Answer:
3120J
Explanation:
Given parameters:
C = Specific heat capacity = 0.8J/g°C
Initial temperature = 20°C
Mass given = 5g
Final temperature = 800°C
Unknown:
Energy given to the mass = ?
Solution:
To find the energy given to the mass, let us simply use the expression below:
H = m c ΔT
H is the unknown, the energy supplied
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
Input the variables;
H = 5 x 0.8 x (800 - 20) = 3120J
Answer:
Explanation:
The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by
where
is the horizontal velocity
t is the time of flight
The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.
In the first situation, the horizontal distance covered is
in the second case, the horizontal velocity is increased to
And so the new distance travelled will be
So, the distance increases linearly with the horizontal velocity.
Answer:
a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N
Explanation:
a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry
sin 30 = Wx / W
cos 30 = Wy / W
Wx = W sin30
Wy = W cos 30
Let's write the equations on each axis
X axis
Wx = ma
Y Axis
N- Wy = 0
N = Wy = mg cos 30
N = 2.0 9.8 cos 30
N = 16.97 N
We calculate the acceleration
a = Wx / m
a = mg sin 30 / m
a = g sin 30
a =9.8 sin 30
a = 4.9 m / s²
b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component
F -Wx = 0
F = Wx
F = m g sin 30
F = 2.0 9.8 sin 30
F = 9.8 N
Answer: 
Explanation:
This problem can be solved by the following equation:
Where:
is the change in kinetic energy
is the electric potential difference
is the electric charge
Finding :
Finally:
