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larisa86 [58]
3 years ago
5

Name one process that requires a force in the human body.

Physics
2 answers:
vova2212 [387]3 years ago
8 0
Non-contact force, which Is a force that acts like an object without coming in physically contact with you
allsm [11]3 years ago
5 0

muscle contraction, blood circulation

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What is the purpose of each of the components (each shown with leader and line) of the circuit shown is the diagram?
NemiM [27]

The wires is what is needed to put together the whole thing, kinda like glue when you're gluing a piece of paper on it.

Anyways, the battery is the main source and main energy per say.

That energy that comes from the battery, thanks to the wires, it can transfer that said energy to both the switch and light bulb.

And as you flick the switch, it depends of how you put it together, there's two options, turning the light bulb on or turning it off.

Though it doesn't mean that since the light bulb is connected to the battery makes the bulb turn on no matter what since the switch can cancel the main source's energy.

- Ouma :>

5 0
3 years ago
What's the value of 1,152 Btu in joules? A. 1,964,445 J B. 2,485,664 J C. 987,875 J D. 1,215,360 J
OLga [1]

Answer: Option (d) is correct.

Explanation:

Given,  1,152 British thermal units

1 British thermal unit = 1055.06 joules

So, in 1,152 British thermal units there will be :

1152\times 1055.06 Joules=1215429.12 Joules=1.21542912\times 10^{8} Joules

Hence, from the given options the closest answer is of option (d). So, option (d) is correct.

4 0
4 years ago
Read 2 more answers
Compare tidal range to spring tide
makkiz [27]

Spring tides have higher high tides and lower low tides whereas neap tides have lower high tides and higher low tides. Hence, the range is much larger in a spring tide than in a low tide.

7 0
3 years ago
The turnbuckle is tightened until the tension in the cable AB equals 2.3 kN. Determine the vector expression for the tension T a
brilliants [131]

Answer:

a) 0.83984 i + 0.41992 j - 2.0996 k KN

b) T_ac = 1.972888 KN

Explanation:

Given:

- The tension in cable AB = 2.3 KN

Find:

a) Determine the vector expression for the tension T as a force acting on member AD.

b) Also find the magnitude of the projection of T along the line AC.

Solution:

part a)

- Find unit vector AB:

                            vector (AB) = 2 i + j - 5 k

                             mag (AB) = sqrt (2^2 + 1^2 + 5^2)

                             mag (AB) = sqrt(30)

                             unit (AB) =  ( 1 / sqrt(30) )* ( 2 i + j - 5 k )

- Find Tension vector:

                             vector (T) = unit(AB)* 2.3 KN

                                              = 0.83984 i + 0.41992 j - 2.0996 k

- The projection of T onto AC can be found from the dot product of vector T to unit vector (AC)

- For unit vector (AC)

                               vector (AC) = 2 i - 2 j - 5 k

                               mag (AC) = sqrt (2^2 + 2^2 + 5^2)

                               mag (AC) = sqrt(33)

                               unit (AC) =  ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

- Compute the projection:

                                T_ac = vector T . unit (AC)

T_ac = (0.83984 i + 0.41992 j - 2.0996 k)  . ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

                                T_ac = 0.2923947572 - 0.146973786 - 1.827467232

                                T_ac = 1.972888 KN

4 0
3 years ago
Please help me with this question ASAP.
victus00 [196]

Answer:

i. 6.923 V

ii. The e.m.f. = 22.5 V

Explanation:

i. The given parameters are;

Length of potentiometer = 1 m

The resistance of the potentiometer = 10 Ω

The e. m. f. of the attached cell = 9 V

The current, I flowing in the circuit = e. m. f/(Total resistance)

The current, I flowing in the circuit = 9 V/(10 + 3) = 9/13 A

The potential difference, p.d. across the 1 m potentiometer wire = I × Resistance of the potentiometer wire

The p.d. across the potentiometer wire = 9/13×10 = 90/13 = 6.923 V

ii) Given that the 1 m potentiometer wire has a resistance of 10 Ω, 75 cm which is 0.75 m will have an e.m.f. given by the following relation;

\dfrac{E}{R_{balance}} = \dfrac{V}{R_{cell}}

Where:

E = e.m.f. of the balance point cell

R_{balance} = Resistance of 75 cm of potentiometer wire  = 0.75×10 = 7.5 Ω

R_{cell} = Resistance of the cell in the circuit = 3 Ω

V = e.m.f. attached cell = 9 V

\dfrac{E}{7.5} = \dfrac{9}{3}

E = 7.5*3 = 22.5 V

The e.m.f. = 22.5 V

7 0
4 years ago
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