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UNO [17]
3 years ago
13

A runner starts slowing down as she gets tired. At 10 seconds, she 8m/a, and slows down to 2m/a at 13 seconds. What is her accel

eration.
Physics
1 answer:
ira [324]3 years ago
3 0

Magnitude of acceleration = (change in speed) / (time for the change) .

Change in speed = (end speed) - (start speed)
                            =        (2 m/a)  -  (8 m/a)  =  -6 m/a .

Time for the change = (end time) - (start time)
                                 =   (13 sec)  -  (10 sec)  =  3 sec .

Magnitude of acceleration =  (-6 m/a) / (3 sec)

                                          =       -2 m/a per second .
 
(That's the best I can do with the ambiguous units.
What unit of speed is 'm/a' ?  Is that "miles per afternoon" ?)

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Mademuasel [1]
I might have did mistake with calculations but this is how you should do.

6 0
3 years ago
Light rays bend as they pass from air into water at an angle (not 90 degrees). this is refraction. air water normal incident ray
xxTIMURxx [149]

When light ray pass from air into water, its speed and wavelength change only the frequency of the light doesn't change.

Light travels slower in a medium of higher refractive index. It bends because of this change in speed. The wavelength of light also changes in order to maintain the constant frequency.

7 0
3 years ago
Suppose a radar gun sends out radio waves with a frequency of 2,000,000.0 Hz. The waves bounce off a moving car and return with
Igoryamba

Answer:

300,000,030m/s

Explanation:

Using the relationship between frequency and speed of a wave;

F ∝ V

F = kV

k = F/V

F1/V1 = F2/V2 = k

Let F1 be the frequency of the radio wave = 2,000,000.0Hz

V1 be the speed of light = 300,000,000m/s

F2 = frequency produced by the car = 2,000,000.2Hz

V2 be the velocity of the moving car

Substituting this values in the equation above;

2,000,000/300,000,000 = 2,000,000.2/V2

Cross multiplying

2,000,000V2 = 300,000,000×2,000,000.2

2V2 = 300×2,000,000.2

V2 = 150×2,000,000.2

V2 = 300,000,030m/s

The velocity of the car is 300,000,030m/s

6 0
3 years ago
Joshua was driving to a friend’s house to study. During his trip, he started on pavement. At one point, he hit an ice patch on t
Tems11 [23]

Answer:

b. Friction decreased when he went from pavement to ice and then increased two more times.

Explanation:

Frictional force depends on the normal force of the surface and a friction coefficient.

F_{f} = -\mu N

Since we're talking about the same car, the value of N will remain constant whereas μ will represent the change in the frictional coefficient of the surface. Now we consider the different surfaces, cars will slide in an icy road which means that the frictional coefficient is smaller than the pavement.

After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.

7 0
3 years ago
Read 2 more answers
A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.
sasho [114]

Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

Normal force (N) = mg CosΘ

μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

tan Θ = 0.29

Θ = 16.17°

b) Let the acceleration be a.

Coefficient of kinetic friction = μ = 0.26

Now, the equation of net force will be:

mg sinΘ - μ mg CosΘ = ma

a = g SinΘ - μg CosΘ

Plugging the values

a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

a = 0.278 m/s^2

Hence, the acceleration is 0.278 m/s^2

7 0
3 years ago
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