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3 years ago

When do you add forces

Physics

1 answer:

san4es73 [151]3 years ago

5 0

When we need to calculate the resultant value of them.......

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Pls help me i have to submit this due tomorrow :'D

aniked [119]

$\textsf {a) As there is no change in the initial and final positions, the displacement is}\\\textsf {equal to 0.}$

b) Finding total distance :

Distance travelled from 0 s to 5 s :

1/2 x 5 x 30 = 75 m

Distance travelled from 5 s to 10 s :

5 x 30 = 150 m

Distance travelled from 10 s to 15 s :

150 + 1/2 x 5 x 5
150 + 12.5 = 162.5 m

Distance travelled from 15 s to 20 s :

36 x 5 + 1/2 x 5 x 4
180 + 10 = 190 m

Distance travelled from 20 s to 25 s :

45 x 5 + 1/2 x 5 x 5
225 + 12.5 = 237.5 m

Distance travelled from 25 s to 30 s :

40 x 5 + 1/2 x 10 x 5
200 + 25 = 225 m

Distance travelled from 30 s to 35 s :

20 x 5 + 1/2 x 20 x 5
100 + 50 = 150 m

Distance travelled from 35 s to 40 s :

1/2 x 20 x 5
50 m

Total = 75 + 150 + 162.5 + 190 + 237.5 + 225 + 150 + 50

Total = 1240 m

c) velocity at t = 15 s

36/15
12/5
2.4 m/s

d) average velocity

0 m/s (as displacement is equal to 0)

e) average speed

1240/40
31 m/s

f) Part d uses displacement whereas part e uses distance

3 0

2 years ago

If you press as hard as you can against a brick wall until you begin to sweating and breathing hard did you do any work on the w

Reika [66]

No, as long as nothing moves you aren't doing work, since W=F.s. You have no s, so no W

4 0

3 years ago

What is strength of the electric field between two parallel conducting plates separated by 1.00 cm and having an electric potent

Alex17521 [72]

Answer:

Electric field will be $1.5\times 10^{6}V/m$

Explanation:

We have given potential difference between the two plates = 15000 volt

Distance between the plates = $d=1cm=10^{-2}m$

We have to find the electric field strength between the plates

Potential difference between the plates is equal to $V=Ed$

So electric field strength will be $E=\frac{V}{d}=\frac{15000}{10^{-2}}=1.5\times 10^{6}V/m$

So electric field will be $1.5\times 10^{6}V/m$

6 0

3 years ago

AKS 8a - Phenomena-Based Question: Use the data and the graph to make a claim as to which person represents each letter on the g

raketka [301]

Answer: what’s the answer

Explanation:

8 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