Oil having a density of 923 kg/m^3 floats on water. A rectangular block of wood 4.81 cm high and with a density of 961 kg/m^3 fl
1 answer:
Answer:
x = 2.37 cm
Explanation:
given,
density of oil = 923 kg/m³
height of the block = 4.81 cm
density of the block = 961 Kg/m³
density of water = 1000 Kg/m³
Let x be the depth of the block in the water.
now,
Weight of block will be acting downward.
Buoyancy force will be acting on the block because of the oil and water.
At equilibrium position both weight and the buoyant force will balance.
x = 0.0237 m
x = 2.37 cm
Hence, the block is 2.37 cm in the water.
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Answer:
His launching angle was 14.72°
Explanation:
Please, see the figure for a graphic representation of the problem.
In a parabolic movement, the velocity and displacement vectors are two-component vectors because the object moves along the horizontal and vertical axis.
The horizontal component of the velocity is constant, while the vertical component has a negative acceleration due to gravity. Then, the velocity can be written as follows:
v = (vx, vy)
where vx is the component of v in the horizontal and vy is the component of v in the vertical.
In terms of the launch angle, each component of the initial velocity can be written using the trigonometric rules of a right triangle (see attached figure):
sin angle = opposite / hypotenuse
cos angle = adjacent / hypotenuse
In our case, the side opposite the angle is the module of v0y and the side adjacent to the angle is the module of vx. The hypotenuse is the module of the initial velocity (v0). Then:
sin angle = v0y / v0 then: v0y = v0 * sin angle
In the same way for vx:
vx = v0 * cos angle
Using the equation for velocity in the x-axis we can find the equation for the horizontal position:
dx / dt = v0 * cos angle
dx = (v0 * cos angle) dt (integrating from initial position, x0, to position at time t and from t = 0 and t = t)
x - x0 = v0 t cos angle
x = x0 + v0 t cos angle
For the displacement in the y-axis, the velocity is not constant because the acceleration of the gravity:
dvy / dt = g ( separating variables and integrating from v0y and vy and from t = 0 and t)
vy -v0y = g t
vy = v0y + g t
vy = v0 * sin angle + g t
The position will be:
dy/dt = v0 * sin angle + g t
dy = v0 sin angle dt + g t dt (integrating from y = y0 and y and from t = 0 and t)
y = y0 + v0 t sin angle + 1/2 g t²
The displacement vector at a time "t" will be:
r = (x0 + v0 t cos angle, y0 + v0 t sin angle + 1/2 g t²)
If the launching and landing positions are at the same height, then the displacement vector, when the object lands, will be (see figure)
r = (x0 + v0 t cos angle, 0)
The module of this vector will be the the total displacement (65 m)
module of r =
65 m = x0 + v0 t cos angle ( x0 = 0)
65 m / v0 cos angle = t
Then, using the equation for the position in the y-axis:
y = y0 + v0 t sin angle + 1/2 g t²
0 = y0 + v0 t sin angle + 1/2 g t²
replacing t = 65 m / v0 cos angle and y0 = 0
0 = 65m (v0 sin angle / v0 cos angle) + 1/2 g (65m / v0 cos angle)²
cancelating v0:
0 = 65m (sin angle / cos angle) + 1/2 g * (65m)² / (v0² cos² angle)
-65m (sin angle / cos angle) = 1/2 g * (65m)² / (v0² cos² angle)
using g = -9.8 m/s²
-(sin angle / cos angle) * (cos² angle) = -318.5 m²/ s² / v0²
sin angle * cos angle = 318.5 m²/ s² / (36 m/s)²
(using trigonometric identity: sin x cos x = sin (2x) / 2
sin (2* angle) /2 = 0.25
sin (2* angle) = 0.49
2 * angle = 29.44
<u>angle = 14.72°</u>
Answer:
(a)
(b)
(c)
Explanation:
Given that,
(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :
Where
m and e are the mass of and charge on an electron
On solving,
V = 100 V
(b) V = 1 kV = 1000 V
(c) If
Hence, this is the required solution.
Answer:
a) I=35mA
b) P=1.73W
Explanation:
a) The max emf obtained in a rotating coil of N turns is given by:
where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.
By calculating A and replacing in the formula (1G=10^{-4}T) we get:
Finally, the peak current is given by:
b)
we have that
hope this helps!!