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rosijanka [135]
3 years ago
5

A water bath in a physical chemistry lab is 1.55 m long, 0.710 m wide, and 0.570 m deep (high). If it is filled to within 3.55 i

n from the top, how many liters of water are in it?
Physics
1 answer:
Lesechka [4]3 years ago
7 0

Answer:

528 liter.

Explanation:

Volume of the tank(cuboid) = l*b*h

But volume of the water = l*b*h

Where

l= length of the tank

b = width of the tank

h = the length from the bottom of the tank,

3.55 in to m,

0.09017m

Length of the water in the tank = 0.570 - 0.09017

= 0.47983 m.

Volume = 0.47983*0.710*1.55

= 0.528 m3.

1 m3 = 1000 liter.

0.528 m3 = 0.528*1000

= 528 liter

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Block A can slide relative to block B which, in turn, can slide on a perfectly smooth horizontal plane. If the initial velocity
Anna11 [10]

Answer:

the final velocity of the two blocks is v = \frac{mv_o}{m+M}

the distance that A slides relative to B is S = \frac{v_o^2M}{2 \mu g (M+m)}  

Explanation:

From the diagram below;

acceleration of A relative to B is : a = - ( \mu g  + \frac{ \mu mg}{M})

where

v = u + at

0 = v_o + ( - \mu g - \frac{\mu m g }{M})t

Making t the subject of the formula; we have:

t = \frac{v_o M}{(\mu g )(M+m)}

v^2 = u^2 +2 as\\\\0^2 = v_o^2 - 2 (\mu g ) (\frac{M+m}{M})S\\\\

S = \frac{v_o^2M}{2 \mu g (M+m)}  which implies the distance that A slides relative to B.

The final velocities of the two blocks can be determined as follows:

v = u + at

v = v_o - \mu g \frac{v_oM}{\mu g (M+m)}\\\\v = \frac{\mu g mv_o}{m+M}\\\\

v = \frac{mv_o}{m+M}

Thus, the final velocity of the two blocks is v = \frac{mv_o}{m+M}

4 0
3 years ago
4) A racing car undergoing constant acceleration covers 140 m in 3.6 s. (a) If it’s moving at 53 m/s at the end of this interval
Scorpion4ik [409]

Answer

given,

distance = 140 m

time, t = 3.6 s

moving speed = 53 m/s

a) distance = (average velocity) x time

    D = \dfrac{v_0 + v_1}{2}\times t

    140 = \dfrac{v_0 + 53}{2}\times 3.6

       v₀ + 53 = 77.78

        v₀ = 24.78 m/s or 25 m/s

b) a = \dfrac{v-u}{t}

   a = \dfrac{53-25}{3.6}

         a = 7.8 m/s²

using equation of motion

  v₀² = v₁² + 2 a s

  53² = 0²+ 2 x 7.8 x s

  s = 180 m

3 0
3 years ago
Please help!!!!
Dafna11 [192]

The intensity of the electric field is 30,000 N/C

Explanation:

The strength of the electric field produced by a single-point charge is given by the equation

E=k\frac{q}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

q=3\cdot 10^{-9}C is the magnitude of the charge

r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity

Substituting, we find:

E=(8.99\cdot 10^9)\frac{3\cdot 10^{-9}}{(0.03)^2}=30,000 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
10. A satellites is in a circular orbit around the earth at a height of 360 km above the earth’s surface. What is its time perio
Afina-wow [57]

Answer:

Orbital speed=8102.39m/s

Time period=2935.98seconds

Explanation:

For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)

V2R+h=g(R2(R+h)2)

V=√g(R2R+h)

V= sqrt(9.8 × (6371000)^2/(6371000+360000)

V= sqrt(9.8× (4.059×10^13/6731000)

V=sqrt(65648789.18)

V= 8102.39m/s

Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)

T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)

T=sqrt(3.40×10^21)/ (3.99×10^14)

T= sqrt(0.862×10^7)

T= 2935.98seconds

4 0
3 years ago
The magnitude of the electric force between two protons is 2.30 10-26 n. how far apart are they?
SCORPION-xisa [38]
The electric force between two charge objects is calculated through the Coulomb's law.
                               F = kq₁q₂/d²
The value of k is 9.0 x 10^9 Nm²/C² and the charge of proton is 1.602 x10^-19 C. Substituting the known values from the given,
                           2.30x10^-26 = (9.0 x 10^9 Nm²/C²)(1.602 x10^-19C)²/d²
The value of d is equal to 0.10 m. 
7 0
3 years ago
Read 2 more answers
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