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rjkz [21]
3 years ago
12

If the speed of an object doubles, how does that affect its kinetic energy? A. Halves B. Doubles C. Quarters D. Quadruples

Physics
2 answers:
kvasek [131]3 years ago
7 0

Answer is :

D. Quadruples

wlad13 [49]3 years ago
6 0

Answer:

D. Quadruples

Explanation:

If the speed of an object doubles, it affects its kinetic energy because it quadruples.

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A boat floats in water a. Archimedes b. Bernoulli c. Density d. Pascal e. Pressure
Ivenika [448]
The answer is Density !, Do you also need an example ?

Rate this the brainliest answer , Thank youuu !
7 0
3 years ago
Read 2 more answers
How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 C?
fomenos

Answer:The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.

Explanation:

7 0
3 years ago
A solid disk of radius 1.4 cm and mass 430 g is attached by a wire to one of its circular faces. It is twisted through an angle
Neporo4naja [7]

Answer:

    f= 4,186  10²  Hz

Explanation:

El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por

             w = √ k/I

donde ka es constante de torsion de hilo e I es el momento de inercia del disco

El  momento de inercia de indican que giran un eje que pasa                 por enronqueces

           I= ½ M R2  

reduzcamos las cantidades al sistema SI

         R= 1,4 cm = 0,014  m

         M= 430 g = 0,430 kg

substituimos

           w= √ (2 k/M R2)

calculemos  

           w = RA ( 2 370 / (0,430  0,014 2)

           w = 2,963 103 rad/s

la velocidad angular esta relacionada con la frecuencia por

            w =2pi f

            f= w/2π

            f= 2,963 10³/ (2π)

            f= 4,186  10²  Hz

5 0
3 years ago
Imagine a system where a block rests on an inclined plane. The block is then given an initial push so that it starts sliding dow
Helen [10]

Answer:

statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.

Explanation:

The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.

The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction

here, the downward direction signifies the downward motion parallel to the inclined plane.

Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.

Hence, for the block to stop sliding the the above statement should be true.

6 0
3 years ago
An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
just olya [345]

Answer:

2.78\times 10^{-35}\ \text{kg m/s}

6.178\times 10^{-34}\ \text{m/s}

0.31\times 10^{-4}\ \text{m/s}

Explanation:

\Delta x = Uncertainty in position = 1.9 m

\Delta p = Uncertainty in momentum

h = Planck's constant = 6.626\times 10^{-34}\ \text{Js}

m = Mass of object

From Heisenberg's uncertainty principle we know

\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}

The minimum uncertainty in the momentum of the object is 2.78\times 10^{-35}\ \text{kg m/s}

Golf ball minimum uncertainty in the momentum of the object

m=0.045\ \text{kg}

Uncertainty in velocity is given by

\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}

The minimum uncertainty in the object's velocity is 6.178\times 10^{-34}\ \text{m/s}

Electron

m=9.11\times 10^{-31}\ \text{kg}

\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}

The minimum uncertainty in the object's velocity is 0.31\times 10^{-4}\ \text{m/s}.

6 0
2 years ago
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