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2 years ago

A 37 N object is lifted to a height of 3 meters, What is the potential energy of this object?

Physics

2 answers:

prisoha [69]2 years ago

6 0

P.e = mgh

37N × 9.8ms⁻² ˣ 3m

= 1087.8 .

Lostsunrise [7]2 years ago

5 0

Answer:

Potential energy of the object is 111 Joules.

Explanation:

It is given that,

Force of gravity acting on the object, F =W=mg= 37 N, W is its weight

It is lifted to a height of 3 meters, h = 3 m

We need to find the potential energy of this object. It is given by :

$E=m\times g\times h$

Since, mg = 37 N

So,

$E=37\ N\times 3\ m$

E = 111 Joules

So, the potential energy of this object is 111 Joules. Hence, this is the required solution.

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sasho [114]

A close representation of the moons motion around earth is that it moves around the earth once in 27.3 days. the moons average movement is 13.2 degrees per day so around 92 per week.

Hope this helps!

5 0

3 years ago

Help help hep help???????????

Sergeu [11.5K]

Momentum

mava + mbvb = mava '+ mbvb'

(300 x 10)+(150 x 0) = (300 x 4.12)+(150 x vb')

3000=1236+150vb'

1764 = 150vb'

vb'=+11.76 m/s ≈ +11.8 m/s (positive sign, to the right)

3 0

1 year ago

A metal ring 4.30 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend

CaHeK987 [17]

Answer:

A)0.00966 N/C

B) counterclockwise direction

Explanation:

We are given;

Diameter of the metal ring; d = 4.3 cm

Radius;r = 2.15 cm = 0.021- m

Initial magnetic field, B = 1.12 T

Rate of decrease of the magnetic field;dB/dt = 0.23 T/s

Now, as a result of change in magnetic field, an emf will be induced in it. Thus, , electric field is induced and given by the formula :

∫E•dr = d/dt∫B.A •dA

This gives;

E(2πr) = dB/dt(πr²)

Gives;. 2E = dB/dt(r)

E = dB/dt × 2r

We are given;

E = 0.23 × 2(0.021)

E = 0.00966 N/C

The magnitude of the electric field induced in the ring has a magnitude of 0.00966 N/C

B) The direction of electric field will be in a counterclock wise direction when viewed by someone on the south pole of the magnet

6 0

2 years ago

In a lightning bolt, a large amount of charge flows during a time of 1.2 x 10-3 s. Assume that the bolt can be represented as a

ohaa [14]

Answer: 10.58 C has flowed during the lightning bolt

Explanation:

Given that;

Time of flow t = 1.2 × 10⁻³

perpendicular distance r = 21 m

Magnetic field B = 8.4 x 10⁻⁵ T

Now lets consider the expression for magnetic field;

B = u₀I / 2πr

the current flow is;

I = ( B × 2πr ) / u₀

so we substitute

I = ( (8.4 x 10⁻⁵) × 2 × 3.14 × 21 ) / 4π ×10⁻⁷

= 0.01107792 / 0.000001256

= 8820 A

Hence the charge flows during lightning bolt will be;

q = It

so we substitute

q = 8820 × 1.2 × 10⁻³

q = 10.58 C

therefore 10.58 C has flowed during the lightning bolt

6 0

3 years ago

In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.

Murrr4er [49]

The central force acting on the electron as it revolves in a circular orbit is $9.52 \times 10^{-8} \ N$ .

The given parameters;

speed of electron, v = 2.2 x 10⁶ m/s
radius of the circle, r = 4.63 x 10⁻¹¹ m

The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

$F = \frac{M_e v^2}{r} \\\\$

where;

$M_e$ is mass of electron = 9.11 x 10⁻³¹ kg

$F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N$

Thus, the central force acting on the electron as it revolves in a circular orbit is $9.52 \times 10^{-8} \ N$ .

Learn more about centripetal force here:brainly.com/question/20905151

8 0

2 years ago