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3 years ago

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A 2.0-kg object is lifted vertically through 3.00 m by a 150-N force. How much work is done on the object by gravity during this

process?

2 answers:

noname [10]3 years ago

6 0

Answer:

-58.8 J

Explanation:

The work done by a force is given by:

$W=Fdcos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement.

In this problem, we are asked to find the work done by gravity, so we must calculate the magnitude of the force of gravity first, which is equal to the weight of the object:

$F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N$

The displacement of the object is d = 3.00 m, while $\theta=180^{\circ}$ , because the displacement is upward, while the force of gravity is downward; therefore, the work done by gravity is

$W=Fdcos \theta=(19.6 N)(3.00 m)(cos 180^{\circ})=-58.8 J$

And the work done is negative, because it is done against the motion of the object.

Nana76 [90]3 years ago

3 0

Answer:

-58.8 J

Explanation:

Work done on the object by gravity = Force due to gravity x Displacement

Force due to gravity = weight of the object

= m g

= (2.0 kg) (9.8 m/s²)

= 19.6 N

The force due to gravity acts in downwards direction opposite the direction of the displacement. Thus, the angle between the force and the displacement is 180°.

Work done on the object by gravity

= 19.6 N x 3.00 m x cos 180° = -58.8 J

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Vinil7 [7]

Answer:

0.25223 seconds.

Explanation:

$m_1$ = Mass of bullet = 0.0146 kg

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g = Acceleration due to gravity = 9.81 m/s²

As linear momentum is conserved

$m_1u_1 + m_2u_2 =(m_1 + m_2)v$

Now $u_2=v$ as the block (with the bullet in it) reverses direction and rises,

$m_1u_1 + m_2v =(m_1 + m_2)v\\\Rightarrow m_1u_1=(m_1 + m_2)v-m_2v\\\Rightarrow 0.0146\times 816=(0.0146 + 2.55)v-(-2.55v)\\\Rightarrow 11.9136=2.2646v+2.55v\\\Rightarrow 11.9136=4.8146v\\\Rightarrow v=\frac{11.9136}{4.8146}\\\Rightarrow v=2.47447\ m/s$

Equation of motion

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{2.47447-0}{9.81}\\\Rightarrow t=0.25223\ s$

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3 years ago

A pendulum has a 0.35\ \text{kg}0.35 kg0, point, 35, space, start text, k, g, end text mass oscillating at a small angle from a

expeople1 [14]

Answer:

The frequency of oscillation of the simple pendulum is 0.49 Hz.

Explanation:

Given that,

Mass of the simple pendulum, m = 0.35 kg

Length of the string to which it is attached, l = 1 m

We need to find the frequency of oscillation. The frequency of oscillation of the simple pendulum is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{9.8}{1}} \\\\f=0.49\ Hz$

So, the frequency of oscillation of the simple pendulum is 0.49 Hz. Hence, this is the required solution.

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3 years ago

The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

$v=36.62 m/s$ This is the speed of the jogger

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In conclusion, although the video is not found here, the sentence makes reference to the transmission spectrum of colored filters.

Learn more about the transmission spectrum here:

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2 years ago

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