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AleksAgata [21]

3 years ago

9

David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2

.0 m/s2 at the instant when David passes. How far does Tina drive before passing David?

1 answer:

dangina [55]3 years ago

5 0

A. 441 m B: 46.0 m/s

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You pull on a crate using a rope as in (Figure 1), except the rope is at an angle of 20.0 ∘ above the horizontal. The weight of

Nezavi [6.7K]

Hi there!

For an object on an incline with friction being pulled, the following forces are present.

Force due to Gravity
Force due to Friction
Force due to tension

The force due to friction opposes the force due to gravity which would cause the object to slide down. (The force due to friction acts up the incline). Additionally, the force due to the rope is also upward.

Let up the incline be positive, and down the incline be negative.

Doing a summation of forces:
$\Sigma F = F_T + F_f - F_g$

For the crate to be moving at a constant velocity, there is NO net force acting on the crate, so:
$0 = F_T + F_f - F_g$

Now, we can express each force as an equation.

Force due to tension:

Must be solved for.

Force due to gravity:

On an incline, this is equivalent to the SINE component of its weight. (Force of gravity is STILL THE WEIGHT, but on an incline, it contains a horizontal component that contributes to the net force)

This is expressed as:
$F_g = Mgsin\theta$

Force due to friction:

Equivalent to the normal force and coefficient of friction. The normal force is the VERTICAL component of the object's weight, so:

$N = Mgcos\theta$

$F_f = \mu Mgcos\theta$

Now, plug these expressions into the above equation.

$0 = F_T + \mu Mgcos\theta - Mgsin\theta\\\\Mgsin\theta - \mu Mgcos\theta = F_T$

Mg = 245 N (weight). Plug in all values:
$245sin(20) - 0.270(245)cos(20) = F_T\\\\F_T = \boxed{21.634 N}$

The normal force is equivalent to the vertical component (PERPENDICULAR TO THE INCLINE) of the weight (cosine), so:

$N = Mgcos\theta\\\\N = 245cos(20) = \boxed{230.225 N}$

5 0

2 years ago

A railroad car containing an angry bull is standing onthe

nata0808 [166]

Answer:

$v_{f}$ = - M / m v

Explanation:

We must define a system formed by the wagon and the torro, in this case the forces of the movement are internal, so the moment is preserved, write the moment in two moments

Initial. Before the bull's movement, in this case the two are still

p₀ = 0

Final. The bull is moving

$p_{f}$ = M v + m $v_{f}$

p₀ = $p_{f}$

0 = M v + m $v_{f}$

We cleared the train speed

$v_{f}$ = - M / m v

The negative sign indicates that the train moves in the opposite direction of the bull

7 0

3 years ago

in a historical movie, two knights on horseback start from rest at 88.0 m spartans ride directly toward each other to do battle.

Tema [17]

The distance formula:d = v i · t + a · t² / 2Sir G.: ( v i = 0 )
d G = 0 + 0.3 · t² / 2Sir A. : d A = 0 + 0.2 t² / 288 m = 0.3 t² / 2 + 0.2 t² / 2 / · 2 ( multiple both sides by 2 )176 = 0.3 t² + 0.2 t²176 = 0.5 t²t² = 176 : 0.5t² = 352t = √352t = 18.76 sd G. = 0.3 · 18.76² / 2 = 0.3 · 352 / 2 = 52.8 mAnswer: The knights collide at 52.8 m relative to Sir George`s starting point.

7 0

3 years ago

A rope swing is hung from a tree right at the edge of a small creek. The rope is 5.0 m long; the creek is 3.0 m wide.

irina [24]

Answer:

1) T = 4.5 s

2) T = 4.5 s

3) v = 9.9 m/s

Explanation:

We can use the equation

T = 2π√(L/g)

1) T = 2π√(5m/9.81 m/s²) = 4.5 s

2) T = 2π√(L/g)

T = 2π√(5m/9.81 m/s²) = 4.5 s

3) v = √(2gR)

v = √(2(9.81 m/s²)(5m))

v = 9.9 m/s

8 0

4 years ago

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umka2103 [35]

Green. This is because chlorophyll reflects the color green, which is why we see it as this color.

7 0

3 years ago