What would make oppositely charged objects attract each other more?

Ali’s family has a farm near the coast. Winds are constantly carrying away soil during the winter when no crops are growing. Wha

Novay_Z [31]

Answer:

a

Explanation:

the other answers wouldnt help

4 0

3 years ago

Help?? This is for my Physics class

Fittoniya [83]

For distance vs time graphs, the slope stands for speed, as speed is a scalar, along with distance.
For displacement (position) vs time graphs, the slope stands for velocity, as velocity is a vector, along with displacement.

7 0

3 years ago

A 100kg couch is being pushed with 196N of force. As it slides along the ground it experiences a coefficient of friction of 0.1.

QveST [7]

Answer:98

Explanation:hope this helps!

4 0

3 years ago

When we do dimensional analysis, we do something analogous to stoichiometry, but with multiplying instead of adding. Consider th

Ksivusya [100]

Answer:

The dimension is $D = L ^{2} T^{-1}$

Explanation:

From the question we are told that

$J = -D \frac{dn}{dx}$

Here $[J] = \frac{1}{L^2 T}$

$[n] =\frac{1}{L^3}$

$[x] = L$

So

$\frac{1}{L^2 T} = -D \frac{d(\frac{1}{L^3})}{d[L]}$

Given that the dimension represent the unites of n and x then the differential will not effect on them

So

$\frac{1}{L^2 T} = -D \frac{(\frac{1}{L^3})}{[L]}$

=> $D = \frac{L^{-2} T^{-1} * L }{L^{-3}}$

=> $D = L ^{2} T^{-1}$

5 0

3 years ago

A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same di

navik [9.2K]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

$X(van)=5.65t+154$

$X(driver)=34.4t+\frac{(-2)t^{2} }{2}$

or by rearanging the drivers equation.

$X(driver)=34.4t+t^{2}$

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

$X(van)=X(driver)$

$5.65t+154=34.4t-t^{2}$

$0=t^{2} -(34.4-5.65)t+154$ $0=t^{2} -28.75t+154$

To solve this equation we use the following formulas

$t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}$

Where a=1; b=-28.75; c=154

So we get:

$t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s$ $t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s$

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

$V(driver)=V_{0} +at$ }

$V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}$ $V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}$

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer A).

Best of luck

8 0

3 years ago