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TEA [102]
3 years ago
6

Ok ths is my last one i swear this is for u sheystar

Physics
1 answer:
tino4ka555 [31]3 years ago
4 0

Answer:

given below.

Explanation:

a = Indirect

b = Not Indirect

c = Indirect

d = Not Indirect

e = Not Indirect

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explain why applying force to an object does not always result in work being done. 20 POINTS WILL MARK BRAINLIEST
fiasKO [112]
Applying force all the time is bad for you body, the object, and the environment and how it looks at you. First of all, you could strain, break, fracture, you name it, to your bones and muscles. Second of all, you could manage to break that object, not even move it. For example, say you have to move a heavy object to save the town from almost flooding and you strain or break your arm. To society or the people watching you do this, you look like a wimp or you're not strong. Then you have a bad reputation in the people's eyes and you get picked on or bullied because of that. You don't have move that object. You could find a way to work around the object and get simpler ways that only you thought of at the time.
7 0
3 years ago
Select the options that best complete the statement.Positively charged particle trajectories(always, never, the same as)follow e
SpyIntel [72]

Answer:

a) always. b) electric field lines are defined by the path positive test charges travel.

Explanation:

By convention, field lines always follow the direction that it would take a positive test charge  (small enough so it can´t disrupt the field created by a charge distribution), under the influence of an electric field, at the same point where the test charge is located.

So any positive charge, subject to an electric field influence, moves along the field line that passes through its current position, in the same way that a positive test charge would.

We could say also that the electric force on a positively charged particle is in the same direction as the electric field that produces that force (due to some charge distribution) , which is true, but it doesn´t explain why.

3 0
4 years ago
Choose the correct statement concerning units of power or energy. a. A kilojoule (kJ) is a unit of power. b. A gigawatt (GW) is
sergiy2304 [10]

Answer:

A kilowatt (kW) is a unit of power.

Explanation:

The power of an object is given by :

P=\dfrac{E}{t}

Here,

E is the energy required

t is time

The SI unit of power is Watts and the SI unit of energy is Joule. the commercial unit of energy is kilowatt per hour.    

Option (1) :  A kilojoule (kJ) is a unit of power is incorrect.

Option (2) : A gigawatt (GW) is a unit of energy is incorrect.

Option (3) : A watt (W) is a unit of energy is incorrect.

Option (4) : A kilowatt x hour per year (kWh/yr) is a unit of energy is incorrect.

Option (4) : A kilowatt (kW) is a unit of power is correct.

Hence, the correct option is (d).

3 0
3 years ago
Which statements accurately describe Ernest Rutherford’s experiment? Check all that apply.
Dimas [21]

Answer:

Option (1), option (4) and option (5)

Explanation:

The main observations of Ernest Rutherford's experiment are given below:

1. most of the positively charged particles pass straight, it means there is an empty space in the atom.

2. Very few positively charged particles retraces their path.

So,

The positively charged particles were deflected because like charges repel, that means they are deflected by protons.

Almost all the positively charge concentrate in a very small part which is called nucleus.

7 0
4 years ago
A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh
daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is B=2.958\times10^{-5}T

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \  \overrightarrow B_s\perp \overrightarrow B_w

Angle of net magnetic field from axial direction is given by:

tan\  \theta=\frac{B_w}{B_s},

Field due to solenoid:

B_s=\mu_onI_s,  \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m

Field due to wire:

B_w=\frac{\mu_oI_w}{2\pi r}

Therefore, r:

tan\  \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi  nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T

Hence, the magnetic field is B=2.958\times10^{-5}T

7 0
4 years ago
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