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Vladimir [108]
3 years ago
15

A 1000-kg car traveling at 70 m/s takes 3 m to stop under full braking. the same car under similar road conditions, traveling at

140 m/s, takes ______________ m to stop under full braking.
Physics
1 answer:
azamat3 years ago
5 0
We assume a=const (acceleration is constant. We apply the equation
v^2=v0^2+2as where s is the distance to stop v=0(m/s). We find the acceleration from this equation
a=-v0^2/(2s)=-70^2/(2*3) =-816.7 (m/s^2)

We know the acceleration, thus we find the distance necesssary to stop when initial speed is v=140 (m/s)
s=-v0^2/(2a) =140^2/(2*816.7)=12 (m)

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ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, qE \times S + mg \times S = 0.5 \times mv^{2}

     1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight  and the above equation will be as follows.

   (1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}

         v = sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}

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Since, the electron is travelling downwards it means that it looses the potential energy.

8 0
2 years ago
You connected the 5 Ω, 10 Ω, 15 Ω resistors in series with a 90 V battery. What is the current?​
kykrilka [37]

Answer:

3A

Explanation:

Rtoal=R1+R2+R3=5+10+15=30

I=V/R 90/30

I=3

3 0
2 years ago
A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.
34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated (Q), measured in joules, by the cylindrical resistor is the power multiplied by operation time (\Delta t), measured in hours. That is:

Q = \dot Q \cdot \Delta t (1)

If we know that \dot Q = 1.2\,W and \Delta t = 86400\,s, then the amount of heat dissipated by the resistor is:

Q = (1.2\,W)\cdot (86400\,s)

Q = 103680\,J

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux (Q'), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder (A), measured in square meters:

Q' = \frac{\dot Q}{A} (2)

Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h } (3)

Where:

D - Diameter, measured in meters.

h - Length, measured in meters.

If we know that \dot Q = 1.2\,W, D = 4\times 10^{-3}\,m and h = 2\times 10^{-2}\,m, the heat flux of the resistor is:

Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }

Q' \approx 4340.589\,\frac{W}{m^{2}}

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces (r), no unit, is the ratio of the top and bottom surfaces to total surface:

r = \frac{\frac{\pi}{2}\cdot D^{2}}{A} (3)

If we know that A \approx 2.765\times 10^{-4}\,m^{2} and D = 4\times 10^{-3}\,m, then the fraction is:

r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}

r = 0.045

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0
3 years ago
A 2500-ohm is connected to a 110v power supply. What is the current through the resistor
Assoli18 [71]

Answer:

0.044 amps

Explanation:

Givens

R = 2500 ohms

E = 110 volts

I = ?

Equation

I = E/R

Solution

I = 110/2500

I = 0.044 amps

7 0
3 years ago
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Artyom0805 [142]
Answer: coil of wire
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