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3 years ago

A tool for studying motion is called?

Physics

1 answer:

kakasveta [241]3 years ago

8 0

Answer:

phytochemical

Explanation:phytochemical is a tool for studying motion

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Which describes a force acting on an object?

weeeeeb [17]

Answer:

Describing a Force:

To fully describe the force acting upon an object, you must describe both its magnitude and direction. Thus, 10 Newtons of force is not a complete description of the force acting on an object. 10 Newtons, downwards is a complete description of the force acting upon an object.

Explanation:

3 0

2 years ago

A circuit is constructed with six resistors and two batteries as shown. the battery voltages are v1 = 18 v and v2 = 12 v. the po

VladimirAG [237]

Answer:

V4=9.197v

Explanation:

Given:

V1= 18v ,V2= 12v ,r1=r5=58ohms ,r2=r6=124ohms , r3=47ohms ,r4= 125ohms

V4= I4R4 = V2/(R4 + R5)×R4

V4= 12×125 /(125 + 58)

V4=1500/183 =9.197v

5 0

2 years ago

When a roller coaster gets to the bottom of a descent, describe the energy transfers and changes to energy stores that happen if

Feliz [49]

Answer:

its B

Explanation:

It's B & A at the same time because A. a roller coaster uses brakes to slow down and stop. B is the most reasonable answer. Because all roller coasters go up and over a second time over the hill, but they also slow down. But go with B.

TELL ME IF I´M RITE

6 0

2 years ago

Two ions with masses of 4.39×10−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic field is

Ilia_Sergeevich [38]

Answer:

7.2 cm

Explanation:

magnetic field, B = 0.301 T

speed, v = 7.92 x 10^5 m/s

mass, m = 4.39 x 10^-27 kg

q = 1.6 x 10^-19 C

The radius of singly changed ion is given by

$r = \frac{mv}{Bq}$

where, m is the mass of ion, v be the speed of ion, B is the magnetic field and q be the charge

$r = \frac{4.39\times 10^{-27}\times 7.92 \times 10^{5}}{0.301\times 1.6\times 10^{-19}}$

r = 0.072 m

r = 7.2 cm

5 0

2 years ago

wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the

lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

$E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)$

The electric field due to charge QB at P is EB leftwards

$E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)$

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0

3 years ago