1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
inna [77]
3 years ago
8

The buoyant force on an object placed in a liquid is (a) always equal to the volume of the liquid displaced. (b) always equal to

the weight of the object. (c) always equal to the weight of the liquid displaced. (d) always less than the volume of the liquid displaced.
Physics
1 answer:
Nana76 [90]3 years ago
7 0

Answer:

(c) always equal to the weight of the liquid displaced.

Explanation:

Archimedes principle (also called physical law of buoyancy) states that when an object is completely or partially immersed in a fluid (liquid, e.t.c), it experiences an upthrust (or buoyant force) whose magnitude is equal to the weight of the fluid displaced by that object.

Therefore, from this principle the best option is C - always equal to the weight of the liquid displaced.

You might be interested in
A particle leaves the origin with an initial velocity v → = (3.00iˆ) m/s and a constant acceleration a → = (−1.00iˆ − 0.500jˆ) m
tatiyna

Answer:

the position vector (x,y) will be (1.5 m,-2.25 m) and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s) when x reaches its maximum x coordinate

Explanation:

Since the velocity is related with the acceleration and coordinates through

vx²=v₀x²+2*ax*x

where

vx = velocity in the x direction

v₀x = initial velocity in the x direction = 3 m/s

ax = acceleration in the x direction = −1.00 m/s²

x= coordinates in the x-axis

when x reaches its maximum coordinate , then vx=0

thus

vx²=v₀x²+2*ax*x

0 = (3 m/s)² + 2* (−1.00 m/s²)*x

x= 1.5 m

also for the time t

vx = v₀x + ax*t → t= (vx-v₀x)/ax = (0- 3 m/s)/  (−1.00 m/s²) = 3 seconds

for the y coordinates

y = y₀+v₀y*t + 1/2 ay*t²

where

v₀y = initial velocity in the y direction = 0 m/s

ay = acceleration in the x direction = −0.5 m/s²

y= coordinates in the y-axis

y₀= initial coordinate in the y-axis =0

then since y₀=0 and v₀y=0

y = 1/2*ay*t²

y = 1/2*ay*t² = 1/2*(−0.5 m/s²)*(3 s)² = -2.25 m

and

vy=v₀y+ ay*t= 0+(−0.5 m/s²)*(3 s)= (-1.5 m/s)

therefore the position vector (x,y) will be (1.5 m,-2.25 m)

and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s)

7 0
3 years ago
A cheetah can accelerate from rest at the rate of 4m /s
lesya [120]
Acceleration of cheetah (a) = 4m/s²
time = 10s
initial velocity(u) = 0
final velocity = v
distance travelled = s

v = u +at = 0 + 10×4 = 40m/s
s = (v²-u²)/2a = 40²/(2×4) = 1600/8 = 200m
   
6 0
3 years ago
Ayuda por favor (archivo adjunto) con un ejercicio de expresión sobre periodo de oscilación de esta figura:
vodka [1.7K]

Answer:

nolo se

Explanation:

no lo se

8 0
3 years ago
200 pages about string thirory
joja [24]

Answer:

do not know the answer

Explanation:

7 0
3 years ago
Read 2 more answers
A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force (F_{B}), as follows:

F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB (1)

<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

5 0
3 years ago
Other questions:
  • If it takes a ball dropped from rest 2.069 s to fall to the ground, from what height h was it released?
    5·1 answer
  • You place a box weighing 200.2 N on a inclined plane that makes a 37.1 angle with the horizontal. Compute the component of the g
    14·1 answer
  • Una pelota rueda por una pendiente con aceleración constante. Cuando pasa por un punto P su velocidad es 1,2 ms-1 y al cabo de u
    13·1 answer
  • Which of these would have particles with the highest kinetic energy? Ice Water Water Vapor
    5·2 answers
  • A baby carriage is sitting at the top of a hill that is 21m high. the carriage with the baby weigh's 12 kg. the carriage has ene
    6·1 answer
  • If the pressing force on the slidding surfaces is greater then friction will be
    8·1 answer
  • Hi please zoom in to see it clearly, uh you don’t have to answer them all but it would be nice !!! (no links please) :D
    10·1 answer
  • It’s a true or false question , needing help with it
    6·1 answer
  • If copper (E⁰ = ⁺0.34) and lithium (E⁰ = ⁻3.05) are used to make a battery, will electrons flow from lithium to copper or vice v
    12·1 answer
  • A 5.0 kg block hangs from the ceiling by a mass-less rope. A Second block with a mass of 10.0 kg is attached to the first block
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!