A rolling ball with a mass of 1.01 kg crashes into a wall. The wall exerts a force of 2.40 Newtons and stops the ball. What was
1 answer:
Answer:
Acceleration, a = 2.38m/s²
Explanation:
Given the following data;
Mass = 1.01 kg
Force = 2.4N
To find the acceleration;
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
Where;
- F represents force.
- m represents the mass of an object.
- a represents acceleration.
Making acceleration (a) the subject, we have;
Substituting into the equation, we have;
<em>Acceleration, a = 2.38m/s²</em>
<em>Therefore, the magnitude of the acceleration of the ball at the time of the crash is 2.38 meters per seconds square. </em>
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Answer:
if the intensity of photons is constant then number of ejected electrons will remain same
Explanation:
As per photoelectric effect we know that when light of sufficient frequency fall on the surface of metal then electrons get ejected out of the surface with certain kinetic energy
Here the energy of photons is used to eject out the electrons from metal surface and to give the kinetic energy to the ejected electrons
so we have
here W = work function of metal which shows the energy required to eject out electrons from metal surface
KE = kinetic energy of ejected electrons
now if we increase the frequency of the photons that incident on the metal surface then in that case the incident energy will increase
So the electrons will eject out with more kinetic energy while if the number of photon is constant or the intensity of photons is constant then number of ejected electrons will remain same
Work done can be computed using the formula:
Where:
W = work (J)
F = Force (N)
d = Distance (d)
Looking at the given, you know that you do not have a value for force, so you will have to solve for it.
Where:
F = Force
m = mass
a = acceleration
Because the object is being lifted, the acceleration will rely on gravity. Acceleration due to gravity is a constant 9.8 m/s^2. Let's list our given first:
F = ?
m = 100kg
a = 9.8m/s^2
Put that into our equation and solve:
Our force is then 980 N.
Now that we have force we can solve for Work. The given for work is as follows:
F= 980N
d = 1.4m
Put that into our formula and solve:
The work done is 1,372J.
Where:
W = work (J)
F = Force (N)
d = Distance (d)
Looking at the given, you know that you do not have a value for force, so you will have to solve for it.
Where:
F = Force
m = mass
a = acceleration
Because the object is being lifted, the acceleration will rely on gravity. Acceleration due to gravity is a constant 9.8 m/s^2. Let's list our given first:
F = ?
m = 100kg
a = 9.8m/s^2
Put that into our equation and solve:
Our force is then 980 N.
Now that we have force we can solve for Work. The given for work is as follows:
F= 980N
d = 1.4m
Put that into our formula and solve:
The work done is 1,372J.
To solve the problem, we can use Kepler's third law, which states that the ratio between the square of the orbital period and the cube of the radius of the orbit is constant. So, we can rewrite it as follows:
where T are the periods, R the radius of the orbits (so, the distance of the satellites from Jupiter), and where the label c refers to Callisto and i to Io.
The problem says that the distance of Callisto from Jupiter is 4.5 times the distance of Io from Jupiter, i.e.:
If we substitute this into the previous equation, we can find a relationship between the two orbital periods Tc and Ti:
and so we have
that we can rewrite as
so, the orbital period of Callisto is 9.5 times the orbital period of Io.
where T are the periods, R the radius of the orbits (so, the distance of the satellites from Jupiter), and where the label c refers to Callisto and i to Io.
The problem says that the distance of Callisto from Jupiter is 4.5 times the distance of Io from Jupiter, i.e.:
If we substitute this into the previous equation, we can find a relationship between the two orbital periods Tc and Ti:
and so we have
that we can rewrite as
so, the orbital period of Callisto is 9.5 times the orbital period of Io.