How to calculate specific heat.

) Determining the order of events and the relative age of rocks by examining the position of

tigry1 [53]

The answer is c; relative dating

5 0

3 years ago

You watch a distant lady driving nails into her front porch at a regular rate of 2 strokes per second. You hear the sound of the

Harlamova29_29 [7]

Since the lady is capable of delivering 2 strokes in 1 second, we know that it takes 0.5 seconds for one stroke of the hammer.

It is given that one more blow is heard even after she is seen as stopping hammering. This means that the sound from the last blow reached us a little later: 0.5 seconds after the last blow.

So, we know that time taken for the sound from the last blow to reach us t = 0.5 s

Speed of Sound is given as V = 340 m/s

Distance of the lady from us D = ?

Using the equation D = V.t, we get D = (340)(0.5) = 170 m

Thus, we can understand that the lady is 170 meters away from us.

3 0

3 years ago

Which of these activities performed by a chemist is primarily based on an understanding of physics?

Yanka [14]

Predicting the volume of a gas, given its temperature and pressure p<span>roperties.</span>

4 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

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

(d) The acceleration vector stays constant throughout, so

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

4 0

3 years ago

A coin is resting on the bottom of an empty container. The container is then fi lled to the brim three times, each time with a d

ahrayia [7]

when we placed in the container of real depth assuming = d

if the container filled with liquid A then its apparent depth d' = 7cm

so the refractive index nA = real depth / apparent depth

= d/7cm = 0.1428d

if the container filled with liquid B then its apparent depth d' = 6cm

so the refractive index nB = real depth / apparent depth

= d/6cm = 0.166d

if the container filled with liquid C then its apparent depth d' = 5cm

so the refractive index nC = real depth / apparent depth

= d/5cm = 0.2d

Since the refractive index is inversely proportional to the apparent depth

then the refractive indices are nC > nB > nA

3 0

3 years ago