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madreJ [45]
3 years ago
5

What two fundamental units that are not vector quantities

Physics
2 answers:
Serggg [28]3 years ago
8 0
Mass (kilogram) and time (second) are not vector quantities.
insens350 [35]3 years ago
5 0
There are seven fundamental units, out of which two examples of non-vector quantities could be:  

1) meter (unit of length)
2) second (unit of time)

Hope this helps!
You might be interested in
2 Points
yulyashka [42]

The  force applied to lift the crate is 171 N

Explanation:

The lever works on the principle of equilibrium of moments, so we can write:

F_i d_i = F_o d_o

where

F_i is the  force in input

d_i is the arm of the input force

F_o is the output force

d_o is the arm of the output force

For the lever in this problem, we have:

d_i = 0.25 m

d_o = 0.19 m

F_i = 130 N (force applied)

Solving the equation for F_o, we find the force applied to lift the crate:

F_o = \frac{F_i d_i}{d_o}=\frac{(130)(0.25)}{0.19}=171 N

Learn more about levers:

brainly.com/question/5352966

#LearnwithBrainly

6 0
3 years ago
A force of 6600 N is exerted on a piston that has an area of 0.010 m2
sveticcg [70]

Answer:

Choice A: approximately 0.015\; \rm m^2, assuming that the two pistons are connected via some confined liquid to form a simple machine.

Explanation:

Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:

\displaystyle P_1 = \frac{F_1}{A_1} = \frac{6.600\times 10^3\; \rm N}{1.0\times 10^{-2}\; \rm m^{2}} = 6.6\times 10^{5}\; \rm N \cdot m^{-2}.

By Pascal's Principle, because the first piston exerted a pressure of 6.6\times 10^{5}\; \rm N \cdot m^{-2} on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.

Assume that the second piston is part of that wall. The pressure on the second piston will also be 6.6\times 10^{5}\; \rm N \cdot m^{-2}. In other words:

P_2 = P_1 = 6.6\times 10^{5}\; \rm N \cdot m^{-2}.

To achieve a force of 9.900 \times 10^3\; \rm N, the surface area of the second piston should be:

\displaystyle A_2 = \frac{F_2}{P_2} = \frac{9.900\times 10^{3}\; \rm N}{6.6\times 10^5\; \rm N \cdot m^{-2}} \approx 0.015\; \rm m^{2}.

4 0
3 years ago
How much force is required to pull a spring 3.0 cm from
avanturin [10]

Answer:

I know that T= kx where T is the tension which equaka the force og gravity = mg = 1.37 * 10 = 13.7 x is the elongation of the spring so the length after dangling the object minus the original length.

I hope it helps

plz let me know if it is wrong or right.

4 0
3 years ago
A 1.0-kg block of aluminum is at a temperature of 50 Celsius. How much thermal energy will it lose when it’s temperature is redu
Ipatiy [6.2K]

Answer:

The lose of thermal energy is, Q = 22500 J

Explanation:

Given data,

The mass of aluminium block, m = 1.0 kg

The initial temperature of block, T = 50° C

The final temperature of the block, T' = 25° C

The change in temperature, ΔT = 50° C - 25° C

                                                     = 25° C

The specific heat capacity of aluminium, c = 900  J/kg°C

The formula for thermal energy,

                             <em>Q = mcΔT</em>

                                 = 1.0 x 900 x 25

                                 = 22500 J

Hence, the lose of thermal energy is, Q = 22500 J

7 0
3 years ago
A non-_____ rock has interlocking grains with no specific pattern.
Kazeer [188]
A non <span>foliated </span>rock has interlocking grains with no specific pattern.
3 0
3 years ago
Read 2 more answers
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