This is the same question as the one previously but with more details, so I will just use my previous answer.
1800 to 1820 is 20 minutes.1830 to 1838 is 8 minutes.1840 to 1905 is 25 minutes.
The total time travelled is 20+8+25 = 53 minutes = 3180 seconds.
The distance between Glasgow and Edinburgh is 28 + 12 + 34 = 74 km = 74000 m.
So, the average speed is 74000m/3180s = 23.27 m/s (4 s.f.)
Answer:
a= g = - 9.81 m/s2.
The following equations will be helpful:
a = (vf - vo)/t d = vot + 1/2 at2 vf2 = vo2 + 2ad
When you substitute the specific acceleration due to gravity (g), the equations are as follows:
g = (vf - vo)/t d = vot + 1/2 gt2 vf2 = vo2 + 2gd
If the object is dropped from rest, the initial velocity ("vi") is zero. This further simplifies the equations to these:
g = vf /t d = 1/2 gt2 vf2 = 2gd
The sign convention that we will use for direction is this: "down" is the negative direction. If you are given a velocity such as -5.0 m/s, we will assume that the direction of the velocity vector is down. Also if you are told that an object falls with a velocity of 5.0 m/s, you would substitute -5.0 m/s in your equations. The sign convention would also apply to the acceleration due to gravity as shown above. The direction of the acceleration vector is down (-9.81 m/s2) because the gravitational force causing the acceleration is directed downward.
hope this info helps you out!
Answer:
You can determine if the ship is moving by lying down and measuring your height.
Explanation:
Answer:
Kf > Ka = Kb > Kc > Kd > Ke
Explanation:
We can apply
E₀ = E₁
where
E₀: Mechanical energy at the beginning of the motion (top of the incline)
E₁: Mechanical energy at the end (bottom of the incline)
then
K₀ + U₀ = K₁ + U₁
If v₀ = 0 ⇒ K₀
and h₁ = 0 ⇒ U₁ = 0
we get
U₀ = K₁
U₀ = m*g*h₀ = K₁
we apply the same equation in each case
a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J
d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J
e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J
f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J
finally, we can say that
Kf > Ka = Kb > Kc > Kd > Ke
The time taken, and the distance travelled by the car before stop, will be 5.3 second and -53.3 m west.
<h3>What is the distance?</h3>
The length of the path traveled by the body is known as the distance covered by the body.
Distance is a 1-dimensional phenomenon. its unit is a meter(m). The distance can be found by the product of velocity and time.
The given data in the problem will be
u is the initial velocity =20m/sec
t is the time =?
d is the distance =?
From the Newtons second law;
The distance travelled before the car stop is,
Hence,the time taken, and the distance travelled by the car before stop, will be 5.3 second and -53.3 m west.
To learn more about the distance, refer to the link;
brainly.com/question/989117
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