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
Effectus [21]
3 years ago
14

What is the maximum mass that can hang without sinking from a 20-cm diameter Styrofoam sphere in water? Assume the volume of the

mass is negligible compared to that of the sphere.
Physics
1 answer:
alex41 [277]3 years ago
3 0

Answer:

the maximum mass that can hang without sinking is 2.93 kg

Explanation:

Given: details:

sphere diameter  d = 20 cm

so, radius r = 10 cm  = 0.10 m

density of the Styrofoam sphere D = 300 kg/m3

sphere volume  V = \frac{4}{3} \pi r^3

                                                   =\frac{4}{3} \pi 0.10^3

                                                   =4.18*10^{-3} m^3

we know that

Density = \frac{Mass}{Volume}

mass  M = Density * Volume

                                  = (300)(4.18*10^{-3} m3)

                                  =1.25 kg

mass of the water displace = volume *density  of water

                                                 = 4.18*10^{-3} m3 * 1000

                                                 = 4.18 kg

The difference between the mass of water and mass of styrofoam is the amount of mass that the sphere can support

=4.18 kg  -1.25 kg

= 2.93 kg

You might be interested in
A(n) 636 kg elevator starts from rest. It moves upward for 4.5 s with a constant acceleration until it reaches its cruising spee
zmey [24]

Answer:

The average power delivered by the elevator motor during this period is 6.686 kW.

Explanation:

Given;

mass of the elevator, m = 636 kg

initial speed of the elevator, u = 0

time of motion, t = 4.5 s

final speed of the elevator, v = 2.05 m/s

The upward force of the elevator is calculated as;

F = m(a + g)

where;

m is mass of the elevator

a is the constant acceleration of the elevator

g is acceleration due to gravity = 9.8 m/s²

a = \frac{v-u}{t} \\\\a = \frac{2.05 -0}{4.5} \\\\a = 0.456 \ m/s^2

F = (636)(0.456 + 9.8)

F = (636)(10.256)

F = 6522.816 N

The average power delivered by the elevator is calculated as;

P_{avg} = \frac{1}{2} (Fv)\\\\P_{avg} = \frac{1}{2} (6522.816 \ \times \ 2.05)\\\\P_{avg} = 6685.89 \ W\\\\P_{avg} =  6.68589 \ kW\\\\P_{avg} =  6.686 \ k W

Therefore, the average power delivered by the elevator motor during this period is 6.686 kW.

3 0
3 years ago
Which subatomic particle has a positive charge?
VMariaS [17]
The answer is protons
Electrons have negative charge and neutrons have 0 charge
6 0
3 years ago
Read 2 more answers
Calculate the magnitude of the angular momentum of the earth due its daily rotation about its own axis
ioda
L=Iw
I_{earth} =  \frac{2}{5} r^{2}m= \frac{2}{5} (6371000^2m)(5.97*10^{24}kg)=9.71*10^ {37}
L_{earth}=Iw=(9.71*10^ {37})(7.27*10^{-5})=7.07*10^{33}<span>[/tex]</span>

7 0
3 years ago
A person pushes horizontally with a force of 220. N on a 61.0 kg crate to move it across a level floor. The coefficient of kinet
Ede4ka [16]

Answer:

(a) 161.57 N

(b) 0.958 m/s^2

Explanation:

Force applied, F = 220 N

mass of crate, m = 61 kg

μ = 0.27

(a) The magnitude of the frictional force,

f = μ N

where, N is the normal reaction

N = m x g = 61 x 9.81 = 598.41 N

So, the frictional force, f = 0.27 x 598.41

f = 161.57 N

(b) Let a be the acceleration of the crate.

Fnet = F - f = 220 - 161.57

Fnet = 58.43 N

According to newton's second law

Fnet = mass x acceleration

58.43 = 61 x a

a = 0.958 m/s^2

Thus, the acceleration of the crate is 0.958 m/s^2.  

7 0
3 years ago
A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a
Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

\Delta x=v_0t+\frac{gt^2}{2}

In this case, the sphere starts from rest, so v_0=0. Replacing the given values and solving for g':

g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}

The acceleration due to gravity near Earth's surface is g=9.8\frac{m}{s^2}. So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0
3 years ago
Other questions:
  • A sample of gold has a mass of 38.6 grams and a volume of 2 cm3 what is the density of gold
    9·2 answers
  • The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0 above the horizon
    11·1 answer
  • 4. What happens during fertilization that makes the offspring unique from the original cells?
    5·1 answer
  • G A dragster starts from rest and accelerates at 35 m/s2 m / s 2 . How fast is it going after t t
    11·1 answer
  • Jack and Jill exercise in a 25.0 m long swimming pool. Jack swims 9 lengths of the pool in 156.9 s ( 2 min and 36.9 s ) , wherea
    5·1 answer
  • a force 10N drags a mass 10 kg on a horizontal table with acceleration 0.2m\s. If the acceleration due to gravity is 10m\s2, a c
    7·1 answer
  • A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.570 diopters. Since these
    12·1 answer
  • A car initially traveling at 21.4 m/s accelerates at a rate of 4.4 m/s2 for 7.5 seconds. What is the final velocity of the car?
    6·1 answer
  • A testable prediction is a(n) ______. Group of answer choices a. hypothesis b. experiment c. exercise d. variable
    14·2 answers
  • The image shows devices that convert wind energy into electrical energy. What is one advantage of using this type of device in p
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!