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3 years ago

A push or pull on an object

Physics

1 answer:

Leno4ka [110]3 years ago

7 0

A pull because you can get the object

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A little girl is going on the merry-go-round for the first time, and wants her 50kg mother to stand near to her on the ride 2.1m

aliya0001 [1]

Angular momemtum : mass * tangential speed * distance to the center = 50*2.1*3.6=37800 J.s

3 0

3 years ago

The current through a 10 ohm resistor connected to a 120 volt power supply is

Leno4ka [110]

Answer:I=12 A

Explanation:

Given

Resistance $R=10 \Omega$

Voltage $V=120 V$

According to ohm's law current through a conductor is directly proportional to the voltage applied.

$V\propto I$

$V=IR$

where V=Voltage

I=Current

R=resistance

$I=\frac{V}{R}$

$I=\frac{120}{10}$

$I=12 A$

4 0

3 years ago

Singly charged uranium-238 ions are accelerated through a potential difference of 2.90 kV and enter a uniform magnetic field of

Lera25 [3.4K]

Answer: 0.091 m

Explanation:

r = 1/B * √(2mV/e), where

r = radius of their circular path

B = magnitude of magnetic field = 1.29 T

m = mass of Uranium -238 ion = 238 * amu = 238 * 1.6*10^-27 kg

V = potential difference = 2.9 kV

e = charge of the Uranium -238 ion = 1.6*10^-19 C

r = 1/1.29 * √[(2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]

r = 1/1.29 * √(2.21*10^-21 / 1.6*10^-19)

r = 1/1.29 * √0.0138

r = 1/1.29 * 0.117

r = 0.091 m

Therefore, the radius of their circular path is 0.091 m

6 0

3 years ago

A rectangular coil of wire (a = 22.0 cm, b = 46.0 cm) containing a single turn is placed in a uniform 4.60 T magnetic field, as

frez [133]

Explanation:

Below is an attachment containing the solution.

4 0

3 years ago

A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm

Galina-37 [17]

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

$v =\sqrt{\dfrac{T}{\mu}}$

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

$v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}$

v = 1074.49 m/s

The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

λ=2 L

λ=2 x 0.577 = 1.154 m

we now,

v = f λ

$f = \dfrac{v}{\lambda}$

$f = \dfrac{1074.49}{1.154}$

f = 931.1 Hz

The frequency for first harmonic is equal to f = 931.1 Hz

7 0

3 years ago