4}  }\\T_{1} =16216d_{1} ^{4}

The same way to θ₂:

\frac{750*5}{G(\frac{\pi }{32})*0.5^{4}  } =\frac{T_{2}*6.3 }{G(\frac{\pi }{32})*d_{2} ^{4}  } \\T_{2} =9523.8d_{2} ^{4}

From equation 2, we have:

d₁ = 0.587 * d₂

From equation 3, we have:

d₂ = 0.505 in

d₁ = 0.29 in

7 0
3 years ago
A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward
algol [13]

Answer:

a) v_{1}=14.29m/s\\v_{2}=9.25m/s\\v_{3}=6.36m/s

b) v=+9.97m/s

Explanation:

From the exercise we know that

x_{1} =15m, t_{1}=3s

x_{2} =-3m, t_{1}=1.74s

x_{3} =29m, t_{3}=5.20s

From dynamics we know that the formula for average velocity is:

v=\frac{x_{2}-x_{1}  }{t_{2}-x_{1}  }

a) For the three intervals:

v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(-3-15)m}{(1.74-3)s}=14.29m/s

v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(29-(-3))m}{(5.20-1.74)s}=9.25m/s

v_{3}=\frac{x_{3}-x_{1}  }{t_{3}-t_{1}  }=\frac{(29-15)m}{(5.20-3)s}=6.36m/s

b) The average velocity for the entire motion can be calculate by the following formula:

v=\frac{v_{1}+v_{2}+v_{3}   }{n} =\frac{(14.29+9.25+6.36)m/s}{3}=+9.97m/s

8 0
3 years ago
Which of the following can penetrate the deepest (Please explain)
harkovskaia [24]

Answer: 3MeV electron

Explanation:

m_e={9.1\times 10^{-31}      m_α=4\times m_e    m_a={9.1\times 10^{-31}

m_p=1.67\times 10^{-27}

(a)  K.E. Energy of electron =\frac{1}{2}\times{m_e}\times{v_{e} ^{2}}=3MeV

v_{e} ^{2}=\frac{2\times3\times1.6\times10^{-19}\times10^{6}  }{9.1\times 10^{-31} }=1.05\times10^{18}

v_e=\sqrt{1.05\times10^{18} } = 1.025\times10^{9}\frac{m}{s}

(b) K.E. Energy of alpha particle =\frac{1}{2}\times{m_\alpha}\times{v_{\alpha} ^{2}}=10MeV

v_{\alpha} ^{2}= \frac{2\times10\times1.6\times10^{-19}\times10^{6}  }{4\times9.1\times 10^{-31} }=0.88\times10^{18}

v_\alpha=\sqrt{0.88\times10^{18} } =.94\times10^{9}\frac{m}{s}

(c) K.E. Energy of auger particle =\frac{1}{2}\times{m_a}\times{v_{a} ^{2}}=0.1MeV

v_{a} ^{2}=\frac{2\times0.1\times1.6\times10^{-19}\times10^{6}  }{9.1\times 10^{-31} }=0.035\times10^{18}

v_a=\sqrt{0.035\times10^{18} } =.19\times10^{9}\frac{m}{s}

(d)  K.E. Energy of proton particle =\frac{1}{2}\times{m_p}\times{v_{p} ^{2}}=400keV

v_{p} ^{2}=\frac{2\times400\times1.6\times10^{-19}\times10^{3}  }{1.67\times 10^{-27} }=0.766\times10^{14}

v_p=\sqrt{0.766\times10^{14} } =0.875\times10^{7}[tex]\frac{m}{s}

from (a),(b),(c),and (d) we can clearly say that the velocity of the electron is more so the penetration of the electron will be deepest.

6 0
2 years ago
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