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3 years ago

10. A boat traveling at 9.5 m/s reduces its acceleration at a rate of 2.5 m/s2. What is the final speed of the boat

Physics

1 answer:

Fiesta28 [93]3 years ago

7 0

Answer:

6.75m/s

Explanation:

using the second equation of motion, the time is calculated.

and with the formula a= (v - u)/t

where a is acceleration but in this case it's deceleration (and should be negated as you solve it ) .

v is final velocity

u is initial velocity

t is time taken

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A 1100 kgkg safe is 2.4 mm above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compres

Drupady [299]

Answer:

191.36 N/m

Explanation:

From the question,

The Potential Energy of the safe = Energy of the spring when it was compressed.

mgh = 1/2ke²............... Equation 1

Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression

Making k the subject of the equation,

k =2mgh/e²................ Equation 2

Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m

Constant: g = 9.8 m/s²

Substitute into equation 2

k = 2(1100)(9.8)(0.0024)/0.52²

k = 51.744/0.2704

k = 191.36 N/m

Hence the spring constant of the heavy-duty spring = 191.36 N/m

3 0

3 years ago

A car traveling 34 mi/h accelerates uniformly for 4 s, covering 615 ft in this time. What was its acceleration? Round your answe

Contact [7]

Answer:

51.94 ft/s²

257.63 ft/s

Explanation:

t = Time taken = 4 s

u = Initial velocity = 34 mi/h

v = Final velocity

s = Displacement = 615 ft

a = Acceleration

Converting velocity to ft/s

$34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s$

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2$

Acceleration is 51.94 ft/s²

$v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s$

Final velocity at this time is 257.63 ft/s

5 0

3 years ago

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the drag force?

timurjin [86]

Answer: The drag force goes up by a factor of 4

Explanation:

The Drag Force equation is:

$F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2}$ (1)

Where:

$F_{D}$ is the Drag Force

$C_{D}$ is the Drag coefficient, which depends on the material

$\rho$ is the density of the fluid where the bicycle is moving (air in this case)

$A_{D}$ is the transversal area of the body or object

$V$ the bicycle's velocity

Now, if we assume $C_{D}$ , $\rho$ and $A_{D}$ do not change, we can rewrite (1) as:

$F_{D}=C.V^{2}$ (2)

Where $C$ groups all these coefficients.

So, if we have a new velocity $V_{n}$ , which is the double of the former velocity:

$V_{n}=2V$ (3)

Equation (2) is written as:

$F_{D}=C.V_{n}^{2}=C.(2V)^{2}$

$F_{D}=4CV^{2}$ (4)

Comparing (2) and (4) we can conclude the Drag force is four times greater when the speed is doubled.

7 0

3 years ago

قوة الجذب المركزي تكون في اتجاه

pochemuha

Answer:

تكون دائمًا متعامدة مع سرعة الجسم وتكون دائمًا في اتجاه مركز انحناء المسار

Explanation:

6 0

3 years ago

Read 2 more answers

a wall switch is connected to a wall outlet. A lamp is plugged into the same outlet. You turn the switch on, but the lamp stays

jolli1 [7]

1: only half the outlet is switched and the lamp is in the other half
2: the lamp is turned off.
3: The light bulb is burned out
4: the switch might be broken
5: the fuse might be blown
6: the electricity might be off

5 0

3 years ago