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3 years ago

A floating boat has a weight of 100 N. What is the buoyant force on the boat?

1 answer:

4 0

If the boat is floating, then it's just sitting there, and not accelerating
up or down. That means the vertical forces on it must be balanced.

So if its weight (acting downward) is 100 newtons, then the buoyant
force on it (acting upward) must also be 100 newtons.

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I need help in physics, pls hurry! I will give Brainliest for correct answers!!

Nana76 [90]

Top left: slowing down
Top right: not moving
Bottom left: moving at a constant speed
Bottom right: speeding up

6 0

2 years ago

The displacement vector A and B when added together , give the resultant vector R so that R= A+B use the data in the drawing and

Salsk061 [2.6K]

The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.

Let:

b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°

We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.

Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.

In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.

5 0

3 years ago

Effects of the earths orbit around the sun

makvit [3.9K]

<h3><u>Effects of the earths orbit around the sun:</u></h3>

The earth moves around the sun in an elliptical orbit, Johannes Kepler, a "German mathematician, and astronomer" described this elliptical orbit first. The orbit is close to being a circle but not a circle. Earth orbiting the sun mainly effects on seasons on earth.

Earth's four seasons are determined when Earth is tilted 23.4 degrees on the vertical axis, which is called as “axial tilt”. When a "southern hemisphere is tilted towards the sun", it experiences summer and northern hemisphere experiences winter, exactly opposite happens when northern hemisphere tilts towards Sun and this climate change goes on in all countries.

7 0

3 years ago

A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image

Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

$=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}$

$=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}$

Re-arrange to get i

$\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m$

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

$m=\frac{h'}{h} =-\frac{i}{o}$

substitute the values to get the height of image h'

$\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm$

5 0

3 years ago

Whar are codes of behavior in the gym that should be followed

Sunny_sXe [5.5K]

Answer: Always ask. ...

Put your weights back. ...

Be mindful of your surroundings. ...

Bring a towel. ...

Clean up after yourself (and, if you have to, other patrons). ...

Selfies: Try to be discreet about them. ...

Personal space: respect it. ...

Don't hog the equipment.

5 0

3 years ago