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masya89 [10]
3 years ago
5

. A 2.00 kg ball is attached to a ceiling by a string. The distance from the ceiling to the center of the ball is 1.00 m, and th

e height of the room is 3.00 m. What is the gravitational potential energy associated with the ball relative to each of the following? (a) the ceiling (b) the floor (c) a point at the same elevation as the ball
Physics
1 answer:
cupoosta [38]3 years ago
4 0
<h2>a) Potential energy of ball relative to ceiling is 19.62 J</h2><h2>b) Potential energy of ball relative to floor is 39.24 J</h2><h2>c) Potential energy of ball relative to same elevation is 0 J</h2>

Explanation:

Mass of ball, m = 2 kg

Acceleration due to gravity, g = 9.81 m/s²

Height of ball from ground = 3 - 1 = 2 m

Potential energy, PE = mgh

Potential energy of ball, PE = 2 x 9.81 x 2 = 39.24 J

a) Potential energy of ball at ceiling = 2 x 9.81 x 3 = 58.86 J

Potential energy of ball relative to ceiling = 58.86 - 39.24 = 19.62 J

b) Potential energy of ball at floor = 2 x 9.81 x 0 = 0 J

Potential energy of ball relative to floor = 39.24 - 0 = 39.24 J

c) Potential energy of ball at same elevation = 2 x 9.81 x 2 = 39.24 J

Potential energy of ball relative to same elevation = 39.24 - 39.24 = 0 J

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M = 15 kg, a = 2 m/s2, F =
Lisa [10]

Answer:

<h2>30 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 15 × 2

We have the final answer as

<h3>30 N</h3>

Hope this helps you

7 0
3 years ago
Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5
pishuonlain [190]

Answer

given,

force per unit length = 350 µN/m

current, I = 22.5 A

y = y = 0.420 m

\dfrac{F}{L}= \dfrac{KI_1I_2}{d}

I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}

I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}

    I₂ = 32.67 A

distance where the magnetic field is zero

\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}

y_1 = 0.248\ m

there the distance at which the magnetic field is zero in the two wire is at 0.248 m.

3 0
3 years ago
On a 100km track , a train travels the first 30km with a speed of 30km/h . How fast the train travel the next 70 km if the avera
nirvana33 [79]

Solution :-

Given :

Distance 1 = 30 km

Distance 2 = 70 km

We know that speed = distance/time

and, Average speed = total distance/total time taken

When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour

Average speed = 9distance 1 + distance 2)/(time 1 + time 2)

AS time 2 or t2 is time taken for the second part of the journey of 70 km

⇒ 40 = 100/(1 + t2)

⇒ 40 + 40t2 = 100

⇒ 40t2 = 100 - 40

⇒ 40t2 = 60

⇒ t2 = 60/40

⇒ t2 = 1.5

So, t2 or time taken to travel the second part of the journey is 1.5 hours.

Speed of the second part of the journey = distance 2/time 2

⇒ 70/1.5

⇒ 46.666 km/hr or 46.7 km/hr.

Hence the answer is = 46.666 km/hr or 46.7 km/hr.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

:)

3 0
3 years ago
A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the h
dybincka [34]

Given,

A player kicks a soccer hits at an angle of 30° at a speed of 26 m/s

We can resolute the trajectory of soccer into horizontal and vertical components.(Please see the attached file)

We can have,

Horizontal velocity component of ball= 26cos(30°)  = 26×(√3÷2) = 22.51 m/s

And vertical velocity component of ball = 26sin(26°) = 26×(1÷2) = 13 m/s


6 0
3 years ago
Weak magnetic fields can be measured at the surface of the brain. Although the currents causing these fields are quite complicat
STALIN [3.7K]

To develop this problem it is necessary to apply the concepts related to a magnetic field in spheres.

By definition we know that the magnetic field in a sphere can be described as

B = \frac{\mu_0}{2}\frac{Ia^2}{(z^2+a^2)^{3/2}}

Where,

a = Radius

z = Distance to the magnetic field

I = Current

\mu_0 = Permeability constant in free space

Our values are given as

D=2a = 16cm \rightarrow diameter of the sphere then,

a = 0.08m

Thus z = a

B = \frac{\mu_0}{2}\frac{Ia^2}{(a^2+a^2)^{3/2}}

B = \frac{\mu_0I}{2(2^{3/2})a}

B = \frac{\mu_0 I}{2^{5/2}a}

Re-arrange to find I,

I = \frac{2^{5/2}Ba}{\mu_0}

I = \frac{2^{5/2}(3*10^{-12})(8*10^{-2})}{4\pi*10^{-7}}

I = 1.08*10^{-6}A

Therefore the current at the pole of this sphere is 1.08*10^{-6}A

5 0
3 years ago
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