calculate the velocity of an object moved around a circle with a radius of 1.65m and an acceleration of 3.5 m/s^2
Answer:
2.4m/s
Explanation:
The graph shows the speeds of two bicyclists on their way to school. what is the speed of bicycle B
Answer: 0.3
Explanation: Jusy took it and got a 100
Answer: 0.3
Explanation: This is right trust me I answered this and it was correct!!
A car turns from a road into a parking lot and into an available parking space. The car's initial velocity is 4 m/s [E 45° N]. The car's velocity just before the driver decreases speed is 4 m/s [E 10° N]. The turn takes 3s. What's the average acceleration of the car during the turn? The answer should have directions with an angle.
Answer:
Explanation:
From the given information:
The car's initial velocity = 4 m/s in the direction of east 45° due north
We can therefore express the vector of this component form as:
v₁ = (4 m/s) (cos(45º)i + sin(45º)j)
v₁ = (2.83 m/s)i + (2.83 m/s)j
Similarly, the car's final velocity = 4 m/s in the direction of the east side 10º north
∴
v₂ = (4 m/s) (cos(10º) i + sin(10º) j)
v₂ = (3.94 m/s) i + (0.695 m/s) j
From the first equation of motion
v = u + at
Making acceleration "a" the subject of the formula, we have:
a = (v - u )/t
a = (v₂ - v₁)/t
a = (0.370 m/s²) + (-0.711 m/s²)
The magnitude of the avg. acceleration is:
[tex]|| a||= \sqrt{(0.370 m/s^2)^2 + (-0.711 m/s^2)^2)[/tex]
[tex]|| a||= 0.8015 \ m/s^2[/tex]
And;
The direction can be determined by taking the tangent of the acceleration:
i.e.
[tex]tan(\theta) = \dfrac{-0.711 m/s^2}{ 0.370 m/s^2}[/tex]
[tex]tan(\theta) = -1.9216[/tex]
[tex]\theta = tan^{-1} ( -1.9216 )[/tex]
[tex]\mathbf{\theta = -62.51 ^0}[/tex]
Thus, the direction of the angle is approximately S 62.51º E
What is also known as watered carbons
Answer:
The name carbohydrate means "watered carbon" or carbon with attached water molecules. Many carbohydrates have empirical formuli which would imply about equal numbers of carbon and water molecules. For example, the glucose formula C6H12O6 suggests six carbon atoms and six water molecules.
A projectile is launched horizontally at a speed of 40 meters per second from a platform located a vertical distance h above the ground. The projectile strikes the ground after time at horizontal distance from the base of the platform. [Neglect friction.]
Calculate the vertical distance, h, if the projectile's total time of flight is 4.5 seconds. You must show work to earn full credit with the minimum 3 steps:
1. Equation used
2. Substitution with correct units
3. Final answer with correct units
Answer:1
Explanation:t=rad2h/g
The launch of projectiles allows to find the height of the body launched horizontally is:
The height is 99.2 m
Projectile launching is an application of kinematics where there is no acceleration on the x-axis and the y-axis is the gravity acceleration.
In the attachment we see a diagram of the movement, as the body is thrown horizontally, its initial vertical speed is zero.
y = y₀ + [tex]v_o_y[/tex] t - ½ g t²
Where y, y₀i are the current and initial heights, respectively, [tex]v_{oy}[/tex] the vertical initial velocity, g the acceleration of gravity and t the time.
0 =y₀ + 0 - ½ g t²
y₀ = ½ g t²
let's calculate
y₀ = ½ 9.8 4.5²
y₀ = 99.2 m
y₀ = h = 99.2 m
This is the initial height of the object when it is thrown.
In conclusion using the projectile launch we can find the height of the horizontally launched body is:
The height is 99.2 m
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