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3 years ago

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Physics

1 answer:

Blizzard [7]3 years ago

6 0

Answer:

0.2 J

Explanation:

The pendulum forms a right triangle, with hypotenuse of 50 cm and base of 30 cm. The height of this triangle can be found with Pythagorean theorem:

c² = a² + b²

(50 cm)² = a² + (30 cm)²

a = 40 cm

The height of the triangle is 40 cm. The height of the pendulum when it is at the bottom is 50 cm. So the end of the pendulum is lifted by 10 cm. Assuming the mass is concentrated at the end of the pendulum, the potential energy is:

PE = mgh

PE = (0.200 kg) (9.8 N/kg) (0.10 m)

PE = 0.196 J

Rounding to one significant figure, the potential energy is 0.2 J.

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A child whirls a 3.00 kg ball on a string .50 m from the axis of rotation in a horizontal circle. The ball makes 1 revolution in

melamori03 [73]

Answers:

a) $0.5 m/s^{2}$

b) $1.5 N$

Explanation:

a) The centripetal acceleration $a_{c}$ of an object moving in a uniform circular motion is given by the following equation:

$a_{c}=\omega^{2} r$

Where:

$\omega=1 \frac{rev}{s}$ is the angular velocity of the ball

$r=0.5 m$ is the radius of the circular motion, which is equal to the length of the string

Then:

$a_{c}=(1 \frac{rev}{s})^{2} 0.5 m$

$a_{c}=0.5 m/s^{2}$ This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force $F$ ) that is directed towards the center and is equal to the tension ( $T$ ) in the string:

$F=T=m. a_{c}$

Where $m=3 kg$ is the mass of the ball

Hence:

$T=(3 kg)(0.5 m/s^{2})$

$T=1.5 N$ This is the tension in the string

7 0

3 years ago

A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal ener

Elina [12.6K]

Answer:

Explanation:

From the given information:

The initial PE $(PE)_i$ = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = $P.E_f -P.E_i$

ΔP.E = 0 - $(PE)_i$

ΔP.E = $-P.E_i$

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

$\Delta P.E + \Delta K.E + \Delta U = 0$

$\Delta U = -\Delta P.E - \Delta K.E$

this can be re-written as:

$\Delta U =- (-\Delta P.E_i) - \Delta K.E$

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

$\Delta U =(70\%) \Delta P.E_i-0$

$\Delta U =(0.70) (490.5)$

ΔU = 343.35 J

Thus, the magnitude of the increase is = 343.35 J

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3 years ago

The nucleus of some isotopes will spontaneously undergo nuclear decay. These isotopes are said to be radioactive. What causes th

Dmitry_Shevchenko [17]

Answer:

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2 years ago

If an electron is released at PP , what is the magnitude of the net force that these rods exert on it?

pishuonlain [190]

The magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.

Given values:

Length of non-conducting rod, l = 1.20 m

Charge on positive rod, +Q = +2.50 μC = +2.50 × 10⁻⁶ C

Charge on negative rod, -Q = -2.50 μC = -2.50 × 10⁻⁶ C

Distance from point P of each rod, x = 60 cm = 0.60 m

Calculation of Net electric force exerted on point P:

Consider an electron released at point P, then the net electric force exerted will be given as:

F = e. E_net - ( 1 )

Step 1:

The net electric field value is given as:

E_net = E₁ cos Φ + E₂ cos Φ

= 2E₁ cos Φ -( 2 )

where, E₁ & E₂ are electric fields due to positive and negative rod

respectively.

Φ is phase angle

Step 2:

The electric field due to positive rod is given as:

E₁ = k (λ/r) - ( 3 )

where, k is Coulomb's force constant

λ is linear charge density

r is distance between point P and half of the rod.

Now, the linear charge density is given as:

λ = Charge/length = Q/x

The value of r is given as:

r = √x²-a²

where, x is length of rod

a is half length of rod

Applying values in above equation, we get:

r = √x²-(x/2)²

r = √(1.20 m)²-(1.20/2)²

= √1.08

= 1.04 m

Substituting all the determined values in equation 3 we get:

E₁ = k (λ/r)

= k [(Q/x)/r]

= k [ Q/xr ]

= (9×10⁹ Nm²/C²) [ |+2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

= 1.803×10⁴ N/C

Step 3:

Similarly, the electric field due to negative rod is given as:

E₂ = k [ Q/xr ]

= (9×10⁹ Nm²/C²) [ |-2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

= 1.803×10⁴ N/C

Step 4:

Consider equation 2:

E_net = 2E₁ cos Φ

From the figure we get the phase angle as:

Φ = tan⁻¹ (0.60 m/0.60 m)

= tan⁻¹ ( 1 )

= π/4

Now, the electric field produced due to each rod is equal and mutually perpendicular. Thus, the net electric field after applying values can be calculated as:

E_net = 2(1.803×10⁴ N/C) cos π/4

= 2(1.803×10⁴ N/C) (0.5)

= 18030 N/C

Step 5:

Consider equation 1 :

F = e. E_net

where, e is charge on an electron

Applying values in above equation we get:

F = (1.6 × 10⁻¹⁹ C)(18030 N/C)

= 2.885 × 10⁻¹⁵ N

Therefore, the magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.

Learn more about electric force here:

<u>brainly.com/question/1634182</u>

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