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givi [52]
3 years ago
5

How many squares for three traits?

Physics
1 answer:
Studentka2010 [4]3 years ago
4 0
You just need one square but with 16 little boxes 
list out all the possible combination for each set of traits 
add then to the each side of the box and match them up 
You might be interested in
An object has a kinetic energy of 175 J and a momentum of magnitude 25.0 kg m/s. Find the
DedPeter [7]

Answer:14 m/s

Explanation:

Kinetic energy(ke)=175J

Momentum(M)=25kgm/s

Speed=v

Mass=m

Ke=(m x v x v)/2

175=(mv^2)/2

Cross multiply

175 x 2=mv^2

350=mv^2

Momentum=mass x velocity

25=mv

m=25/v

Substitute m=25/v in 350=mv^2

350=25/v x v^2

350=25v^2/v

v^2/v=v

350=25v

v=350/25

v=14 m/s

5 0
4 years ago
Pick the answer that would make the following statement true.
Darina [25.2K]

Explanation:

option A is the correct answer, if the gravitational acceleration is taken 10m/s²(rounding of 9.8/ms²).

hope this helps you.

4 0
3 years ago
Which is most likely the length of a student’s textbook?
UkoKoshka [18]

For this case, what we must do is to rewrite these measurements in the same unit in order to compare them.

By writing the measurements in meters we have:

30 mm = (30) * (\frac{1}{1000}) = 0.030 m

30 cm = (30) * (\frac{1}{100}) = 0.30 m

30 dm = (30) * (\frac{1}{10}) = 3 m\\30 Hm = (30) * (100) = 3000 m

Therefore, physically the correct measure is:

0.30 m = 30 cm

Answer:

the length of a student's textbook most likely is:

30 centimeters

3 0
3 years ago
Read 2 more answers
A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure P1 to pressure P2 while los
mafiozo [28]

Answer:

Option B

Change in entropy of the process is \Delta S= Rln(\frac{P_{1}}{P_{2}})

Explanation:

The entropy of a system is a measure of the degree of disorderliness of the system.

The entropy of a system moving from process 1 to 2 is given as

\Delta S = \int\limits^2_1 {\frac{\delta q}{T}}

recall from first law, \delta q =du +Pdv

hence we have, \Delta S = \int\limits^2_1 {\frac{du +Pdv}{T}}

since the process is isothermal, du= 0

this gives us \Delta S = \int\limits^2_1 {\frac{Pdv}{T}}

integrating within the limits of 1 and 2, will give us

\Delta S = R ln (\frac{V_{2}}{V_{1}})

also from ideal gas laws,

\frac{V_{2}}{V_{1}}=\frac{P_{1}}{P_{2}}

hence we have    \Delta S = R ln (\frac{P_{1}}{P_{2}})

This makes the correct option B

4 0
3 years ago
Read 2 more answers
A car rounds a curve. The radius of curvature of the road is R, the banking angle with respect to the horizontal is θ and the co
exis [7]

Answer:

v = √[gR (sin θ - μcos θ)]

Explanation:

The free body diagram for the car is presented in the attached image to this answer.

The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.

Resolving the weight into the axis parallel and perpendicular to the inclined plane,

N = mg cos θ

And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ

Frictional force = Fr = μN = μmg cos θ

Centripetal force responsible for keeping the car in circular motion = (mv²/R)

So, a force balance in the plane parallel to the inclined plane shows that

Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)

(mv²/R) = (mg sin θ - μmg cos θ)

v² = R(g sin θ - μg cos θ)

v² = gR (sin θ - μcos θ)

v = √[gR (sin θ - μcos θ)]

Hope this Helps!!!

5 0
3 years ago
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