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Dovator [93]
2 years ago
8

If the force squeezing two surfaces together is decreased, the force of dry sliding friction between the two surfaces will most

likely: a. increase b. decrease c. stay about the same ​
Physics
1 answer:
Anastasy [175]2 years ago
7 0

<em>If the force squeezing two surfaces together is decreased, the force of dry sliding friction between the two surfaces will most likely decrease. </em>

<u>therefore your answer is B)</u><u>d</u><u>e</u><u>crease </u>

Hope this helps you- have a good day bro cya)

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The sales price of a product is $2 per unit; the variable cost is $1 per unit; and fixed costs total $1,000. how many units must
irakobra [83]

Answer:

<u><em>1000 units for breakeven</em></u>

Explanation:

Let x be the number of units sold at breakeven.

The total sales at the point would be $2x.

Variable costs would be $1x and fixed costs are $1000.

Total costs are = $1x + $1000

At breakeven: Sales = Costs

Sales =m Costs

$2x = $1x + $1000

$1x = $1000

x = 1000 units.

At 1000 units the sales are equal to the costs ("breakeven").

4 0
1 year ago
Help me pleaseeee I will give you 42 points!<br> (I am dumb so that’s why)
Anit [1.1K]
Search each one of them up on GOOGLE it’s easier trust me
7 0
3 years ago
Read 2 more answers
Which unit of measurement is included in the International System of Unit! (si)?
ozzi

Answer:

See the explanation below

Explanation:

There are several measures for the international system of measures. Let's name some and their representation symbol.

meter = [m]

time = [s] = seconds

mass = [kg] = kilograms

Temperature = [°C] = celcius degrees

Power = [W] = watts.

Force = [N] = Newtons

3 0
3 years ago
How much force is needed to accelerate a bicycle and rider with a total mass of 50 kg at a rate of 2.0 m/s2 ?
True [87]

Answer:

The answer is D 100 newton

Explanation:

2.0m/s2 is d acceleration while the 50kg is the mass. Force = mass x acceleration. So f=50x2.so force is 100 newton

8 0
3 years ago
Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
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