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2 years ago

Please help me, this is a physics test.

Physics

1 answer:

sweet-ann [11.9K]2 years ago

4 0

Answer:

a = 2 [m/s²]

Explanation:

To be able to solve this problem we must make it clear that the starting point when the time is equal to zero, the velocity is 5 [m/s] and when three seconds have passed the velocity is 11 [m/s], this point is the final point or the final velocity.

We can use the following equation.

$v_{f}=v_{o}+a*t\\$

where:

Vf = final velocity = 11 [m/s]

Vo = initial velocity = 5 [m/s]

a = acceleration [m/s²]

t = time = 3 [s]

$11 = 5 + a*3\\6=3*a\\a= 2[m/s^{2} ]$

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Which of the following statements best describes energy conservation in heat engines?

Lorico [155]

B is the correct answer

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2 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

3 years ago

A stationary boat in the ocean is experiencing waves from a storm. The waves move at 59 km/h and have a wavelength of 145 m . Th

krek1111 [17]

Answer:

The time elapses until the boat is first at the trough of a wave is 4.46 seconds.

Explanation:

Speed of the wave, v = 59 km/h = 16.38 m/s

Wavelength of the wave, $\lambda=145\ m$

If f is the frequency of the wave. The frequency of a wave is given by :

$v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{16.38\ m/s}{145\ m}\\\\f=0.112\ Hz$

The time period of the wave is given by :

$T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.112\ Hz}\\\\T=8.92\ s$

We need to find the time elapses until the boat is first at the trough of a wave. So, the time will be half of the time period of the wave.

$T=\dfrac{8.92}{2}\\\\T=4.46\ s$

Hence, this is the required solution.

5 0

3 years ago

Describe the full water cycle

sergey [27]

The water cycle (hydro-logic cycle), explains the constant motion of water on, above and below the surface of the Earth. The mass of water on Earth stays equally constant over time, although the severe portion of the water goes into the major reservoirs of ice, fresh water, saline water and atmospheric water is variable depending on a wide range of climatic variables. The water moves from one reservoir to another, such as from river to ocean, or from the ocean to the atmosphere, by the physical processes of evaporation, condensation, precipitation, infiltration, surface runoff, and subsurface flow. In doing so, the water goes through different forms: liquid, solid (ice) and vapor.

5 0

3 years ago

Your iclicker operates at a frequency of approximately 900 mhz (900x106 hz). what is the approximate wavelength of the em wave p

Umnica [9.8K]

The clicker emits EM (electromagnetic) wave which travels at the speed of light, that is
v = 3 x 10⁸ m/s

The frequency is
f = 900mHz = 9 x 10⁸ Hz

Because velocity = frequency * wavelength, the wavelength, λ, is given by
fλ = v
λ = v/f
= (3 x 10⁸ m/s) / (9 x 10⁸ 1/s)
= 1/3 m

Answer: 1/3 m

5 0

3 years ago