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neonofarm [45]
3 years ago
12

A force of 6.0 N is applied horizontally to a 3.0 kg crate initially at rest on a horizontal frictionless surface. After the cra

te is pushed for 1.5 seconds, it has a gelocity of 3.0 m/s. Does this support Newton’s second law of motion, that force equals mass times acceleration?
Physics
1 answer:
monitta3 years ago
5 0
Your answer to your question is 6.9 m/s
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A 129-kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 65.
LUCKY_DIMON [66]

Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

Radius of the disk rotating about vertical axis = 1.61 m

Mass of the person  standing on platform = 65.7 kg

Distance from the center of platform = 1.07 m

Mass of the dog on the platform = 27.3 kg

Distance from center of platform = 1.31 m

We have to calculate the moment of inertia.

Formula:

MOI of disk = \frac{MR^2}{2}

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia (I_y_y )  of the system with respect to the axis yy.

⇒ I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g

⇒ I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2

⇒ I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2

⇒ I_y_y=289.088\ kg.m^2

The moment of inertia of the system is 289.088 kg.m^2

7 0
3 years ago
If a train travels 500 kilometers from Stockholm to
o-na [289]

Hello,

Average speed is total distance divided by total time. From the problem, our total distance is given as 500 kilometers and given time is 5 hours. Therefore, the average speed is:

\displaystyle{v_\text{average}=\sum_{i=1}^n \dfrac{s_i}{t_i}}\\\\\displaystyle{v=\dfrac{500\ \text{km}}{5 \ \text{h}}}\\\\\displaystyle{v=100 \ \text{km/h}}

Therefore, the average speed is 100 km/h. Please let me know if you have any questions!

4 0
9 months ago
Could someone please help me with this problem? I’m thinking that they both will have the same force because they have the same
Simora [160]
You would be correct
7 0
3 years ago
A sample of ice at 0°C has a mass of 23 g and a volume of 25 cm^3. Why does ice float on water? The density of water is 1 g.
vlabodo [156]

Answer: ice float in water due to density differences. Water has a higher density than ice.less dense object in liquid with high density

Explanation: from density formulae.

D=M/V

Density of ice =Mass of ice/Volume of ice

D=23/25

D=0.92g/cm^3

7 0
3 years ago
A rotating flywheel has moment of inertia 18.0 kg⋅m^2 for an axis along the axle about which the wheel is rotating. Initially th
timama [110]

Answer:

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

Explanation:

By the Principle of Energy Conservation and the Work-Energy Theorem we know that flywheel slow down due to the action of non-conservative forces (i.e. friction), the energy losses are equal to the change in the rotational kinetic energy. That is:

\Delta E = K_{1}-K_{2} (1)

Where:

\Delta E - Energy losses, measured in joules.

K_{1}, K_{2} - Initial and final rotational kinetic energies, measured in joules.

By definition of rotational kinetic energy, we expand the equation above:

\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2}) (2)

Where:

I - Moment of inertia of the flywheel, measured in kilograms per square meter.

\omega_{1}, \omega_{2} - Initial and final angular speed, measured in radians per second.

If we know that K_{1} = 30\,J, K_{2} = 15\,J and I = 18\,kg\cdot m^{2}, then the initial angular speed is:

K_{1} = \frac{1}{2}\cdot I \cdot \omega_{1}^{2} (3)

\omega_{1}=\sqrt{\frac{2\cdot K_{1}}{I} }

\omega_{1} = \sqrt{\frac{2\cdot (30\,J)}{18\,kg\cdot m^{2}} }

\omega_{1} \approx 1.825\,\frac{rad}{s}

\omega_{1}\approx 0.291\,\frac{rev}{s}

K_{2} = \frac{1}{2}\cdot I \cdot \omega_{2}^{2} (4)

\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }

\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }

\omega_{2} \approx 1.291\,\frac{rad}{s}

\omega_{2} \approx 0.205\,\frac{rev}{s}

Under the assumption that flywheel is decelerating uniformly, we get that the time taken for the flywheel to slowdown is:

t = \frac{\omega_{2}-\omega_{1}}{\alpha} (5)

If we know that \omega_{1}\approx 0.291\,\frac{rev}{s}, \omega_{2} \approx 0.205\,\frac{rev}{s} and \alpha = -0.200\,\frac{rev}{s^{2}}, then the time needed is:

t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }

t = 0.43\,s

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

6 0
3 years ago
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