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kkurt [141]
1 year ago
12

What is the kinetic energy of a 2.0 kg object moving at 5 m/s?

Physics
2 answers:
Zolol [24]1 year ago
3 0

Let's put the given values into the formula below and get the result;

  • K.E=\frac{1}{2}mv^2

We know the numerical value of speed. We also know the mass. We can jump straight to the conclusion.

  • K.E=\frac{1}{2}(2kg)(5m/s)^2
  • =\frac{1}{2}(2kg)(25m^2/s^2)
  • =25m^2/s^2=25J
alexira [117]1 year ago
3 0

The Kinetic Energy is 25J.

If we want to accelerate an object, we must apply force on it, after applying Force some work has to be done in which energy should be transferred to the object. The energy is known as kinetic energy. The kinetic energy always depends on the mass and the velocity. It is denoted as K and the Si unit is Joules (J).

To calculate the Kinetic Energy,

       

           K= 1/2 mv^2

m = mass  

v = velocity

K = kinetic energy

In solving the above equation,

K = 1/2 x2 x 5^2

K = 5x5

K = 25

 

The Kinetic Energy is 25J.

To learn about Kinetic Energy:

brainly.com/question/26472013

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What atoms has most mass?<br>​
Ipatiy [6.2K]
Oganesson has the highest atomic number and highest atomic mass of all know elements.
4 0
2 years ago
the mass of a brick is 4 kg, find the mass of water displaced by it when completely immersed in water.​
vova2212 [387]

Answer:

The correct answer is = 1.6

Explanation:

Density of water = 1000kg/m³ = d₁

Mass of brick = 4kg = m

Density of brick = 2.5 g/cm³ = 2.5 × 1000 =2500 kg/m³ = d₂

Volume of brick = m/d₂ = 4/2500 =16/10000 = 0.0016 L = v

Buoyant Force = v × d₁ × g          (g= acceleration due to gravity =9.8m/s²)

= 0.0016 × 1000 × 9.8 = 15.68 Newtons

By the Archimedes' Principle, the buoyant force is equal to the weight of the liquid displaced by an object.

Weight of the water displaced=Buoyant Force

=Mass of water displaced × g,

as weight = mass × acceleration due to gravity

15.68= mass of brick × 9.8

15.68/9.8 =Mass of water displaced

1.6 kg = Mass of water displaced

4 0
3 years ago
You've recently read about a chemical laser that generates a 20.0-cm-diameter, 26.0 MW laser beam. One day, after physics class,
aksik [14]

Answer :

(a). The speed of the block is 0.395 m/s.

(b). No

Explanation :

Given that,

Diameter = 20.0 cm

Power = 26.0 MW

Mass = 110 kg

diameter = 20.0 cm

Distance = 100 m

We need to calculate the pressure due to laser

Using formula of pressure

P_{r}=\dfrac{I}{c}

P_{r}=\dfrac{P}{Ac}Put the value into the formula[tex]P_{r}=\dfrac{26.0\times10^{6}}{\pi\times(10\times10^{-2})^2\times3\times10^{8}}

P_{r}=2.75\ N/m^2

We need to calculate the force

Using formula of force

F=P\times A

F=P\times \pi r^2

Put the value into the formula

F=2.75\times\pi (0.01)^2

F=0.086\ N

We need to calculate the acceleration

Using formula of force

F=ma

Put the value into the formula

0.086=110\times a

a=\dfrac{0.086}{110}

a=0.000781\ m/s^2

a=7.81\times10^{-4}\ m/s^2

(a). We need to calculate speed of the block

Using equation of motion

v^2=u^2+2ad

Put the value into the formula

v=\sqrt{2\times7.81\times10^{-4}\times100}

v=0.395\ m/s

(b). No because the velocity is very less.

Hence, (a). The speed of the block is 0.395 m/s.

(b). No

8 0
3 years ago
What is the minimum force required to increase the energy of a car by 69 J over a distance of 42 m? Assume the force is constant
sineoko [7]
ANSWER: 1.6N (forth option)



EXPLANATION:
E= Force x displacement
69=F(42)
69/42 = F
F=1.64 N

Answer : 1.6N
3 0
3 years ago
At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counsel
sleet_krkn [62]

Answer:

47 m

Explanation:

Data obtained from the question include the following:

Length of dry leg 1 (L1) = 40 m

Length of dry leg 2 (L2) = 25 m

Length of swimming course (L) =..?

The length of the swimming course can be obtained by using pythagoras theory as shown below:

L² = L1² + L2²

L² = 40² + 25²

L² = 1600 + 625

L² = 2225

Take the square root of both side.

L = √2225

L = 47.1 ≈ 47 m

Therefore, the length of the swimming course is approximately 47 m.

7 0
3 years ago
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