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3 years ago

Whenever a net force acts on an object, there is a change in the object's _______?

1 answer:

4 0

Whenever a net force acts on an object, it means object is under acceleration which implies that there is change in the object's velocity.

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The magnitude of the electrostatic force between two identical ions that are separated by a distance of 5.0A is 3.7×10^-9N.a) wh

jeyben [28]

Explanation:

Given that,

Electrostatic force, $F=3.7\times 10^{-9}\ N$

Distance, $r=5\ A=5\times 10^{-10}\ m$

(a) $F=\dfrac{kq^2}{r^2}$ , q is the charge on the ion

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{3.7\times 10^{-9}\times (5\times 10^{-10})^2}{9\times 10^9}}$

$q=3.2\times 10^{-19}\ C$

(b) Let n is the number of electrons are missing from each ion. It can be calculated as :

$n=\dfrac{q}{e}$

$n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}$

n = 2

Hence, this is the required solution.

8 0

3 years ago

Equal masses of he and ne are placed in a sealed container. What is the partial pressure of he if the total pressure in the cont

EastWind [94]

Answer:

6 atm.

Explanation:

Let the mass of both be m

Then moles of He = m/ 4

Moles of Ne = m/ 20

mole fraction of He = Moles of He/ Total moles = m/4/ (m/4 + m/20) = 0.25 m/0.3m = 0.83

Pressure of He = Mole fraction×total pressure = 0.83 × 6 atm = 5 atm

6 0

3 years ago

What are the laws of Physics

uranmaximum [27]

Answer:

thermodynamics

Explanation:

The laws of thermodynamics define a group of physical quantities, such as temperature, energy, and entropy, that characterize thermodynamic systems in thermodynamic equilibrium.

8 0

3 years ago

The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?

Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight = 60 .3 (9.81) = 591.54 N

Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift

60.3 ( 9.81 - L ) = 564

solve for L = .457 m/s^2 DOWNWARD

so L = - .457 m/s^2

4 0

2 years ago

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm. The explorer finds that

Natasha2012 [34]

Answer:

g = 11.2 m/s²

Explanation:

First, we will calculate the time period of the pendulum:

$T = \frac{t}{n}$

where,

T = Time period = ?

t = time taken = 135 s

n = no. of swings in given time = 98

Therefore,

$T = \frac{135\ s}{98}$

T = 1.38 s

Now, we utilize the second formula for the time period of the simple pendulum, given as follows:

$T = 2\pi \sqrt{\frac{l}{g}}$

where,

l = length of pendulum = 54 cm = 0.54 m

g = acceleration due to gravity on the planet = ?

Therefore,

$(1.38\ s)^2 = 4\pi^2(\frac{0.54\ m}{g} )\\\\g = \frac{4\pi^2(0.54\ m)}{(1.38\ s)^2}$

3 0

3 years ago