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Lisa [10]
3 years ago
12

Which of the following is equal to the boyant force of a rock placed in a beaker of water?

Physics
2 answers:
Elina [12.6K]3 years ago
8 0
B)Weight of displaced fluid
algol133 years ago
4 0
B) Weight of the displaced fluid 

There you go.
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.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c
ohaa [14]

Answer:

 a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

          I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

        F t = 2 (m v)

therefore the average force is

         F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

        Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

        Em_f = K = ½ m v²

energy is conserved in this system

        Em₀ = Em_f

        m g h = ½ m v²

        v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

        F = 2m √(2gh)   /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

          F = m a

          a = F / m

 

          a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

         a = 2 √ (2 9.8 1.5) / 10⁻³

         a = 1.1 10⁵ m / s²

6 0
3 years ago
This is for physical science, need help :/
marusya05 [52]

1).  From the frame of reference of a passenger on the airplane looking out of his window, the tree appears to be moving, at roughly 300 miles per hour toward the left of the picture.

2).  The SI unit best suited to measuring the height of a building is the meter.

3).  'Displacement' is the straight-line distance and direction from the start-point to the end-point, regardless of the path that was followed to get there.  

The ball started out in the child's hand, and it ended up 2 meters away from her in the direction of the wall.  So the displacement of the ball from the beginning to the end of the story is:  2 meters toward the wall.


6 0
3 years ago
A 50 kg object traveling at 100 m/s collides (perfectly elastic) with a 50 kg object initially at rest.
tino4ka555 [31]

Answer:

a

Explanation:

8 0
3 years ago
It is easier to see clothes with pointed needle than a blunt one​
topjm [15]

Answer:

Explanation:

It is easier to see clothes with pointed needle than a blunt one because pressure exerted is more in a pointed needle as it occupies less space compared to blunt needle, A blunt has more surface area so the pressure exerted will less as compared to a pointed needle.

IF YOU FOUND MY ANSWER USEFUL THEN PLEASE MARK ME BRAINLIEST.

7 0
3 years ago
A ball is thrown horizontally from the top of a building 37.5 m high. The ball strikes the ground at a point 80.3 m from the bas
trasher [3.6K]

Answer:

t = 2.77 s

Explanation:

The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:

Equation of movement of the ball in the X axis

X = v₀x*t   Equation  (1)

Equation of movement of the ball in the Y axis

Y = y₀+ v₀y*t -½ g*t² Equation  (2)

Where

X : horizontal position in meters (m)  

Y : vertical  position in meters (m)  

y₀ : initial vertical  position in meters (m)

v₀x :  X-initial speed in m/s  

v₀y :  Y-initial speed in m/s  

g: acceleration due to gravity in m/s²

t : time to position (X,Y)

Data

y₀ = 37.5 m

v₀y = 0

g = 9.8 m/s²

Problem development

The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0

We apply the Equation (2):

Y = y₀+ (v₀y)*t - (½) g*t²

0 =  37.5 +(0)*t- (1/2)*(g)*t²

0 =  37.5 - (1/2)*(9.8)*t²

(1/2)*(9.8)*t²  = 37.5

t² = (2)(37.5)/(9.8)

t= \sqrt{\frac{2*37.5}{9.8} }

t = 2.77 s

5 0
3 years ago
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