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dexar [7]
3 years ago
5

William was swimming at a speed of 3 m/s when he started to slow down. He slowed to a stop in 2 seconds. What is his acceleratio

n?
Physics
1 answer:
schepotkina [342]3 years ago
3 0

Answer:

-1.5\;\rm m \cdot s^{-2} on average.

Explanation:

The acceleration of an object is the rate of change in its velocity. The average acceleration of an object is equal to the change in velocity over the time required for that change to take place.

For the William in this question:

  • The initial speed of William is 3\; \rm m \cdot s^{-1}.
  • Because William came to a complete stop after 2\; \rm s, his speed at that time would be 0\; \rm m \cdot s^{-1}.

Had William not came to a complete stop, it would be necessary to consider whether there is a change to his orientation. However, because in this question William came to a stop, the change in his velocity would be:

\begin{aligned} &\text{Change in velocity} \\&= \text{Final velocity} - \text{Initial velocity}\\ &= 0\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1} = -3\; \rm m \cdot s^{-1}\end{aligned}.

That change in velocity happened over 2\; \rm s. Therefore, the average acceleration would be:

\begin{aligned}&\text{Average Acceleration} = \frac{\text{Change in velocity}}{\text{Time taken}} = \frac{-3\; \rm m \cdot s^{-1}}{2\; \rm s} \approx -1.5\; \rm m \cdot s^{-1}\end{aligned}.

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A projectile is fired vertically upwards and reaches a height of 78.4 m. Find the velocity of projection and the time it takes t
Musya8 [376]

Answer:

1.) U = 39.2 m/s

2.) t = 4s

Explanation: Given that the

height H = 78.4m

The projectile is fired vertically upwards under the acceleration due to gravity g = 9.8 m/s^2

Let's assume that the maximum height = 78.4m. And at maximum height, final velocity V = 0

Velocity of projections can be achieved by using the formula

V^2 = U^2 - 2gH

g will be negative as the object is moving against the gravity

0 = U^2 - 2 × 9.8 × 78.4

U^2 = 1536.64

U = sqrt( 1536.64 )

U = 39.2 m/s

The time it takes to reach its highest point can be calculated by using the formula;

V = U - gt

Where V = 0

Substitute U and t into the formula

0 = 39.2 - 9.8 × t

9.8t = 39.2

t = 39.2/9.8

t = 4 seconds.

7 0
4 years ago
You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb
mafiozo [28]

Answer:

The angle it subtend on the retina is  \theta_z = 0.44586^o    

Explanation:

From the question we are told that

     The length of the warbler is  L = 14cm = \frac{14}{100} = 0.14m

      The distance from the binoculars is    d = 18cm = \frac{18}{100} = 0.18m

        The magnification of the binoculars is  M =8

Without the 8 X binoculars the  angle made with the angular size of the object  is mathematically represented as

          \theta = \frac{L}{d}

        \theta  = \frac{0.14}{0.18}

           = 0.007778 rad

Now magnification can be represented mathematically as

         M = \frac{\theta _z}{\theta}

Where \theta_z is the angle the image of the warbler subtend on your retina when the   binoculars i.e the  binoculars zoom.

So

      \theta_z = M * \theta

=>    \theta_z =8 * 0.007778

            = 0.0622222224

Generally the conversion to degrees can be mathematically evaluated as

             \theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )

              \theta_z = 0.44586^o  

7 0
3 years ago
What is the connection between angle of incidence and angle of reflection
Radda [10]

Answer:

The normal line divides the angle between the incident ray and the reflected ray into two equal angles. The angle between the incident ray and the normal is known as the angle of incidence. The angle between the reflected ray and the normal is known as the angle of reflection.

8 0
3 years ago
Two identical 0.400 kg masses are pressed against opposite ends of a light spring of force constant 1.75 N/cm, compressing the s
nlexa [21]

Answer:

0.853 m/s

Explanation:

Total energy stored in the spring = Total kinetic energy of the masses.

1/2ke² = 1/2m'v².................... Equation 1

Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.

make v the subject of the equation,

v = e[√(k/m')].................... Equation 2

Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.

Substitute into equation 2

v = 0.39[√(1.75/0.8)

v = 0.39[2.1875]

v = 0.853 m/s

Hence the speed of each mass = 0.853 m/s

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