All categories

3 years ago

A 2.80 kg mass is dropped from a

2 answers:

8 0

Mark Brainliest please

Answer : 41.2 J

Explanation

The energy that the ball had instantaneously prior to being dropped was Potential Energy (EP) only and gravitational potential energy is calculated as Ep = mgh = (2.80)(9.81)(4.50) = 123.606 Joules.

Because no outside energy was added and no work was done, the energy instantaneously prior to being dropped is equal to the energy when the mass is 3 m above the ground only at that point, the energy consists of two parts, gravitational Potential Energy (EP) and Kinetic Energy (EK). So, considering the point just prior to the drop as point 1 and the point 3 m above the ground as point 2, we can say:
EP1 = EP2 + EK2
so EK2 = EP1 - EP2
and EP2 = (2.80)(9.81)(3.00) = 82.404
So EK2 = 123.606 - 82.404 = 41.2 Joules

inn [45]3 years ago

7 0

Answer:

PE = 82.32J

Explanation:

PE = m*g*h

PE = 2.80kg*9.8m/s²*3m

PE = 82.32J

You might be interested in

In an early attempt to understand atomic structure, Niels Bohr modeled the hydrogen atom as an electron in uniform circular moti

AleksandrR [38]

Answer:

The answer is below

Explanation:

Using Coulomb's law of electric field which is:

$F=k\frac{q_1q_1}{r^2}\\\\ k =constant=9*10^9\ Nm^2/C^2,q_1=q_2=1.6*10^{-19}C,r=5.29*10^{-11}m\\\\Substituting\ gives:\\\\F=9*10^9*\frac{(1.6*10^{-19})*(1.6*10^{-19})}{(5.29*10^{-11})^2} =8.22*10^{-8}N\\\\Both\ centripetal\ force\ is\ given\ by:\\\\F=m\frac{v^2}{r} \\\\m = mass\ of \ electron=9.11*10^{-31}g,v=speed\ of\ electron\\\\F=m\frac{v^2}{r} \\\\v=\sqrt{\frac{F*r}{m} } \\\\subsituting:\\\\v=\sqrt{\frac{8.22*10^{-8}*5.29*10^{-11}}{9.11*10^{-31}} } \\\\v=2.18*10^6\ m/s\\\\$

$But\ \omega=\frac{v}{r}=\frac{2.18*10^6}{5.29*10^{-11}} =4.13*10^{17}\\\\\omega=2\pi f; f=frequency\\\\f=\frac{\omega}{2\pi} =\frac{4.13*10^{17}}{2\pi} \\\\f=6.57*10^{15}\ Hz$

6 0

3 years ago

If a 2kg ball has and initial velocity of

lakkis [162]

Answer:

$\boxed {\boxed {\sf 12 \ Newtons }}$

Explanation:

Force is equal to the product of mass and acceleration.

$F=m*a$

We know the mass, but not the acceleration. Therefore, we must calculate it before we can calculate force.

1. Calculate Acceleration

Acceleration is the change in velocity over the change in time.

$a=\frac{V_f-V_i}{t}$

The final velocity is 10 meters per second and the initial velocity is 4 meters per second. The time is 1 second.

$V_f=10 \ m/s \\V_i= 4 \ m/s \\t= 1 \ s$

Substitute the values into the formula.

$a=\frac{10 \ m/s-4 \ m/s }{1 \ s}$

Solve the numerator.

$a=\frac{6 \ m/s}{1 \ s }$

Divide.

$a= 6 \ m/s/s=6 \ m/s^2$

2. Calculate Force

Now we know the acceleration and the mass.

$m= 2 \ kg \\a= 6 \ m/s^2$

Substitute the values into the fore formula.

$F= 2 \ kg * 6 \ m/s^2$

Multiply.

$F= 12 \ kg*m/s^2$

1 kilogram meter per square second is equal to 1 Newton.
Our answer of 12 kg*m/s² is equal to 12 Newtons

$F= 12 \ N$

The force applies to the ball was <u>12 Newtons.</u>

8 0

3 years ago

A ball of mass 0.6 kg flies through the air at low speed, so that air resistance is negligible. What is the net force acting on

Anettt [7]

To solve this problem we will apply Newton's second law. The second law says that mass by acceleration is equivalent to the Force exerted on a body. Mathematically this

$F = ma$