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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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In 1939 or 1940, Emanuel Zacchini took his human-cannonball act to an extreme. After being shot from a cannon, he soared over th

Arlecino [84]

We first determine the vertex by using the formula,<span>-b/2a = vertex, in order to get the values for the t-coordinate. That is why we got
</span>
v_y=26.5 sin(53)=21.163v_x=26.5 cos(53)=15.948
then

let x=0since you are going to land on a 3m tally=-.5(9.8)t^2+ 21.163*t
y=0=-4.9t+21.163t=4.31
vx*4.31= total distance travelled=68.88m
Then for the first wheel, you have 15.948m=vxdetermine the time when he reaches 23 meters, that is

23/15.948=1.44218 sec

substitute t with1.44218 sec, then determine the height.
h(1.44218)=20.329

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34.438m away from cannon

8 0

3 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

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3 years ago

A ball is attached to a vertical spring. The ball is initially supported at a height y so that the spring is neither stretched n

scoundrel [369]

Answer:

All the three quantities will have non zero joules.

Explanation:

At the initial position of rest the system will have only gravitational potential energy while the other 2 quantities will be zero.

when the system reaches a height (y-h) only kinetic energy will be zero while the other 2 quantities will be non zero

At the position of (y-h/2) all the 3 quantities will be non zero.

3 0

3 years ago

Which statement is correct? (2 points) Select one:

Olin [163]

<span>The correct answer is B - Light can travel in a vacuum, and its speed is constant if the source is moving or stationary.</span>

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3 years ago

Definition: A potential path for the flow of electricity with at least one gap in the path that prevents this flow. The gap is u

Andreas93 [3]

Answer:

Open circuit

Explanation:

An open circuit is simply an electrical circuit that is not complete. In such a circuit, there is a gap and this will not allow the electric current to pass through.

Despite all the elements being complete in the circuit, an open circuit will halt the flow of electric current and will not do deliver the necessary energy it is supposed to.

In such a circuit, the wires are cut of and not connected properly.

The reverse is a closed circuit.

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3 years ago

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