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RUDIKE [14]
3 years ago
9

Alan leaves Los Angeles at 8:00 A.M. to drive to San Francisco 400 mi away. He manages to travel at a steady 50 mph in spite of

traffic. Beth leaves Los Angeles at 9:00 A.M. and surprisingly manages to also drive at a constant speed, in this case 60 mph. (Knight 2.1) a. Who gets to San Francisco first? (Beth) b. How long does the first to arrive have to wait for the second? (20 minutes)
Physics
1 answer:
Fudgin [204]3 years ago
3 0

Answer:

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

Explanation:

let 'd' be distance b/w Los Angeles and San  Francisco i.e 400 mi

considering ,

Alan's speed v_A=50mph

Beth's speed v_B=60mph

->For Alan:

The time required t_A= d/v_A= 400/50 => 8h

-> For beth:

The time required t_B=\frac{d}{v_B} =\frac{400}{60} =>6\frac{2}{3} h => 6h 40m

Alan will reach at 8:00 a.m +8h = 4:00p.m.

Beth will reach at 9:00 a.m +6h 40m= 3:40p.m.

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

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A firefighter with a weight of 707 N slides down a vertical pole with an acceleration of 2.79 m/s2, directed downward. (a) What
DedPeter [7]

Answer:

The vertical force acting on the firefighter = 908.27 N

Explanation:

Force: Force of a body is defined as the product of mass and its acceleration. The S.I unit of force is Newton (N)

The vertical force acting on the firefighter = Force due to the weight of the firefighter + force due to acceleration.

Ft = Fw - Fa

Where Ft = The vertical force acting on the firefighter, Fw = Force due to the weight of the firefighter, Fa = force due to acceleration.

Fw = mg

Making m the subject of formula in the equation above

m = Fw/g................... Equation 1

Where m = mass of the firefighter, g = acceleration due to gravity,

<em>Given: Fw = 707 N, </em>

<em> Constant: g = 9.8 m/s²</em>

Substituting these values into eqaution 1

m = 707/9.8

m = 72.14 kg.

But, Fa = ma

Where a = acceleration of the firefighter.

<em>Given: a = 2.79 m/s², </em>

<em>And m = 72.14 kg</em>

Fa = 72.14 × 2.79

Fa = 201.27 N

Therefore, Ft = 707 + 201.27  = 908.27 N

Ft = 908.27 N

The vertical force acting on the firefighter = 908.27 N

7 0
3 years ago
Is the classification for an instrument that produces sound whne a string or strings stretched between two points is plucked?
Yuki888 [10]
The correct answer for the question is Chordophone 
Chordophone is an instrument in which a stretched, vibrating string produces the initial sound. Strings instruments produce sound through the vibration of strings. The length, tightedness, and thickness determines the sound produced by the strings. 
5 0
3 years ago
You have a string with a mass of 0.0133 kg. You stretch the string with a force of 8.89 N, giving it a length of 1.97 m. Then, y
Kazeer [188]

Answer:

(i) The wavelength is 0.985 m

(ii) The frequency of the wave is 36.84 Hz

Explanation:

Given;

mass of the string, m = 0.0133 kg

tensional force on the string, T = 8.89 N

length of the string, L = 1.97 m

Velocity of the wave is:

V = \sqrt{\frac{F_T}{M/L} } \\\\V = \sqrt{\frac{8.89}{0.0133/1.97} } \ = 36.29 \ m/s

(i) The wavelength:

Fourth harmonic of a string with two nodes, the wavelength is given as,

L = 2λ

λ = L/2

λ = 1.97 / 2

λ = 0.985 m

(ii) Frequency of the wave is:

v = fλ

f = v / λ

f = 36.29 / 0.985

f = 36.84 Hz

3 0
3 years ago
Please help asap! Thank you.
olganol [36]

Answer:

direct current

Explanation:

it has a direct path to go down to reach the specific point

4 0
3 years ago
Read 2 more answers
If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0
3 years ago
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