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horsena [70]
3 years ago
10

Which of the following would be a valid method to increase the buoyant force acting on an object?

Physics
2 answers:
sergiy2304 [10]3 years ago
5 0
C or B jshsbnsnsnsmaanwjkqjwbs
shepuryov [24]3 years ago
4 0
I think it’s B hope this helps.
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A car accelerates from 20.0 m/s to 28.0 m/s over a distance of 50.0 m. What is the car’s acceleration?
Blababa [14]

Answer:

Explanation:

V^2=V^2_o+2a(x_f-x_i)

28^2=20^2+2a(50)\\784=400+100a\\384=100a\\a=3.84m/s^2

5 0
3 years ago
How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?
Anestetic [448]

Answer:

21 m

Explanation:

The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

d=vt

where

d is the distance covered

v is the speed

t is the time

The frog in this problem has a speed of

v = 2.1 m/s

and therefore, after t = 10 s, the distance it covered is

d=(2.1)(10)=21 m

3 0
4 years ago
Mars moves in an elliptical orbit around the sun, and the mass of mars is much less than the mass of the sun. at the instant sho
Bess [88]
I believe the answer would be that the system potential energy will increase; while the Mar's angular momentum will remain the same. 
Mars rotates on its axis, completing one revolution every 24.6 hours. The axis of mars is tilted at 25 degrees and 12 minutes relative to its orbital plane about the Sun. This produces seasons on the surfaces of Mars, similar to the seasons on Earth. 
4 0
3 years ago
At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball
Dmitriy789 [7]

Answer:

390 N

Explanation:

The net acceleration is:

a² = aₓ² + aᵧ²

a² = (900 m/s²)² + (870 m/s²)²

a = 1250 m/s²

So the net force is:

F = ma

F = (0.31 kg) (1250 m/s²)

F = 390 N

5 0
3 years ago
An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha
LenaWriter [7]

Answer:

t=6.4534 s

Explanation:

This is an exercise where you need to use the concepts of <em>free fall objects</em>

Our <u>knowable variables</u> are initial high, initial velocity and the acceleration due to gravity:

y_{0}=75m

v_{oy} =20m/s

g=9.8 m/s^{2}

At the end of the motion, the <u><em>rock hits the ground</em></u> making the final high y=0m

y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}

If we <em>evaluate the equation</em>:

0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}

This is a classic form of <u><em>Quadratic Formula</em></u>, we can solve it using:

t=\frac{-b ± \sqrt{b^{2}-4ac } }{2a}

a=-4.9\\b=20\\c=75

t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s

t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s

Since the <u><em>time can not be negative</em></u>, the <em>reasonable answer</em> is

t=6.4534s

8 0
3 years ago
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