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Finger [1]
3 years ago
8

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1,

the rocket engine is shut down and the sled moves with constant velocity v for a time t2. If the total distance traveled by the sled is 18,350 ft and the total time is 90 s, find the following. (a) the times t1 and t2 t1 = s t2 = s (b) the velocity v
Physics
1 answer:
mash [69]3 years ago
3 0

Answer:

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 − t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ − 44 t₁²

18350 = 3960 t₁ − 22 t₁²

9175 = 1980 t₁ − 11 t₁²

11 t₁² − 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.

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The given problem can be exemplified in the following diagram:

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\Sigma F=ma

Replacing the values:

mg\sin 40=ma

We may cancel out the mass:

g\sin 40=a

Using the gravity constant as 9.8 meters per square second:

9.8\frac{m}{s^2}\sin 40=a

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6.3\frac{m}{s^2}=a

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8 0
1 year ago
A block is held at rest against a wall by a force of magnitude F exerted at an angle theta from the horizontal, as shown in the
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Answers:

B.) F cos\theta=F_{n}

C.) F sin\theta=F_{g} \pm F_{f}

Explanation:

The image attached shows the way the force F is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:

Net Force in X:

-F_{n}+F cos\theta=0 (1)

Where:

F_{n} is the Normal force

F is the magnitude of the force exerted on the block

\theta is the angle

Net Force in Y:

F sin\theta \pm F_{f}-F_{g}=0 (2)

Where:

F_{f} is the Friction force (it is expresed with the \pm sign because this force may be up or down, we cannot know because the block is at rest)

F_{g} is the gravity force

Rewrittin (1):

F cos\theta=F_{n} (3) This is according to option B

Rewritting (2):

F sin\theta=F_{g}\pm F_{f} (3) This is according to option C

3 0
3 years ago
The distance around the block is equal to 3000 feet, how many blocks would you have to run to run a total of a mile?
lesya692 [45]
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7 0
3 years ago
An airplane is moving at 350 km/hr. If a bomb is
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Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
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A spatially challenged goldfish swims along the x-axis only. Its initial position is 7.8 m. After swimming back and forth a whil
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Answer:

<em>The fish displacement was -3.4 m</em>

Explanation:

<u>Distance and Displacement</u>

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dtotal=d1+d2+d3+...+dn

This sum is obtained independently of the direction the object moves.

The displacement only takes into consideration the initial and final points of the path defined by the object in its moving. The displacement, unlike distance, is a vectorial magnitude and can be even zero if the object starts and ends the movement at the same point.

The displacement, when the object moves in one axis only is given by:

d = final position - initial position

We know the fist started at 7.8 m from a given reference along the x-axis.

After some undisclosed movements, it ends up at the position 4.4 m. Thus, the displacement is:

d = 4.4 m - 7.8 m = -3.4 m

This means the fish ended up to the left of the position it started from.

The fish displacement was -3.4 m

3 0
3 years ago
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