A tuna fish body is streamlined to reduce friction. Which describes the type of friction the fish must overcome?

almond37 [142]

The displacement is just the distance between the starting point and the
ending point, and you don't care about the route taken or the actual distance
covered along the way.

This can bite you sometimes. For example, some day you'll be given the
diameter of a circle, and you'll be asked for the displacement of an ant that
walks around the circle 17 times and finally stops at the same place it
started from. You might go to work calculating the circumference of the
circle and multiplying it by 17. But if you think about it first, you realize that
if the ant ends up at the place he started from, then his displacement is zero.

6 0

3 years ago

Heat transfer between two substances is affected by specific heat and the

Pie

<span>Heat transfer between two substances is affected by specific heat and the "Temperature difference between them"

Hope this helps!</span>

6 0

3 years ago

Read 2 more answers

Which force acts on the plane at a distance?

shutvik [7]

Answer:

gravity

Explanation:

6 0

3 years ago

A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of

puteri [66]

The frequency and amplitude of the SHM beam is 0.8 Hz and 0.098 m. The frequency of the SHM wave when gravel falls is 0.8 Hz and the amplitude of subsequent SHM beam is 0.4m.

(a) Mass of the spring = 225 Kg

Mass of the sack = 175 Kg

Amplitude of the beam = 40 cm = 0.40 m

Frequency of the beam = F = 0.60 cycles/s

The formula for frequency of oscillation =

= f = (1/2π) X √(k/m)

where, k = 2π²F²m

= k = 2 X (3.14)² X 0.6² X (225 + 175)

= k = 5685.37 N/m

Strength of the spring before gravels fall = x =

= x = mg / k

= x = [ (225 + 175 ) X 9.8 ] / 5685.37

= x = 0.689 m

But, the spring is stretched by the distance of x' which is expressed as,

= X = x - x'

= X = 0.689 - 0.40

= X = 0.289 m

Now, since we know that the gravel falls, thus frequency = f =

= f = (1/2π) X √(k/m)

= f= (1/ 2 X 3.14) X √ 5685.37 / 225

= f = 0.8 Hz

(b) Assuming that the spring is stretched, x = mg/k =

= x = (225 X 9.8) / 5685.37

= x = 0.3878 m

Thus, the amplitude of the sack = A = 0.3878 - 0.289

= A = 0.098 m

(c) If the gravel falls, the speed is maximum hence speed = s =

= s = A X √(k/m)

= s = 0.4 X √(5685.37/400)

= s = 1.508 m/s

The frequency = f' =

= f' = (1/2π) X √(k/m)

= f' = (1/2 X 2.14) X √(5685.37/225)

= f' = 0.8 Hz

(d) New amplitude = A' =

= A' = 0.38 + 0.038 (after calculating the new distance)

= A' = 0.4 m

To know more about Spring:

brainly.com/question/15850235

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8 0

2 years ago

If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what wo

Elena L [17]

Answer: The question is incomplete or some details are missing. Here is the complete question ; (a) The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.55 m/s2 for 4.05 s, making straight skid marks 63.0 m long, all the way to the tree. With what speed (in m/s) does the car then strike the tree? m/s

(b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2

a ) With what speed (in m/s) does the car then strike the tree? m/s = 4.3125m/s

b) then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2 = -5.696m/s2

Explanation:

The detailed steps and calculation is as shown in the attached file.

6 0

3 years ago