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3 years ago

2. The car slows down from 36 m/s to 15 m/s over a 3.0 second time interval. What is its average acceleration?

Physics

2 answers:

stira [4]3 years ago

7 0

Answer:

wendys

Explanation:nutz go in yo mouth

VMariaS [17]3 years ago

6 0

Answer:

9 meter per second square

Explanation:

acceleration equation is final minus initial velocity divided by time in seconds.

36-15/3=9 meter per second square.

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A key falls from a bridge that is 32 m above the water. It falls directly into a model boat, moving with constant velocity, that

FinnZ [79.3K]

Answer:

Speed of the boat, v = 4.31 m/s

Explanation:

Given that,

Height of the bridge, h = 32 m

The model boat is 11 m from the point of impact when the key was released, d = 11 m

Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$t=\sqrt{\dfrac{2s}{g}}$

$t=\sqrt{\dfrac{2\times 32}{9.8}}$

t = 2.55 seconds

Let v is the speed of the boat. It can be calculated as :

$v=\dfrac{d}{t}$

$v=\dfrac{11\ m}{2.55\ s}$

v = 4.31 m/s

So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.

3 0

3 years ago

A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T

Serga [27]

Answer:

a) $k_{e}$ = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m, e) v = 15.23 m / s

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

$k_{e}$ = ½ k x²

$k_{e}$ = ½ 2900 0.80²

$k_{e}$ = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

U = m h and

U = 8 9.8 (-0.80)

U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

K = ½ m v²

K = 0

d) write the energy at two points, maximum compression and maximum height

Em₀ = ke = ½ m x²

$E_{mf}$ = mg y

Emo = $E_{mf}$

½ k x² = m g y

y = ½ k x² / m g

y = ½ 2900 0.8² / (8 9.8)

y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

Y = y- 0.80

Y = 11.8367 -0.80

Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

Emo = ½ k x² = 928 J

$E_{mf}$ = ½ m v²

Emo = $E_{mf}$

Emo = ½ m v²

v =√ 2Emo / m

v = √ (2 928/8)

v = 15.23 m / s

8 0

3 years ago

A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each

irinina [24]

Answer:

1.24 m/s

Explanation:

Metric unit conversion:

9.25 mm = 0.00925 m

5 mm = 0.005 m

The volume rate that flow through the single pipe is

$\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s$

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

$\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s$

So the flow speed of each of the narrower pipe is:

$v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}$

$v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s$

8 0

3 years ago

how much force is needed to cause a 15 kilogram bike to accelerate at a rate of 10 meters per second?

egoroff_w [7]

F = m*a, mass times acceleration.

F = 15*10 = 150 N

8 0

3 years ago

Read 2 more answers

A certain moving electron has a kinetic energy of 0.991 × 10−19 J. Calculate the speed necessary for the electron to have this e

alisha [4.7K]

Answer: The speed necessary for the electron to have this energy is 466462 m/s

Explanation:

Kinetic energy is the energy posessed by an object by virtue of its motion.

$K.E=\frac{1mv^2}{2}$

K.E= kinetic energy = $0.991\times 10^{-19}J$

m= mass of an electron = $9.109\times 10^{-31}kg$

v= velocity of object = ?

Putting in the values in the equation:

$0.991\times 10^{-19}J=\frac{1\times 9.109\times 10^{-31}kg\times v^2}{2}$

$v=466462m/s$

The speed necessary for the electron to have this energy is 466462 m/s

4 0

3 years ago

Read 2 more answers