A car leaves the pit after a refueling stop and accelerates uniformly to a speed of 62.2 m/s in 5.0 s to rejoin the race. What is the car's acceleration during this time?
Answers
Answer:
The car's acceleration is: a = 12.44 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity = 62.2 [m/s]
Vo = initial velocity = 0 (because the car stars from the rest)
a = acceleration [m/s²]
t = time = 5 [s]
62.2 = 0 + a*5
a = 12.44 [m/s²]
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He experimented with photography and learned how to create “cameraless” photos, also known as photograms or photograms, which he termed radiographs. He created them by physically laying things on light-sensitive paper, which he then developed in light. Thus, option A is correct.
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Therefore, One of the most important modernist photographers, filmmakers, and object makers was Man Ray.
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Answers
Here are the following choices.
A. bar graph
B. line graph
C. histogram
D. scatterplot
The correct answer is "Bar Graph"
Answers
That's 1/2 volt.
B. 3.8 m/s2
C. 2.4 m/s2
D. 9.8 m/s2
Answers
Acceleration due to gravity on this planet will be 3.802 m / s^2
What are equations of motion?
Equation of motion are defined as equations that describe the behavior of a physical system in terms of its motion as a function of time
Using equation of motion
u=0
s= 2.3 m
t = 1.1 sec
to find = g (acceleration due to gravity on this planet)
s = u t + 1/2 (a ) (t ^(2))
s = 1/2 (g) (t^2)
2.3 = 1/2 (g) (1.1^2)
g = 2 * 2.3 /(1.1)^2
g = 4.6 /1.21= 3.802
correct answer is b) g = 3.802 m / s^2
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Answers
The velocity of the package just before it hits the ground is 16.7 m/s.
What is velocity?
The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
Initial velocity(u)=2 m/s
V(velocity of hitting)=?
S(distance travelled)= 14 meters
The velocity of the package just before it hits the ground is found as;
From the Newton's third equation of motion;
v²=u²+2as.
v²=(-2)²+2(9.81)(14)
v=16.7 m/s
Hence,the velocity of the package just before it hits the ground is 16.7 m/s.
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Taking downwards as positive, use v²=u²+2as.
v²=(-2)²+2(9.81)(14)
v=16.7 m/s
Answers
Answer:
49 km/h
Explanation:
60 km/h acceleration for 0.75 hours.
60 × 3/4 = 45 km/h
Take the acceleration and add it to the speed the train was already moving at:
45 + 4 = 49km/h