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Anon25 [30]
3 years ago
6

What is the kinetic energy of a 1500 kg object that is moving at a speed of 71 meters per second​

Physics
1 answer:
Fynjy0 [20]3 years ago
4 0

Answer:

3.78075x10^6 kg x meter^2/ second^2

Explanation:

Kinetic energy  K= 1/2 x mass x velocity^2

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Convert 68852 millijoules into Calories. (Write your answer in the decimal form. Do not include units in your answer).
alex41 [277]

Answer: 68852 millijoules = 16.46 calories

Explanation:

Given;

Convert 68852 millijoules to calories.

1 calorie = 4.184J = 4184millijoules

Therefore,

1 millijoule = 1/4184 calories

68852 millijoule = 68852 × 1/4184 calories

= 16.46 calories

6 0
3 years ago
Determine the volume displaced and then calculate the density of this 54 g sample of brass.
inessss [21]

Answer:

DETAILS IN THE QUESTION INSUFFICIENT TO ANSWER

Explanation:

Assuming the liquid to be water ,

the density d_{w} of water is : 1000kgm^{-3}=1gcm^{-3}

Buoyant force exerted by a liquid on an object with V_{imm} of it's volume immersed is :

F_{B}=V_{imm}*d_{l}*g

where ,

  • F_{B} is the buoyant force
  • d_{l} is the density of the liquid
  • g is the acceleration due to gravity

Thus at equilibrium:

m_{brass}*g=V_{imm}*d_{l}*g\\m_{brass}=V_{imm}*d_{l}\\54=V_{imm}*1\\V_{imm}=54cm^{3}

from these , we get the density of brass to be 1gcm^{-3}

which is not possible

7 0
3 years ago
In what ways do people use natural resources? What problems does use of those resources cause?
Lunna [17]

Answer:

The Challenges of Using Natural Resources

Extracting, processing and using natural resources can cause environmental problems such as: air, land and water pollution; disruption or destruction of ecosystems; and a decrease in biodiversity.

4 0
3 years ago
Calculate the Force required to give a bullet of mass 50 g an acceleration of 300 m/s2
NNADVOKAT [17]

Answer:

<h3>The answer is 15 N</h3>

Explanation:

The force acting on an object can be found by using the formula

<h3>Force = mass × acceleration</h3>

From the question

mass = 50 g = 0.05 kg

acceleration = 300 m/s²

We have

force = 0.05 × 300

We have the final answer as

<h3>15 N</h3>

Hope this helps you

7 0
3 years ago
A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
  • ⇒ Kf = 1/2*m*vf²  (2)
  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
7 0
3 years ago
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