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8090 [49]
3 years ago
9

A 350-kg sailboat has an acceleration of 0.62 m/s2 at a angle of 64 degrees north of east find the magnitude and direction of th

e net force that acts on the sail boat
Physics
1 answer:
Kay [80]3 years ago
4 0

Answer:

<em>The net force that acts on the sailboat has a magnitude of 217 N and is applied at an angle of 64° north of east.</em>

Explanation:

<u>Mechanical Force</u>

The second Newton's law states the net force exerted by an external agent on an object of mass m is:

\vec F = m.\vec a

Where \vec a is the acceleration of the object. Note both the force and the acceleration are vectors. The relationship between them is the mass, a scalar.

This means the net force and the acceleration have the same direction.

The sailboat has a mass m= 350 Kg and moves at a=0.62 m/s^2. The magnitude of the net force acting on the boat is:

F=0.62 * 350=217\ N

As stated above its direction is the same as the acceleration, thus:

The net force that acts on the sailboat has a magnitude of 217 N and is applied at an angle of 64° north of east.

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Kay [80]
It would be what 3•15 is so that's 45 m i think
5 0
3 years ago
high school physics, no need detail explain, just give the answer, but you have to make sure thank you
andrey2020 [161]

Answer:

approximately 30 degrees

Explanation:

If it takes the cannonball 2 seconds to reach the maximum height, we can use the analysis of the vertical component of the velocity and the fact that the acceleration of gravity is the one acting opposite to this initial vertical component (v_y) of the velocity. We know as well that at the top of the trajectory, the vertical component of the velocity is zero, and then the movement starts going down in it trajectory. So, the final velocity for the first part of the ascending movement is zero, giving us the following equation for the velocity under an accelerated movement (with acceleration of gravity "g" acting):

v_f=v_i-a\,t\\v_f=v_y-g\,t\\0=v_y-9.8\,*\,2\\v_y=9.8\,*\,2=19.6 \frac{m}{s}

By knowing the vertical component of the initial velocity (19.6 m/s), and the actual magnitude of the total initial velocity (40 m/s), we can calculate what angle was the initial velocity vector forming above the horizontal. We use for such the fact that the sine of the angle relates the opposite side of a right angle triangle with the hypotenuse, and solve for the angle using the arcsin function:

sin(\theta)=\frac{opp}{hyp} \\sin(\theta)=\frac{19.6}{40}\\\theta=arcsin(\frac{19.6}{40})\\\theta=29.34^o

which tells us that the closer answer shown is 30^o

7 0
3 years ago
When a ball is launched from the ground at a 45° angle to the horizontal, it falls back to the ground 50 m from the launch point
Inessa [10]

Answer:

Explanation:

Given

angle through which ball is launched=45^{\circ}

Range of ball=50 m

Range of projectile is =\frac{u^2sin2\theta }{g}

50=\frac{u^2sin90}{9.8}

u=22.136 m/s

If ball is thrown straight upward

v^2-u^2=2as

0-(22.136)^2=2(-9.8)s

s=\frac{22.136^2}{2\times 9.8}

s=25 m

(b)For Projectile time of flight is

t=\frac{2usin\theta }{g}

t=\frac{2\times 22.136\times sin45}{9.8}

t=3.19 s

7 0
3 years ago
Anino acids are important to the body because they build
tino4ka555 [31]

Answer:

Amino acids and proteins are the building blocks of life. When proteins are digested or broken down, amino acids are left. The human body uses amino acids to make proteins to help the body: Break down food.

6 0
3 years ago
Read 2 more answers
After a model rocket reached its maximum height, it then took 5.0 seconds to return to the launch site. what is the approximate
kifflom [539]
Air resistance is ignored.
g = 9.8 m/s².
At maximum height, the vertical velocity is zero.

Let h =  the maximum height reached.
Let u =  the vertical launch velocity.

Because ot takes 5.0 seconds to reach maximum height, therefore
(u m/s) - (9.8 m/s²)*(5 s) = 0 
u = 49 m/s

The maximum height reached is
h = (49 m/s)*(5 s) - (1/2)*(9.8 m/s²)*(5 s)²
   = 122.5 m

Answer: 122.5 m
3 0
3 years ago
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