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Mila [183]
3 years ago
14

An object was placed in 50 ml of water. When the object was then measured in the water. The new volume of the water with the obj

ect was 86 ml. What is the volume of the object?
Physics
1 answer:
slega [8]3 years ago
8 0

Answer:

36 ml

Explanation:

The volume of the water was 50 ml without anything in it. When you put something in water, the amount of volume that it rises is the volume of the object put into it. It rose up to 86 ml.

86-50=36

36 ml

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TC
frosja888 [35]

Answer:

0.08 sin 3nt +

metre. Then calculate-

a time period

(d) Time period

(

Minitial phase

wõisplacement

7

(c) Displacement from mean position at t=

36

sec.

Explanation:

thats the answer

6 0
3 years ago
During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 4.02 s, how high above the point where i
alukav5142 [94]

Answer:

d = 19.796m

Explanation:

Since the ball is in the air for 4.02 seconds, the ball should reach the maximum point from the ground in half the total time, therefore, t=2.01s to reach maximum height. At the maximum height, the velocity in the y-direction is 0.

So we know t=2.01, vi=0, g=a=9.8m/s and we are solving for d.

Next, you look for a kinematic equation that has these parameters and the one you should choose is:

d=vt+\frac{1}{2}at^2

Now by substituting values in, we get

d=(0\frac{m}{s})*(2.01s)+\frac{1}{2}(9.8\frac{m}{s^2})(2.01)^2

d = 19.796m

7 0
3 years ago
A van has a weight of 4000 lb and center of gravity at Gv. It carries a fixed 900 lb load which has a center of gravity at Gl. I
natulia [17]

Answer:

 x = 25 / μ     [ ft]

Explanation:

To solve this exercise we can use Newton's second law.

Let's set a reference system where the x axis is parallel to the road

Y axis  

       N_B + N_A - W_van - W_load = 0

       N_B + N_A = W_van + W_load

X axis

     fr = ma

     a = fr / m

the total mass is

        m = (W_van + W_load) / g

the friction force has the expression

      fr = μ N_{total}

      fr = μy (W_van + W_load)

we substitute

      a = μ (W_van + W_load)    \frac{g}{W_van + W_load}

      a = μ g

taking the acceleration let's use the kinematic relations where the final velocity is zero

       v² = v₀² - 2 a x

       0 = v₀² -2a x

        x = \frac{v_o^2}{2a}

        x = \frac{v_o^2}{2 \mu g}

        x = \frac{40^2}{2 \ 32 \  \mu}

        x = 25 / μ     [ ft]

5 0
3 years ago
A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows: ρ(r)=ρ0(1−r/r) for r
Nadusha1986 [10]

A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² - r³/R) dr 
which when integrated from 0 to r is 
total charge = 4π * ρ0 (r³/3 + r^4/(4R)) 
and when r = R our total charge is 
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3 
and after substituting ρ0 = 3Q / πR³ we have 
total charge = Q ◄ 

B) E = kQ/d² 
since the distribution is symmetric spherically 

C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 - r/R)dr 
so 
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is 
and after substituting for ρ0 is 
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4)) 
which could be expressed other ways. 

D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that 
r = R for a min/max (and we know it's a max since r = 0 is a min). 

<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>

4 0
3 years ago
If the specific gravity of copper is 8.91, it weighs:
denis-greek [22]

Answer:

Option D is the correct answer.

Explanation:

The specific gravity of copper is 8.91.

We have

            \frac{\texttt{Density of copper}}{\texttt{Density of water}}=8.91\\\\\texttt{Density of copper}=8.91\times 1000kg/m^3=8910kg/m^3

We also have

           1 kg = 2.205 lb

            1 m = 3.28 ft

Substituting

            \texttt{Density of copper}=8910kg/m^3=\frac{8910\times 2.205}{3.28^3}=556.76lb/ft^3

Option D is the correct answer.

6 0
3 years ago
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