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mr_godi [17]
4 years ago
8

Which is true about inelastic collisions: a. An inelastic collision does not obey conservation of energy. b. An inelastic collis

ion conserves kinetic energy. c. Objects will stick together upon collision. d. Momentum is not conserved in inelastic collisions..
Physics
1 answer:
Phoenix [80]4 years ago
5 0

Answer:

Option c is correct

Explanation:

There are two types of collisions-elastic collision and inelastic collision.

In elastic collision, both kinetic energy and total momentum are conserved. On the other hand, in inelastic collision, total momentum is conserved but kinetic energy is not conserved. Thus, option b and d are incorrect.

Total energy is always conserved in both types. Thus, option a is incorrect.

In a perfectly inelastic collision, objects stick together. This happens because maximum kinetic energy is dissipated and used in bonding of the two objects. Thus, correct option is c.

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A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di
dimulka [17.4K]

Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

        Στ = 0

        τ = F r

we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

      + W d₁ - w_cat d₂ = 0

      d₂ = W / w d₁

      d₂ = M /m d₁

      d₂ = 5.00 /2.9    0.850

      d₂ = 1.466 m

6 0
4 years ago
A particle on the x-axis is moving to the right at 2 units per second. At a certain instant it is at the point (5,0). How rapidl
koban [17]

Answer:

dr/dt = 1.94 units per second

Explanation:

A particle is moving on the x-axis to the right at 2 u/s.

To know how is changing the distance of the particle respect to the point (0,9), on the y-axis, you first take into account that the distance between charge and a point over the y-axis is given by:

r^2=x^2+y^2          (1)

Next, you calculate implicitly the derivative of the equation (1) respect to t:

\frac{d}{dt}r^2=\frac{d}{dt}[x^2+y^2]\\\\2r\frac{dr}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}       (2)

Next, you solve the previous equation for dr/dx:

\frac{dr}{dt}=\frac{x(dx/dt)+y(dy/dt)}{\sqrt{x^2+y^2}}       (3)

dx/dt: the speed of the particle on the x-axis = 2

dy/dt: speed of the particle on the y-axis = 0

For the instant given in the statement, you have that:

x = 5

y = 9

Then, you replace the values for x, y, dx/dt and dy/dt in the equation (3):

\frac{dr}{dt}=\frac{2(5)(2)+0}{\sqrt{(5)^2+(9)^2}}=1.94

The speed of change of the distance between particle and point (0,9) is 1.94 units per second

6 0
3 years ago
Can someone help me with this​
Gemiola [76]

Answer:

5 . the correct answer is option 3rd , these continents were once single landmass !

6. correct answer is option 2nd ! Relative dating is the science of determining the relative order of past events (i.e., the age of an object in comparison to another), without necessarily determining their absolute age (i.e. estimated age). In geology, rock or superficial deposits, fossils and lithologies can be used to correlate one stratigraphic column with another.

8 0
3 years ago
A beam of light has a wavelength of 549nm in a material of refractive index 1.50. In a different material of refractive index 1.
Rzqust [24]

Explanation:

someone to check if the answer is correct

6 0
3 years ago
One airplane is approaching an airport from the north at 181 kn/hr. A second airplane approaches from the east at 278 km/hr. Fin
umka2103 [35]

Answer:

The value  is  \frac{dR}{dt} =  -286.2 \  km/hr

Explanation:

From the question we are told that  

   The speed of the airplane from the north is \frac{dN}{dt}  =  -181 \  km /hr

The negative sign is because the direction is towards the south

  The speed of the airplane from the east is  \frac{dE}{dt}  =  -278 \  km/hr

The negative sign is because the direction is towards the west

   The distance of the southbound plane from the airport is  N  =  30 \  km

   The distance of the westbound plane is  E =  15 \  km

Generally the distance between the plane is mathematically represented using Pythagoras theorem  as

    R^2  = N^2 + E^2

Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes

So

     2R\frac{dR}{dt} =  2N \frac{dN}{dt} +   2E\frac{dE}{dt}

Here

     R = \sqrt{N^2 + E^2}

=>    R = \sqrt{30^2 + 15^2}

=>    R = \sqrt{30^2 + 15^2}

=>    R =33.54 \ m

    2(33.54) * \frac{dR}{dt} =  2( 30)*(-181)  +   2*15*(-278)

=>   67.08 * \frac{dR}{dt} =  -19200

=>   \frac{dR}{dt} =  -286.2 \  km/hr

4 0
4 years ago
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