A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I

Answer:

The snowball's speed after the impact is 3 m/s

Explanation:

Given the following:

mass of each ball

m₁ = 0.4 Kg

m₂ = 0.6 Kg

initial speed of both balls = v₁ = 15 m/s

Speed of 1 Kg mass post-collision =?

Applying conservation of momentum

m₁ v₁ - m₂ v₁ = (m₁+m₂) V

A negative velocity indicates that the second ball moves in the opposite direction.

0.4 x 15 - 0.6 x 15 = (1) V

Therefore,

V = - 3 m/s

Consequently,

The snowball's speed following the collision is 3 m/s

Answer:

3.4 x 10⁴ m/s

Explanation:

Analyze the circular path of the electron

B = magnetic field = 80 x 10⁻⁶ T

m = mass of an electron = 9.1 x 10⁻³¹ kg

v  = speed in the radial direction

r = radius of the circular trajectory = 2 mm = 0.002 m

q = charge of an electron = 1.6 x 10⁻¹⁹ C

For the electron’s circular movement

qBr = mv

(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v

v = 2.8 x 10⁴ m/s

Now, consider the electron's movement in a straight line:

v' = speed in linear motion

x = distance traveled horizontally = 9 mm = 0.009 m

t = duration = = = 4.5 x 10⁻⁷ sec

Using the formula

x = v' t

0.009 = v' (4.5 x 10⁻⁷)

v' = 20000 m/s

v' = 2 x 10⁴ m/s

The resultant speed is given by

V = sqrt(v² + v'²)

V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)

v = 3.4 x 10⁴ m/s

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To calculate the kinetic energy variation, we can utilize the work-energy theorem.

W = ΔK

∫ F .dx = K - K₀

If the object starts from rest, then K₀ = 0.

So, ∫ F dx cos θ = K.

As the force and displacement directions align, the angle is zero, and hence the cosine is 1.

Now we can substitute and perform integration:

α ∫ x³ dx + β ∫ dx = K.

Thus, α x⁴ / 4 + β x = K.

Next, we evaluate from the limits F = 0 to F:

α (x⁴ / 4 - 0) + β (x - 0) = K.

Consequently, K = αX⁴ / 4 + β x.

This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.

To finalize the computation, we need to ascertain the displacement.

As per the information provided, the car's mass is one-fourth that of the truck

The truck exerts a force of 6000 N onto the car

We aim to determine the force that the car exerts on the truck

According to Newton's third law, every action has an equal and opposite reaction

Hence, the force from the car on the truck is 6000 N in the opposite direction

The question lacks details. Here is the full question.

The accompanying image was captured with a camera capable of shooting between one and two frames per second. A series of photos was merged into this single image, meaning the vehicles depicted are actually the same car, documented at different intervals.

Assuming the camera produced 1.3 frames per second for this image and that the length of the car is approximately 5.3 meters, based on this information and the photo, how fast was the car moving?

Answer: v = 6.5 m/s

Explanation: The problem requires calculating the car's velocity. Velocity can be computed using:

Since the camera captured 7 images of the car and its length is noted as 5.3, the car's displacement is:

Δx = 7(5.3)

Δx = 37.1 m

The camera operates at 1.3 frames per second and recorded 7 images, thus the time driven by the car is:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

v = 6.87 m/s