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antiseptic1488 [7]
3 years ago
10

What is the mass and density of 237 mL of water

Physics
1 answer:
oee [108]3 years ago
8 0

Answer:

<h2><em>V(water)= 237 mL=237×10^-6 m^3</em></h2><h2><em>ρ(water)=1000 kg/m^3</em></h2><h2><em>m=</em><em>ρ×V=(1000)×(237×10^-6)</em></h2><h2><em>m= 237×10^-3 = 0.237 kg</em></h2><h2><em>m= 237 gram.</em></h2>
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When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin
mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}

where n=1,2,3 or ...n

L=Length of string

T=Tension

\mu =Mass per unit length

When string is clamped at mid-point

Effecting length becomes L'=0.5 L

Thus new Frequency becomes

f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz      

4 0
3 years ago
You are working at a company that manufactures electri- cal wire. Gold is the most ductile of all metals: it can be stretched in
Alona [7]

Explanation:

We know that the relation between volume and density is as follows.

      Volume = \frac{\text{mass}}{\text{density}}

So,       V = \frac{10^{-3}}{19.3 \times 10^{3} kg/m^{3}}

               = 5.181 \times 10^{-8} m^{3}

Now, we will calculate the area as follows.

      Area = \frac{\text{volume}}{\text{length}}

               = \frac{5.181 \times 10^{-8} m^{3}}{2.4 \times 10^{3}}

               = 2.15 \times 10^{-11} m^{2}

Formula to calculate the resistance is as follows.

         R = \rho \frac{l}{A}

             = \frac{2.44 \times 10^{-8} \times 2400}{}2.15 \times 10^{-11}}

             = 2.71 \times 10^{6} ohm

Thus, we can conclude that the resistance of given wire is 2.71 \times 10^{6} ohm.

4 0
3 years ago
A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s
Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is:   w_{c} =m_{c} g   and maximum tension is T=2.50 m_{c} g=1.41*10^4N

total mass and weight is :

M =m_{c}+ m_{b} =740kg+550kg=1290 kg

w_{M} =1.2650*10^4N

∑F_{y} =ma_{y}

T-M_{g} =Ma_{y}

a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg

=0.645m/s^2

B)

maximum acceleration

a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0

using y-y_{0} =v_{oy} t+1/2(a_{y} )t^2

to solve for t

t=\sqrt{2(y-y_{0} )/a_{y} }

t=\sqrt{2(119m)/0.645m/s^2} =19.20s

6 0
3 years ago
A person throws a ball from height of 6 feet with an initial vertical velocity of 48 feet per second. Use the vertical motion mo
True [87]

Explanation:

Recall the equation for time is distance divided by speed. Here you can use that to solve for "t".

7 0
3 years ago
The four-wheel-drive all-terrain vehicle has a mass of 385 kg with center of mass G2. The driver has a mass of 75 kg with center
jenyasd209 [6]

The coefficient of friction is missing and it has a value of μ = 0.4

Answer:

a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

Where;

N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

μ(N1 + N2) = (mp + mv) x a

Thus,

0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

Thus,

1150a = 4512.6N

a = 4512.6/1150

a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

7 0
3 years ago
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