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Nostrana [21]
3 years ago
8

Compare the Vf calculated at the point of impact to the horizontal velocity you calculated using Δx and Δy. Were the vf and the

projectile velocity equal? If not what contributed to the difference? Think about the collision.
Physics
1 answer:
34kurt3 years ago
4 0

Answer: Velocities are not equal

Explanation: A projectile usually have a given initial velocity, the velocity must be resolve into components of x and y with respect to the velocity of the projectile in a certain angle. In a projectile trajectory the final velocity for the y- component is always 0m/s since the object reaches its maximum point.

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. A stationary mass explodes into two parts of mass 4 kg and 40 kg . If the K.E of larger mass is 100 J . The K.E of small mass
alisha [4.7K]

Answer:

10J

Explanation:

KE = (1/2)mv²

100J = (.5)(40kg)v²

v²=(100J)/(20kg)

v²= 5

KE = 5(.5)(4kg)

KE = 10J

3 0
3 years ago
List the 3 types of collisions that occur when your vehicle hits an object.
koban [17]

These collisions are: "a Vehicle Collision, a Human Collision, Internal Collision." A vehicle collision is a collision that involves two or more vehicles and is when the vehicles collide against each other creating a unbalanced force since how the force comes from opposite directions. A human collision would involve a vehicle and a human which would also be a unbalanced force but the human wouldn't have much affect of it's speed. A internal collision is when something happens inside the vehicle which decreases, or increases the vehicles speed.

Hope this helps!

7 0
3 years ago
The magnitude of a force vector is 89.6 newtons (N). The x component of this vector is directed along the +x axis and has a magn
insens350 [35]

Answer:

(a) θ = 33.86°

(b) Ay = 49.92 N

Explanation:

You have that the magnitude of a vector is A = 89.6 N

The x component of such a vector is Ax = 74.4 N

(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:

A_x=Acos\theta       (1)

Ax: x component of vector A

A: magnitude of vector A

θ: angle between vector A and the x axis

You solve the equation (1) for θ, by using the inverse of cosine function:

\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°

the angle between the A vector and the x axis is 33.86°

(b) The y component of the vector is given by:

A_y=Asin\theta\\\\A_y=(89.6N)sin(33.86\°)=49.92N

the y comonent of the vecor is Ay = 49.92 N

3 0
3 years ago
A racetrack curve has radius 70.0 m and is banked at an angle of 12.0 ∘. The coefficient of static friction between the tires an
zubka84 [21]

Answer:

See attachment below

Explanation:

7 0
3 years ago
You collect some more data on that horse at a later time interval, but now you are measuring thehorse’s velocity, not its positi
Monica [59]

Answer:

a)  x(t) = 10t + (2/3)*t^3

b) x*(0.1875) = 10.18 m

Explanation:

Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.

Given:

- v(t) = 10 + 2*t^2 (radar gun)

- x*(t) = 10 + 5t^2 + 3t^3  (our coordinate)

Find:

-The position x of horse as a function of time t in radar system.

-The position of the horse at x = 2m in our coordinate system

Solution:

- The position of horse according to radar gun:

                              v(t) = dx / dt = 10 + 2*t^2

- Separate variables:

                              dx = (10 + 2*t^2).dt

- Integrate over interval x = 0 @ t= 0

                             x(t) = 10t + (2/3)*t^3

- time @ x = 2 :

                              2 = 10t + (2/3)*t^3

                              0 = 10t + (2/3)*t^3 + 2

- solve for t:

                              t = 0.1875 s

- Evaluate x* at t = 0.1875 s

                              x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3

                              x*(0.1875) = 10.18 m

3 0
3 years ago
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