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Sauron [17]
3 years ago
14

Please answer this question for me ​

Physics
1 answer:
jenyasd209 [6]3 years ago
4 0

Answer:

1- 66mL

2-16mL

3-7.8mL

Explanation:

they are numbered from left to right, and have the number of the values. Dont forget the measurements, because sometimes if they are forgotten, they are counted wrong.

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A body with a mass of 5 Kg has the same kinetic energy as a second body. The second body has a mass of 10 kg and is moving at a
vitfil [10]
First body:

E_{k_1}=\frac{1}{2}.m.v^2\\
\\
E_{k_1}=\frac{1}{2}.5.v_1^2=\frac{5}{2}.v_1^2

Second body:

E_{k_2}=\frac{1}{2}.m.v_2^2\\
\\
E_{k_2}=\frac{1}{2}.10.(20)^2=2.000 \ J

From description of the task we have:

E_{k_1}=E_{k_2}\\
\\
\frac{5}{2}.v_1^2=2.000 \ J\\
\\
5.v_1^2=4.000 \ J\\
\\
v_{1}^2=800 \ J\\
\\
v_{1}= \sqrt{800} \ J \approx 28,3 \ m/s
7 0
3 years ago
The force of attraction between two like charged table tennis balls is 2.4 × 10-5 newtons. If the charge on the one is 3.8 × 10-
k0ka [10]
To have a force of 3.8x10-8 the two chaarges must be 0.65m apart. 

7 0
4 years ago
Read 2 more answers
Line segment Q R , Line segment R S and Line segment S Q are midsegments of Î"WXY. Triangle R Q S is inside triangle X Y W. Poin
Whitepunk [10]

The perimeter of ΔWXY is : ( D ) 14.5 cm

<u>Calculating the </u><u>perimeter </u><u>of ΔWXY</u>

QR = WY / 2

RS = XW / 2

QS = XY / 2

Given that : QR = 2.93 cm ,  RS = 2.04 cm,  QS = 2.28 cm

Therefore

Perimeter of ΔWXY = ∑ WY + XW + XY  

                                 = 2SR + 2QS + 2QR

                                 = 2(2.04) + 2(2.28) + 2(2.93)

                                 = 14.5 cm

Hence we can conclude that the perimeter of ΔWXY = 14.5 cm

learn more about perimeter calculations : brainly.com/question/24744445

8 0
3 years ago
Read 2 more answers
Two tiny, spherical water drops, with identical charges of −8.00 ✕ 10^(−17) C, have a center-to-center separation of 2.00 cm.
Damm [24]

Answer:

F=1.4384×10⁻¹⁹N

Explanation:

Given Data

Charge q= -8.00×10⁻¹⁷C

Distance r=2.00 cm=0.02 m

To find

Electrostatic force

Solution

The electrostatic force between between them can be calculated from Coulombs law as

F=\frac{kq^{2} }{r^{2} }

Substitute the given values we get

F=\frac{(8.99*10^{9} )*(-8.00*10^{-17} )^{2} }{(0.02)^{2} }\\ F=1.4384*10^{-19} N

7 0
3 years ago
A 1459 kg car is traveling WEST at 43 m/s. A 9755 kg truck is traveling EAST at 11 m/s. They collide head-on, and stick together
otez555 [7]

Answer:

<em>Both vehicles move east at 3.97 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states that the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is:

P=mv.

If we have a system of two bodies, then the total momentum is the sum of both momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

Assume both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

Assuming east direction to be positive, we have an m1=1459 kg car traveling west at v1=-43 m/s. An m2=9755 kg truck is traveling east at v2=11 m/s. They collide head-on and stick together after that.

Computing the resultant velocity after the collision:

\displaystyle v'=\frac{1459*(-43)+9755*11}{1459+9755}

\displaystyle v'=\frac{44568}{11214}

v' = 3.97 m/s

Both vehicles move east at 3.97 m/s

4 0
3 years ago
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