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rewona [7]
3 years ago
6

A jet ski accelerates towards a ramp at 2.5 m/s/s for 35 s until it finally flies off the water. Determine the

Physics
1 answer:
zlopas [31]3 years ago
8 0

Answer:

1531 m

Explanation:

The motion of the jet ski is an uniformly accelerated motion, so we can find the distance travelled by using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For the jet ski in this problem,

a=2.5 m/s^2

t = 35 s

u = 0 (it starts from rest)

Solving for s, we find the distance travelled:

d=0+\frac{1}{2}(2.5)(35)^2=1531 m

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A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a
Rasek [7]

Answer:

W=173.48J

Explanation:

information we know:

Total force: F=45N

Weight: w=100N

distance: 4m

vertical component of the force: F_{y}=12N

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: F_{x}=Fcos\theta

vertical component: F_{y}=Fsen\theta

but from the given information we know that F_{y}=12N

so, equation these two F_{y}=Fsen\theta and F_{y}=12N

Fsen\theta =12N

and we know the force F=45N, thus:

45sen\theta=12

now we clear for \theta

sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

F_{x}=Fcos\theta

F_{x}=45cos(15.466)\\F_{x}=43.37N

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J

the work done is W=173.48J

7 0
3 years ago
PLS HELP, I AM WILLING TO MARK BRAINLIEST TO FIRST ANSWER AND HAS TOO BE RIGHT
Ann [662]
Weak winds that blow for short periods of time with a short fetch.
6 0
3 years ago
Read 2 more answers
2. A can filled with sand has a mass of 0.65kg is swung overhead in a horizontal circle of radius 0.70m at a constant rate of 2.
Aliun [14]
<h3><u>Answer</u>;</h3>

≈ 5 Kgm²/sec

<h3><u>Explanation</u>;</h3>

Angular momentum is given by the formula

L = Iω, where I is the moment of inertia and ω is the angular speed.

I = mr², where m is the mass and r is the radius

 = 0.65 × 0.7²

 = 0.3185

Angular speed, ω = v/r

                              = (2 × 3.142 × r × 2.5) r

                              =  15.71

Therefore;

Angular momentum =  Iω

                                 = 0.3185 × 15.71

                                 = 5.003635

                                 <u>≈ 5 Kgm²/sec</u>

6 0
3 years ago
A satellite’s velocity is 30000m/s. After 60 secs, it’s velocity shows to 15000m/a. What is the satellite’s acceleration?
Orlov [11]

Answer:

Acceleration = 9 × 10^5 m/s^2 ( deceleration )

Explanation:

From the first equation of motion:

V = u + at

15000 = 30000 + 60a

a = ( 15000-30000)/60

a = 9 × 10^5 m/s^2

4 0
3 years ago
People often use simple machines like pulleys, levers, and ramps because they say the machine “makes the work easier.” Which of
dimulka [17.4K]
I think the answer is D.
Hope this help
3 0
2 years ago
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