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3 years ago

What must occur for a process to be a chemical reaction?

Physics

1 answer:

Evgen [1.6K]3 years ago

8 0

The answer is d !!!!!!

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A child has two red wagons, with the rear one tied to the front by a (non-stretching) rope. If the child pushes on the rear wago

Sergeeva-Olga [200]

Answer:

The rear wagon gains the kinetic energy, but the front wagon will remain at rest.

The two-wagon system will gain a kinetic energy $\dfrac{1}{16}$ of the kinetic energy gained by the rear wagon.

Explanation:

Let's consider that the masses of the wagons to be 'M'. When the child pushes the rear wagon let's assume that the velocity of the rear wagon be 'v'.

Therefor the kinetic energy gained by the rear wagon be $K_{r} = \frac{1}{2}Mv^{2}$ .

Now let's assume that the velocity of the centre of mass (C), as shown in the figure, be 'V'. So from momentum conservation law we can write,

$&& Mv = (M + M)V\\\\&or,& V = \frac{v}{2}$

Now the centre of mass ( $M_{C}$ ) is given by

$M_{C} = \frac{M \times M}{M + M} = \frac{M}{2}$

So the kinetic energy ( $K_{C}$ ) of the system will be

$K_{C} = \frac{1}{2}M_{C}V^{2} = \frac{1}{2}\frac{M}{2}(\frac{v}{2})^{2} = \frac{1}{16}Mv^{2}$

5 0

3 years ago

By which method does heat travel from the sun to Earth? A. Conduction B. Convection C. Evaporation D. Radiation

qaws [65]

Answer:

radiation

Explanation:

7 0

3 years ago

Read 2 more answers

Should we focus on the p-value instead of the alpha level?

madreJ [45]

Answer:

Yes, we must focus on p-value along with the alpha level

Explanation:

Alpha and P values are provides two different set of information. While on one hand alpha values sets an extreme value set before rejecting the hypothesis, P value determines the extremeness of the values in the data set. Thus, if p values is lesser than or equal to alpha value, the null hypothesis is rejected, If p value is greater than alpha value, then the null hypothesis is not rejected.

3 0

3 years ago

A person throws a stone from the corner edge of a building. The stone's initial velocity is 28.0 m/s directed at 43.0° above the

Naya [18.7K]

The stone's acceleration, velocity, and position vectors at time $t$ are

$\mathbf a(t)=-g\,\mathbf j$

$\mathbf v(t)=v_{i,x}\,\mathbf i+\left(v_{i,y}-gt\right)\,\mathbf j$

$\mathbf r(t)=v_{i,x}t\,\mathbf i+\left(y_i+v_{i,y}t-\dfrac g2t^2\right)\,\mathbf j$

where

$g=9.80\dfrac{\rm m}{\mathrm s^2}$

$v_{i,x}=\left(28.0\dfrac{\rm m}{\rm s}\right)\cos43.0^\circ\approx20.478\dfrac{\rm m}{\rm s}$

$v_{i,y}=\left(28.0\dfrac{\rm m}{\rm s}\right)\sin43.0^\circ\approx19.096\dfrac{\rm m}{\rm s}$

and $y_i$ is the height of the building and initial height of the rock.

(a) After 6.1 s, the stone has a height of 5 m. Set the vertical component $(\mathbf j)$ of the position vector to 5 m and solve for $y_i$ :

$5\,\mathrm m=y_i+\left(19.096\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.1\,\mathrm s)^2$

$\implies\boxed{y_i\approx70.8\,\mathrm m}$

(b) Evaluate the horizontal component $(\mathbf i)$ of the position vector when $t=6.1\,\mathrm s$ :

$\left(20.478\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)\approx\boxed{124.92\,\mathrm m}$

(c) The rock's velocity vector has a constant horizontal component, so that

$v_{f,x}=v_{i,x}\approx20.478\dfrac{\rm m}{\rm s}$

where $v_{f,x}$

For the vertical component, recall the formula,

${v_{f,y}}^2-{v_{i,y}}^2=2a\Delta y$

where $v_{i,y}$ and $v_{f,y}$ are the initial and final velocities, $a$ is the acceleration, and $\Delta y$ is the change in height.

When the rock hits the ground, it will have height $y_f=0$ . It's thrown from a height of $y_i$ , so $\Delta y=-y_i$ . The rock is effectively in freefall, so $a=-g$ . Solve for $v_{f,y}$ :

${v_{f,y}}^2-\left(19.096\dfrac{\rm m}{\rm s}\right)^2=2(-g)(-124.92\,\mathrm m)$

$\implies v_{f,y}\approx-53.039\dfrac{\rm m}{\rm s}$

(where we took the negative square root because we know that $v_{f,y}$ points in the downward direction)

So at the moment the rock hits the ground, its velocity vector is

$\mathbf v_f=\left(20.478\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(-53.039\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which has a magnitude of

$\|\mathbf v_f\|=\sqrt{\left(20.478\dfrac{\rm m}{\rm s}\right)^2+\left(-53.039\dfrac{\rm m}{\rm s}\right)^2}\approx\boxed{56.855\dfrac{\rm m}{\rm s}}$