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4 years ago

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Physics

2 answers:

pav-90 [236]4 years ago

7 0

Answer:

???

Explanation:

mojhsa [17]4 years ago

4 0

Answer:

the question, plz?

Explanation:

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Calculate the volume of this regular solid,

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2144.7

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3 years ago

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A compound contains two or more different elements.
It also contains two or more atoms.

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A 15 kg dog jumps out a stationary sled which has a mass of 40 kg. If

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3 years ago

Two 2.3 kg bodies, A and B, collide. The velocities before the collision are and . After the collision, . What are (a) the x-com

myrzilka [38]

Missing details in the question:

The velocities before collision are

$u_A=(40i+49j) m/s$

$u_B=(35i+11j) m/s$

After the collision:

$v_A=(14i+21j) m/s$

(a) 61 m/s

We can solve the problem by simply treating separately the x- and the y-components of the motion.

Here we want to analzye the motion along x. We have:

$u_A = 40 m/s$ is the initial velocity of A along the x-direction

$u_B = 35 m/s$ is the initial velocity of B along the x-direction

$v_A = 14 m/s$ is the final velocity of A along the x-direction

$v_B = ?$ is the final velocity of B along the x-direction

Since the total momentum along the x-direction must be conserved, we can write

$mu_A + mu_B = mv_A + mv_B$

where

m = 2.3 kg is the mass of the two bodies. Since the mass is the same, we can eliminate it from the equation,

$u_A + u_B = v_A + v_B$

And so, we find the final velocity of B along the x-direction:

$v_B = u_A + u_B - v_A=40+35-14=61 m/s$

(b) 39 m/s

Similarly to what we did in part a), here we analyze the conservation of momentum along the y-direction.

We have:

$u_A = 49 m/s$ is the initial velocity of A along the y-direction

$u_B = 11 m/s$ is the initial velocity of B along the y-direction

$v_A = 21 m/s$ is the final velocity of A along the y-direction

$v_B = ?$ is the final velocity of B along the y-direction

Since the total momentum along the y-direction must be conserved, we can write

$mu_A + mu_B = mv_A + mv_B$

Since the mass is the same, we can eliminate it from the equation,

$u_A + u_B = v_A + v_B$

And so, we find the final velocity of B along the y-direction:

$v_B = u_A + u_B - v_A=49+11-21=39 m/s$

c) +615 J

Here we have to find the total kinetic energy before and after the collision first.

First, we have to find the speed of each object before and after the collision. We have:

$u_A = \sqrt{40^2+49^2}=63.2 m/s\\u_B = \sqrt{35^2+11^2}=36.7 m/s\\v_A = \sqrt{14^2+21^2}=25.2 m/s\\v_B = \sqrt{61^2+39^2}=72.4 m/s$

So, the total kinetic energy before the collision was

$K_i = \frac{1}{2}mu_A^2+\frac{1}{2}mu_B^2 = \frac{1}{2}(2.3)(63.2)^2+\frac{1}{2}(2.3)(36.7)^2=6143 J$

While after the collision

$K_f = \frac{1}{2}mv_A^2+\frac{1}{2}mv_B^2 = \frac{1}{2}(2.3)(25.2)^2+\frac{1}{2}(2.3)(72.4)^2=6758 J$

So, the change in kinetic energy is

$\Delta K = K_f - K_i = 6758-6143 = +615 J$

(note that the system cannot gain kinetic energy in the collision, unless there is an external force acting on it)

3 0

3 years ago

Advertising claims sometimes state that adding something mechanical to a car's engine will allow it to recover 100 percent of th

Andrei [34K]

Answer:

C. second law of thermodynamics.

Explanation:

One of the consequences of this law is that there is no process of energy transformation 100% efficient. Some energy will always be lost in in the form of heat, which will be used to raise the temperature of some engine component, or its surroundings, and we will not be able to take advantage of it.

6 0

3 years ago