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Molodets [167]
3 years ago
6

A block floats partially submerged in a container of liquid. When the entire container is accelerated upward, which of the follo

wing happens? Assume that both the liquid and the block areincompressible.
(A) The block descends down lower into the liquid.(B) The block ascends up higher in the liquid.(C) The block does not ascend nor descend in the liquid.
(D) The answer depends on the direction of motion of the container.(E) The answer depends on the rate of change of the acceleration
Physics
1 answer:
Nookie1986 [14]3 years ago
7 0

Answer:C

Explanation:

Partially submerged block along with vessel is accelerated upwards .

Initially the block weight is supported by buoyant force such that it is in equilibrium.

when the system start accelerating upwards then the effective gravity will be

g+a where a is the acceleration of the system.

so only net gravity is increased so block will not ascend or descend.

Mathematically

F_b=weight

\rho _L\cdot V\cdot (g+a)=m(g+a)

where \rho _L=density of liquid

V=volume of object inside the water

                         

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Critical angle of glass is 42 .what does it mean?​
elena55 [62]

Answer:

i think..its fraction that its have multiple fractions on it..if you minus the 397 000-355 it should be 381+ so i say if you get the 5 multiply it by 9!! so you will get it!

Explanation:

HOPE IT HELPS!!

6 0
3 years ago
A 1.50x103-kilogram car is traveling east at 30 meters per second.
frez [133]

Answer:

39000\ \text{kg m/s}

West

Explanation:

m = Mass of car = 1.3\times 10^{3}\ \text{kg}

t = Time = 9 seconds

u = Initial velocity = 30 m/s

v = Final velocity = 0

Impulse is given by

J=m(v-u)\\\Rightarrow J=1.3\times 10^3(0-30)\\\Rightarrow J=-39000\ \text{kg m/s}

The magnitude of the total impulse applied to the car to bring it to rest is 39000\ \text{kg m/s}.

The direction is towards west as the sign is negative.

7 0
3 years ago
How far from the base of the platform does she land?
seropon [69]

When Janet leaves the platform, she's moving horizontally at 1.92 m/s.  We assume that there's no air resistance, and gravity has no effect on horizontal motion.  There's no horizontal force acting on Janet to make her move horizontally any faster or slower than 1.92 m/s.

She's in the air for 1.1 second before she hits the water.

Moving horizontally at 1.92 m/s for 1.1 second, she sails out away from the platform

(1.92 m/s) x (1.1 sec) = <em>2.112 meters</em>

3 0
3 years ago
Compute the power output (watts) during one minute of treadmill exercise, given the following: Treadmill grade-10% Horizontal sp
erma4kov [3.2K]

Answer:

c. 981 watts

P=981\ W

Explanation:

Given:

  • horizontal speed of treadmill, v=100\ m.min^{-1}=\frac{5}{3} \ m.s^{-1}
  • weight carried, w=588.6\ N
  • grade of the treadmill, G\%=10\%

<u>Now the power can be given by:</u>

P=v.w

P=588.6\times\frac{5}{3} (where grade is the rise of the front edge per 100 m of the horizontal length)

P=981\ W

6 0
4 years ago
Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t
katrin2010 [14]

Answer:

1.9161676647\times 10^{13}\ m/s^2 or 1.9532799844\times 10^{12}g

23.4843749996 m

Yes

Explanation:

E = Electric field = 2\times 10^5\ N/C

c = Speed of light = 3\times 10^8\ m/s

m = Mass of proton= 1.67\times 10^{-27}\ kg

q = Charge of electron = 1.6\times 10^{-19}\ C

Acceleration is given by

a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2

Dividing by g

\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g

The acceleration is 1.9161676647\times 10^{13}\ m/s^2 or 1.9532799844\times 10^{12}g

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

5 0
3 years ago
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