A roller coaster accelerates from 0 m/s to 25 m/s in 5 seconds. Determine the acceleration of the roller coaster.

Bill and Janet are pulling on opposite sides of a table. Bill pulls with a force of 250 N to the left, and Janet pulls with a fo

Dafna1 [17]

Answer: 75 N to the right

Explanation:

5 0

2 years ago

A centrifuge used in DNA extraction spins at a maximum rate of 7000rpm producing a "g-force" on the sample that is 6000 times th

defon

Answer:

A) a = 73.304 rad/s²

B) Δθ = 3665.2 rad

Explanation:

A) From Newton's first equation of motion, we can say that;

a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.

Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s

Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s

We are given; t = 10 s

Thus;

a = 733.04/10

a = 73.304 rad/s²

B) From Newton's third equation of motion, we can say that;

ω² = ω_o² + 2aΔθ

Where Δθ is angular displacement

Making Δθ the subject;

Δθ = (ω² - ω_o²)/2a

At this point, ω = 0 rad/s while ω_o = 733.04 rad/s

Thus;

Δθ = (0² - 733.04²)/(2 × 73.304)

Δθ = -537347.6416/146.608

Δθ = - 3665.2 rad

We will take the absolute value.

Thus, Δθ = 3665.2 rad

8 0

3 years ago

Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax

kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$