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2 years ago

A stretched rubber band is an example of which type potential energy

Physics

2 answers:

Alenkinab [10]2 years ago

6 0

Answer: elastic potential energy

Explanation:

Kazeer [188]2 years ago

6 0

Answer:

<h3>Elastic potential energy</h3>

Explanation:

A stretched rubber band has the potential to do work or change things. This form of energy is called elastic potential energy.

<h2>❣️✧Jess bregoli ✧❣️</h2>

<h3>#keep learning!!</h3>

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Two objects, one with a mass of and the other with a force of 30.0kg experience a gravitational force of attraction of 7.50 * 10

nydimaria [60]

Answer:

Explanation:

The formula for this is

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the gravitational constant, m1 is the mass of one object and m2 is the mass of the other object. We are looking for r, the distance between the centers of their masses.

Filling in:

$7.5*10^{-8}=\frac{6.67*10^{-11}(90.0)(30.0)}{r^2}$ and moving things around to solve for r:

$r=\sqrt{\frac{6.67*10^{-11}(90.0)(30.0)}{7.5*10^{-8}} }$ Doing all that and rounding to the 3 sig fig's you need gives us a distance of 1.55 m

8 0

3 years ago

1. A uniform magnetic field is directed vertically upwards. In which direction in this field should an alpha particle be project

max2010maxim [7]

Answer:

1. Fleming's left hand rule

2. It must be projected towards the east

Explanation:

Fleming's left-hand rule states that; When a current-carrying conductor is placed in an external magnetic field, the conductor experiences a force perpendicular to both the field and to the direction of the current flow. This rule was first put forward by John Ambrose Fleming in the later part of the nineteenth century.

Hence if the thumb, fore finger and middle finger of the lefthand are held mutually at right angles to each other; the thumb shows the direction of motion, the fore finger shows the direction of the field while the middle finger shows the direction of the current.

Hence, if the alpha particle is projected eastwards(at right angles) to the uniform magnetic field, it will be deflected southwards in the magnetic field.

3 0

3 years ago

11 e) Can a conductor be given limitless charge? Obtain the equivalent resistance of several resistors if (a) they are in series

horsena [70]

Answer:

(e) no

(a) Rs = R' + R'' + R'''

(b) 1/Rp = 1/R' + 1/R'' + 1/R'''

Explanation:

11 e)

Practically it is not possible to give limitless charge to a conductor. It depends to the number of valence electrons.

(a) When the three resistances R'. R'' and R''' is in series combination.

Let they are connected to the voltage V and the current in each resistance is I.

According to Ohm's law

Voltage across R', V' = I R'

Voltage across R'', V'' = I R''

Voltage across R''', V''' = I R'''

So, let the equivalent resistance is Rs.

I Rs = I R' + I R'' + I R'''

Rs = R' + R'' + R'''

(b)

When the three resistances R'. R'' and R''' is in parallel combination.

Let they are connected to the voltage V and the current in each resistance is I', I''. I'''.

Current in R', I' = V/R'

Current in R'', I'' = V/R''

Current in R''', I''' = V/R'''

The equivalent resistance is Rp.

V/Rp = V/R' + V/ R'' + V/R'''

1/Rp = 1/R' + 1/R'' + 1/R'''

6 0

3 years ago

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra

Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

$m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}$

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

$v=\sqrt{\frac{GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s$

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula, we find:

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s$

5 0

3 years ago

A sample of gas with a volume of 750 ml exerts a pressure of 98 kpa at 30◦c. What pressure will the sample exert when it is comp

Tanzania [10]

Answer:

241 kPa

Explanation:

The ideal gas law states that:

$pV=nRT$

where

p is the gas pressure

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

We can rewrite the equation as

$\frac{pV}{T}=nR$

For a fixed amount of gas, n is constant, so we can write

$\frac{pV}{T}=const.$

Therefore, for a gas which undergoes a transformation we have

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where the labels 1 and 2 refer to the initial and final conditions of the gas.

For the sample of gas in this problem we have

$p_1 = 98 kPa=9.8\cdot 10^4 Pa\\V_1 = 750 mL=0.75 L=7.5\cdot 10^{-4}m^3\\T_1 = 30^{\circ}C+273=303 K\\p_2 =?\\V_2 = 250 mL=0.25 L=2.5\cdot 10^{-4} m^3\\T_2 = -25^{\circ}C+273=248 K$

So we can solve the formula for $p_2$ , the final pressure:

$p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(9.8\cdot 10^4 Pa)(7.5\cdot 10^{-4} m^3)(248 K)}{(303 K)(2.5\cdot 10^{-4} m^3)}=2.41\cdot 10^5 Pa = 241 kPa$

4 0

2 years ago