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3 years ago

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A force of 50n is applied to an object at an angle of 50 degrees measured relative to the horizontal. Determine the horizontal a

nd vertical components of the applied force?

1 answer:

kobusy [5.1K]3 years ago

3 0

Answer:

Fx = 32.14 [N]

Fy = 38.3 [N]

Explanation:

To solve this problem we must decompose the force vector, for this we will use the angle of 50 degrees measured from the horizontal component.

F = 50 [N]

Fx = 50*cos(50) = 32.14 [N]

Fy = 50*sin(50) = 38.3 [N]

We can verify this result using the Pythagorean theorem.

$F = \sqrt{(32.14)^{2}+ (38.3)^{2}} \\F = 50 [N]$

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An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0

4 years ago

A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout

nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
So, we can write the following equation:

$E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)$

As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
So, on the outer surface of the shell there must be a charge that be the difference between them:

$Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C$

Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

$E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C$

As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.

6 0

3 years ago

Rotation of the lever OA is controlled by the motion of the contacting circular disk of radius r = 300 mm whose center is given

sergiy2304 [10]

Answer:

The angular velocity is

5.64rad/s

Explanation:

This problem bothers on curvilinear motion

The angular velocity is defined as the rate of change of angular displacement it is expressed in rad/s

We know that the velocity v is given as

v= ωr

Where ω is the angular velocity

r is 300mm to meter = 0.3m

the radius of the circle

described by the level

v=1.64m/s

Making ω subject of the formula and solving we have

ω=v/r

ω=1.64/0.3

ω=5.46 rad/s

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3 years ago

The total surface area of the human body is 1.20 m2 and the surface temperature is 30∘C=303∘K. If the surroundings are at a temp

zalisa [80]

To solve this problem we must consider the expressions of Stefan Boltzmann's law for which the rate of change of the radiation of energy H from a surface must be

$H = Ae\sigma T^4$

Where

A = Surface area

e = Emissivity that characterizes the emitting properties of the surface

$\sigma$ = Universal constant called the Stefan-Boltzmann constant $(5.67*10^{-8}m^{-2}K^{-4})$

T = Absolute temperature

The total heat loss would be then

$Q = H_2 -H_1$

$Q =Ae\sigma T_2^4-Ae\sigma T_1^4$

$Q = Ae\sigma (T_2^4-T_1^4)$

$Q = (1.2)(1)(5.67*10^{-8})(303^4-280^4)$

$Q = 155.29J$

Therefore the net rate of heat loss from the body by radiation is 155.29J

8 0

3 years ago

Write a complete sentence using the scientific meaning of energy

Nitella [24]

Sentence:

"A collision with no energy is transferred."

Or you could put

"Gamma rays often occur at different energies"

...

Scientific energy is the power that an object or force has that can be measured in joules.

7 0

4 years ago