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Sedbober [7]
2 years ago
14

An automobile being tested on a straight road is 400 feet from its starting point when the stopwatch reads 8.0 seconds and is 55

0 feet from the starting point when the stopwatch reads 10.0 seconds.
A. What was the average velocity of the automobile during the interval from t = 10.0 seconds to t = 8.0 seconds
B. What was the average velocity of the automobile during the interval from t - Ostot - 10.0 s? (Assume that the stopwatch read t = 0 and started at the same time as the auto.)
C. If the automobile averages 100 ft/s from t - 10.0 stot - 20.0 s, what distance does it travel during this interval?
D. The automobile has a special speedometer calibrated in feet/s instead of in miles/hour. Att 85 the speedometer reads 65 ft/s; and at t = 10 s it reads 80 ft/s. What is the average acceleration during this interval?
Physics
1 answer:
m_a_m_a [10]2 years ago
4 0

Answer:

a)   v = 75 ft / s , b)  v = 55 ft / s , c)   Δx = 1000 ft

Explanation:

We can solve this exercise with the expressions of kinematics

a) average speed is defined as the distance traveled in a given time interval

        v = (x₂-x₁) / (t₂-t₁)

         v = (550 - 400) / (10 -8)

         v = 75 ft / s

b) we repeat the calculations for this interval

   v = (550 - 0) / (10 -0)

   v = 55 ft / s

c)  we clear the distance from the average velocity equation

     Δx = v (t₂ -t₁)

     Δx = 100 (20-10)

     Δx = 1000 ft

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exis [7]
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7 0
2 years ago
For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser poin
nydimaria [60]

Answer:

The width of slide is 0.092 mm.

Explanation:

Given that,

Wave length = 680 nm

Distance between slit and screen D= 5.4 m

Distance of bright band 2y = 7.9 cm

y =\dfrac{7.9}{2}

y=3.95\ cm

We need to calculate the width of slide

Using formula of width

d=\dfrac{\lambda D}{y}

Where, d = width

D =Distance between slit and screen

y = Distance of bright band

Put the value into the formula

d=\dfrac{680\times10^{-9}\times5.4}{3.95\times10^{-2}}

d=0.092\times10^{-3}\ m

d=0.092\ mm

Hence, The width of slide is 0.092 mm.

4 0
2 years ago
A seaplane flies horizontally over the ocean at 50 meters/second. It releases a buoy, which lands after 21 seconds. What's the v
pantera1 [17]
The motion of the buoy consists of two independent motions on the horizontal and vertical axis.

On the horizontal axis, the motion of the buoy is a uniform motion with constant speed v=50 m/s. On the vertical axis, the motion of the buoy is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2. The vertical position of the buoy at time t is given by
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where h is the initial heigth of the buoy when it is released from the plane. At the time t=21 s, the buoy reaches the ground, so y(21 s)=0. If we substitute these two numbers inside the equation, we can find the value of h, the vertical displacement from the plane to the ocean:
0=h- \frac{1}{2}gt^2
h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(21 s)^2=2163 m
8 0
2 years ago
Waves that move the particles of the medium parallel to the direction in which the waves are traveling are called
Nikolay [14]

1. a. longitudinal waves.

There are two types of waves:

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So, these types of waves are called longitudinal waves.


2. d. a medium

There are two types of waves:

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3. a. AM/FM radio

Analogue signals consist of continuous signals, which vary in a continuous range of values. On the contrary, digital signals consist of discrete signals, which can assume only some discrete values. For AM and FM radios, signals are transmitted by using analogue signals.

5 0
2 years ago
The figure shows two 1.0 kg blocks connected by a rope. a second rope hangs beneath the lower block. both ropes have a mass of 2
lutik1710 [3]
We need first to use the formula  F=m(a+g), m iis the total mass, a is the acceleration, g is gravity pulling the blocks. So the procedure will be 
<span>m=2kg(both blocks)+500g(both ropes) → m=2.5kg </span>
<span>a=3.00m/s^2 </span>
<span>g=9.8m/s^2 </span>
<span>F=m(a+g) → F=2.5kg (3.00m/s^2 + 9.8m/s^2) → F=2.5kg (12.8m/s^2) → F=32 N 
To calculate the tension at the top of rope 1 you need to use the formula </span>T=m(a+g) so it will be <span>T=m(a+g) → T=1.5kg(12.8m/s^2) → T=19.2N 
</span>We can now calculate the tension at the bottom of rope 1 using the formula: <span>T=m(a+g) → T=1.25kg(12.8m/s^2) → T=16N 
</span>Now to find the tension at the top of rope 2 we do it like this: 
<span>T=m(a+g) → T=.25kg(12.8m/s^2) → T=3.2</span>
7 0
2 years ago
Read 2 more answers
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