Answer: The unit of impulse is applied to an object produces an equivalent vector change in its linear momentum, also in the same direction.
Explanation:
Answer:
180° C
Explanation:
First we start by finding the area of the stopper.
A = πd²/4, where d = 1.5 cm = 0.015 m
A = 3.142 * 0.015² * ¼
A = 1.767*10^-4 m²
Next we find the force on the stopper
F = (P - P•)A, where
F = 10 N
P = pressure inside the tube,
P• = 1 atm
10 = (P - 101325) * 1.767*10^-4
P - 101325 = 10/1.767*10^-4
P - 101325 = 56593
P = 56593 + 101325
P = 157918 Pascal
Now, remember, in an ideal gas,
P1V1/T1 = P2V2/T2, where V is constant, then we have
P1/T1 = P2/T2, and when we substitute the values, we have
101325/(273 + 18) = 157918/ T2
101325/291 = 157918/ T2
T2 = (157918 * 291)/101325
T2 = 453 K
T2 = 453 - 273 = 180° C
Answer:
The acceleration would be 3.455.
Answer:
2.73×10¯³⁴ m.
Explanation:
The following data were obtained from the question:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Wavelength (λ) =?
Next, we shall determine the energy of the ball. This can be obtained as follow:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Energy (E) =?
E = ½m²
E = ½ × 0.113 × 43²
E = 0.0565 × 1849
E = 104.4685 J
Next, we shall determine the frequency. This can be obtained as follow:
Energy (E) = 104.4685 J
Planck's constant (h) = 6.63×10¯³⁴ Js
Frequency (f) =?
E = hf
104.4685 = 6.63×10¯³⁴ × f
Divide both side by 6.63×10¯³⁴
f = 104.4685 / 6.63×10¯³⁴
f = 15.76×10³⁴ Hz
Finally, we shall determine the wavelength of the ball. This can be obtained as follow:
Velocity (v) = 43 m/s
Frequency (f) = 15.76×10³⁴ Hz
Wavelength (λ) =?
v = λf
43 = λ × 15.76×10³⁴
Divide both side by 15.76×10³⁴
λ = 43 / 15.76×10³⁴
λ = 2.73×10¯³⁴ m
Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.
<h2>Answer: 10.52m</h2><h2 />
First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).
According to this, the initial velocity
has two components, because the brick was thrown at an angle
:
(1)
(2)
(3)
(4)
As this is a projectile motion, we have two principal equations related:
<h2>
In the x-axis:
</h2>
(5)
Where:
is the distance where the brick landed
is the time in seconds
If we already know
and
, we have to find the time (we will need it for the following equation):
(6)
(7)
<h2>
In the y-axis:
</h2>
(8)
Where:
is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>
is the acceleration due gravity
Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:
(9)
(10)
Multiplying by -1 each side of the equation:
>>>>This is the height of the building