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denis23
1 month ago
7

The acceleration of segment D is m/s2. Rank segments A, B , C from least accelerations to greatest acceleration. Least

Physics
2 answers:
Ostrovityanka [3.2K]1 month ago
8 0

Answer:

D, C, B, A

Explanation:

The derivative from a velocity-time graph provides the acceleration value.

Segment A

\frac{dy}{dx} = \frac{15m/s}{1s} = 15m/s^2

Segment B

\frac{dy}{dx} = \frac{5m/s}{1s} = 5m/s^2

Segment C

\frac{dy}{dx} = \frac{0m/s}{2s} = 0m/s^2

Segment D

\frac{dy}{dx} = \frac{-20m/s}{1s} = -20m/s^2

Sorted from the lowest to the highest acceleration:

D, C, B, A

Keith_Richards [3.2K]1 month ago
7 0

Segment D is -20 m/s squared

C

B

A

I concluded this after completing the calculations.

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Alicia intends to swim to a point straight across a 100 m wide river with a current that flows at 1.2 m/s. She can swim 2.5 m/s
Sav [3153]

Answer:

θ = 61.3°

Alicia must swim at an angle of 61.3°

Explanation:

Parameters given include:

Width of the river = 100 m

Alicia's speed in still water = 2.5 m/s

Speed of river's current = 1.2 m/s

The angle she needs to swim can be determined by combining the velocities, taking into account the current's influence.

Her swimming speed aimed against the current must offset the current's velocity;

2.5cosθ - 1.2 = 0

2.5cosθ = 1.2

cosθ = 1.2/2.5

θ = cosinverse(1.2/2.5)

θ = 61.3°

4 0
2 months ago
A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro
Ostrovityanka [3204]

Answer:

a) v_{o} =16m/s

b) v=9.8m/s

c) \beta =-35.46º

Explanation:

According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.

With this information, we can compute the ball’s starting speed.

a) Let's first assess the horizontal trajectory.

x=v_{ox}t

x=v_{o}cos(60)t

v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)} (1)

This gives us our initial equation.

Next, we need to examine the vertical trajectory.

y=y_{o}+v_{oy}t+\frac{1}{2}gt^2

y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2

Utilizing v_{o} in our first equation (1)

8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2

\frac{1}{2}(9.8)t^2=16tan(60)-8

Now let’s solve for t.

t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s

The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.

v_{o}=\frac{16m}{(2s)cos(60)}=16m/s

b) To determine the velocity magnitude just before impact, we must calculate both x and y components.

v_{x}=v_{ox}+at=16cos(60)=8m/s

v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s

The computed velocity magnitude is:

v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s

c) The ball's angle is:

\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º

4 0
2 months ago
Suppose you analyze standardized test results for a country and discover almost identical distributions of physics scores for fe
kicyunya [3294]

Answer:

consult a teacher

Explanation:

go to your school

locate the teacher

request assistance from him/her

complete the question and you're done:)

7 0
2 months ago
A tennis ball bounces on the floor three times. If each time it loses 22.0% of its energy due to heating, how high does it rise
Ostrovityanka [3204]

Answer:

H = 109.14 cm

Explanation:

Given,                                                            

Assume that the total energy equals 1 unit.                                

Energy remaining after the first collision = 0.78 x 1 unit

Balance after the first impact = 0.78 units

Remaining energy after the second impact = 0.78 ^2 units

Balance after the second impact = 0.6084 units

Remaining energy after the third impact = 0.78 ^3 units

Balance after the third impact = 0.475 units

The height reached after the third collision is equivalent to the remaining energy.

Let H denote the height achieved after three bounces.

0.475 (m g h) = m g H                  

H = 0.475 x h                                    

H = 0.475 x 2.3 m                          

H = 1.0914 m                      

H = 109.14 cm                      

6 0
2 months ago
A small block of mass 200 g starts at rest at A, slides to B where its speed is vB=8.0m/s,vB=8.0m/s, then slides along the horiz
serg [3582]

Answer

Data provided:

mass of the block = 200 g = 0.2 Kg

Velocity at A = 0 m/s

Velocity at B = 8 m/s

distance of slide = 10 m

height of the block = 4 m

calculation for the block's potential energy

    P = m g h

    P = 0.2 x 9.8 x 4

    P = 7.84 J

kinetic energy calculated as

    KE = \dfrac{1}{2}mv^2

    KE = \dfrac{1}{2}\times 0.2 \times 7.84^2

    KE =6.14 J

Work done = P - KE

work = 7.84 - 6.14

work = 1.7 J

b) using the formula v² = u² + 2 a s

   0 = 8² - 2 x a x 10

   a = 3.2 m/s²

ma - μ mg = 0

 \mu = \dfrac{a}{g}

 \mu = \dfrac{3.2}{9.8}

 \mu = 0.327

7 0
2 months ago
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