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Lilit [14]
3 years ago
15

A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding t

he speed limit? Justify your answer.
Physics
1 answer:
Tom [10]3 years ago
5 0

Answer: Yes, he is exceeding the speed limit

Explanation:

Hi!

This is problem about unit conversion

1 mile = 1,609.344 m

Then the speed limit v is:

v = 75 mi/h = 120,700.8 m/h

1 hour = 60 min = 60*60 s = 3,600 s

v = (120,700.8/3,600) m/s = 33.52 m/s

38 m/s is higher than the speed limit v.

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5,6,1,9,21,37,68 ? find missing​
maria [59]

Answer:

Missing numbers are 3,11,13, 15,17,19, 23,27,29 and wrong are 6,68

4 0
3 years ago
Which best describes a difference between laser light and regular light?
Inga [223]
The basic difference is that the ordinary sources are incoherent that means that the discrete frequencies merge up to give an intermediate between the maximum and minimum frequencies. While the laser is coherent containing the single frequency with maximum amplitude. thus travelling far.
4 0
3 years ago
A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k
Veronika [31]

Answer:

14.98\ \text{kg m/s}

45.26^{\circ}

Explanation:

P_1 = Initial momentum of the pin = 13 kg m/s

P_i = Initial momentum of the ball = 18 kg m/s

P_2 = Momentum of the ball after hit

55^{\circ} = Angle ball makes with the horizontal after hitting the pin

\theta = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}

Momentum in the y direction

P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}

(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}

The pin's resultant velocity is 14.98\ \text{kg m/s}

P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}

The pin's resultant direction is 45.26^{\circ} below the horizontal or to the right.

4 0
3 years ago
A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals
slamgirl [31]

Answer:

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI................ Equation 1

Where P = Power, V = Voltage, I = Current.

make I the subject of the equation

I = P/V................ Equation 2

From the question,

Given: P = 0.414 W, V = 1.50 V

Substitute into equation 2

I = 0.414/1.50

I = 0.276 A

Also,

Q = It............... Equation 3

Where Q = amount of charge, t = time

make t the subject of the equation

t = Q/I.................. Equation 4

From the question,

4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs

Q = 7.899×10⁻¹⁹ C

Substitute these value into equation 4

t =  7.899×10⁻¹⁹/0.276

t = 2.86×10⁻¹⁸ seconds

5 0
3 years ago
A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm
Karo-lina-s [1.5K]

Answer:

the thermistor temperature = 325.68 \ ^0 \ C

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K

Resistance of the thermistor R_1 = 20,000 ohms

Material constant \beta = 3650

Resistance of the thermistor R_2 = 500 ohms

Using the equation :

R_1 = R_2  \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

\frac{R_1}{ R_2} =   \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

Taking log of both sides

In \ \frac{R_1}{ R_2} = In \  \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})

\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})

\frac{1}{T_2} =   \frac{1}{T_1}  -          \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}

{T_2} =  \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}

Replacing our values into the above equation :

{T_2} =  \frac{3650*373}{3650 - In (\frac{20000}{500})373}

{T_2} =  \frac{1361450}{3650 - 3.6888*373}

{T_2} =  \frac{1361450}{3650 - 1375.92}

{T_2} =  \frac{1361450}{2274.08}

{T_2} = 598.68 \ K

{T_2} = 325.68 \ ^0 \ C

Thus, the thermistor temperature = 325.68 \ ^0 \ C

4 0
4 years ago
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