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Contact [7]
2 years ago
10

A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex

t, divided by, start text, s, end text, end fraction to the right when the driver turns on the motor. The boat speeds up for 6.0\,\text s6.0s6, point, 0, start text, s, end text with an acceleration of 4.0\,\dfrac{\text m}{\text s^2}4.0 s 2 m ​ 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction leftward.
Physics
1 answer:
love history [14]2 years ago
8 0

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

\Delta x = 30 - 2.36

\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

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When laser light shines on a screen after passing through two closely spaced slits, it becomes
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Answer:

diffracted into semicircular waves. constructive interference occurs where the waves are crest to crest or trough to trough, destructive interference occurs where they are crest to trough. The light that falls on the screen produces bands of light and dark fringes on the screen as a result of these constructive and destructive interferences. This is called the young's slit experiment.

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1 year ago
This is a cell, which is the basic unit of all life. All organs in human bodies are made of cells and require oxygen to survive.
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The two systems that work together to deliver oxygen are D, respiratory and cardiovascular
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2 years ago
In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Suppose that during 115 m of her walk Salvador is observed to
dexar [7]

Answer:

t = 25 seconds

Explanation:

Given that,

Distance, d = 115 m

Initial speed, u = 4.2 m/s

Final speed, v = 5 m/s

We need to find the time taken in increasing the speed.

We know that,

Acceleration, a=\dfrac{v-u}{t} ....(1)

The third equation of kinematics is as follows :

v^2-u^2=2ad\\\\\text{Put the value of a in above equation}\\\\v^2-u^2=2\times \dfrac{v-u}{t}\times d\\\\\because (a^2-b^2)=(a-b)(a+b)\\\\(v-u)(v+u)=\dfrac{2\times (v-u)d}{t}\\\\t=\dfrac{2d}{v+u}\\\\\text{Putting all the values}\\\\t=\dfrac{2\times 115}{4.2+5}\\\\t=25\ s

Hence, it will take 25 seconds to increase the speed.

6 0
2 years ago
American Football Field Uses A Field That Is 100.0 Yd Long, Whereas A Soccer Field Is 100.0m Long. Which Field Is Longer And By
postnew [5]
Note that
1 yd = 0.9144 m

Therefore,
The length of an American Football field is
(100 yds)*(09144 m/yd) = 91.44 m

Because the soccer field is 110 m long, its length exceeds the American Football Field by
100 - 91.44 = 8.56 m
or
(8.56/.9144) =  9.36 yd
This difference is equivalent to (8.56/91.44)*100 = 9.4%

Answer:
The Soccer Field is longer by
8.56 m, or
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4 0
2 years ago
A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of
ivann1987 [24]

Answer:

a rock of 50kg should be placed =drock=0.5m from the pivot point of see saw

Explanation:

τchild=τrock  

Use the equation for torque in this equation.

(F)child(d)child)=(F)rock(d)rock)

The force of each object will be equal to the force of gravity.

(m)childg(d)child)=(m)rockg(d)rock)

Gravity can be canceled from each side of the equation. for simplicity.

 (m)child(d)child)=(m)rock(d)rock)  

Now we can use the mass of the rock and the mass of the child. The total length of the seesaw is two meters, and the child sits at one end. The child's distance from the center of the seesaw will be one meter.

(25kg)(1m)=(50kg)drock

Solve for the distance between the rock and the center of the seesaw.

drock=25kg⋅m50kg

drock=0.5m

6 0
2 years ago
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