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3 years ago

Which of following is an example of deduction?

Physics

1 answer:

Jobisdone [24]3 years ago

6 0

A, because Antoine came to a conclusion that it must have rained based on his visuals.

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A race car accelerates at a constant rate from 20 m/s to 60 m/s in a time of 2 seconds. What is the car’s acceleration?

Vlada [557]

$\huge\fbox{Answer ☘}$

Given ,

initial velocity , u = 20 m/s

final velocity , v = 60 m/s

time taken = 2 seconds

Now ,

$\bold{acceleration = \frac{v - u}{t} } \\ \\ = > \bold{ \frac{60 - 20}{2} } \\ \\ = > \bold{\frac{40}{2} } \\ \\ \bold{ => 20 \: ms {}^{ - 2}}$

hope helpful~

5 0

2 years ago

The mass of 25cm of ivory was found to be 0.045kg. caculate the density of ivory in SI units

solmaris [256]

Answer:

dgdhvjgfjfdcdrtdgxbhjcdtsdhcbgtdytfgiuftrshdryrgbiuwdbcuhdds

Explanation:

,ajsgfgsdfubcygduhdsbcdshyudbckjdshfyendhdndbhfdbuehundweugdjnbfcjdfhewjdnbewqwioehsajcnbsfhhgjhbgjhhbjdhgwbvdwnjjguhbwqhjbsjwqhshwbdjagdasndhsajuwqbdjasnjashdjaskxnjgjjsklx ms,dhwshjksnjdbhjssnj,

7 0

2 years ago

A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 33.0 nC , respecti

matrenka [14]

Explanation:

Formula depicting relation between total flux and total charge Q is as follows.

$\phi = \frac{Q}{\epsilon_{o}}$ (Gauss's Law)

Putting the given values into the above formula as follows.

Q = $\phi \times \epsilon_{o}$

= $-953 Nm^{2}/C \times 8.854 \times 10^{-12}$

= $-8.4 \times 10^{-9} C$

= -8.4 nC

Therefore, when the unknown charge is q then,

-14.0 nC + 33.0 nC + q = -8.4 nC

q = -27.4 nC

Thus, we can conclude that charge on the third object is -27.4 nC.

7 0

3 years ago

Suppose the coefficient of static friction between the road and the tires on a car is 0.638 and the car has no negative lift. Wh

larisa [96]

Answer:

12.6332454263 m/s

Explanation:

m = Mass of car

v = Velocity of the car

$\mu$ = Coefficient of static friction = 0.638

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of turn = 25.5 m

When the car is on the verge of sliding we have the force equation

$\dfrac{mv^2}{r}=\mu mg\\\Rightarrow v=\sqrt{\mu gr}\\\Rightarrow v=\sqrt{0.638\times 9.81\times 25.5}\\\Rightarrow v=12.6332454263\ m/s$

The speed of the car that will put it on the verge of sliding is 12.6332454263 m/s

4 0

3 years ago

Explain the relationship between resistance and the velocity of shortening.

butalik [34]

The velocity of shortening refers to the speed of the contraction from the muscle shortening while lifting a load. The relationship between the resistance and velocity of shortening is inverse. The greater the resistance, the shorter the velocity of shortening and the smaller the resistance, the larger the velocity of shortening.

Hopefully this help :)

7 0

3 years ago