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Levart [38]
2 years ago
15

Velocity is a vector quantity which has both magnitude and direction. Using complete sentences, describe the object's velocity.

Comment on both the magnitude and the direction.
Physics
1 answer:
Shtirlitz [24]2 years ago
3 0

Answer & Explanation:

A girl is driving a bicycle at a velocity of 30m/s at 40 degree North west.

Considering the above sentence, the magnitude of the bicycle velocity is 30 meters per second and the direction of the motion of the bicycle is 40 degree North west. So, the bicycle is having both magnitude and direction, so the velocity is a vector quantity.

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A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
Archy [21]

Answer:

a)  y= 3.5 10³ m, b)   t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

         y₁ = ½ 16 10²

         y₁ = 800 m

At the end of this stage it has a speed

        v₁ = vo + a₁ t₁

        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

        y₂ = 1092.5 m

Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

        v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

         y₃ = v₂² / 2g

         y₃ = 215²/2 9.8

         y₃ = 2358.4 m

The total height is

          y = y₃ + y₂

          y = 2358.4 + 1092.5

          y = 3450.9 m

           y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          t_{s} = t₁ + t₂ + t₃

          t_{s} = 10+ 5+ 21.94

          t_{s} = 36.94 s

The time to descend from the maximum height is

          y = v₀ t - ½ g t²

When it starts to slow down it's zero

         y = - ½ g t_{b}²

         t_{b}  = √-2y / g

         

        t_{b} = √(- 2 (-3450.9) /9.8)

        t_{b} = 26.54 s

Flight time is the rise time plus the descent date

        t = t_{s} + t_{b}

        t = 36.94 + 26.54

        t =63.84 s

        t = 64 s

3 0
3 years ago
An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where ther
jeka94

Answer:

T = 188.5 s, correct is  C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

         

initial instant. Before the crash

        L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

        L_f = I₀ w + m r v

        L₀ = L_f

        m r v₀ = I₀ w + m r v

angular and linear velocity are related

        v = r w

        w = v / r

        m r v₀ = I₀ v / r + m r v

         m r v₀ = (I₀ / r + mr) v

       v = \frac{m}{\frac{I_o}{r}  +mr} \ r v_o

let's calculate

       v = \frac{0.020}{\frac{1.4}{0.6  } + 0.020 \ 0.6  } \ 0.6 \ 4

       v = \frac{0.020}{2.345} \ 2.4

       v = 0.02 m / s

         

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

        v = x / T

         T = x / v

the distance of a circle with radius r = 0.6 m

         x = 2π r

we substitute

         T = 2π r / v

let's calculate

         T = 2π 0.6/0.02

         T = 188.5 s

reduce

         t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is  C

6 0
3 years ago
A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is paralle
Alex73 [517]

Answer:

715 N

Explanation:

Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.

Let g = 9.8 m/s2

Gravity and equalized normal force is:

N = P = mg = 107*9.8 = 1048.6 N

Kinetic friction force and equalized tension force on the rope is

T = F_{\mu} = N\mu = 1048.6 * 0.682 = 715.1452 N

6 0
2 years ago
Does the ke of a car change more when it accelerates from 11 km/h to 21 km/h or when it accelerates from 21 km/h to 31 km/h?
svp [43]
Titty milk I think because it taste amazing so you can go 21km/h
7 0
3 years ago
Teachers are interested in knowing what study techniques their students are utilizing. The researchers randomly select every 10t
uranmaximum [27]

Answer:

Simple Random Sample (SRS)

Explanation:

5 0
3 years ago
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