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posledela
4 years ago
12

Necesito ayuda urgente por favor, gracias 2) Sabiendo que el espacio que recorre un cuerpo en función del tiempo viene dado por

la siguiente ecuación: X(t)=t 2 -8t+6, donde x se mide en metros y t en segundos. a) Realizar la gráfica , posición-tiempo de en el intervalo [0,5] Nota: Tabular para t=0,1,2,3,4,5, ubicar los puntos sobre el plano y luego unir los puntos con una curva. b) Que distancia habrá recorrido en un minuto.
Physics
1 answer:
Viefleur [7K]4 years ago
6 0

Answer:b

Explanation:

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What type of energy travels through space as waves?
Nezavi [6.7K]

Answer:

electromagnetic energy

8 0
3 years ago
Read 2 more answers
Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 64.7 N64.7 N , Ji
WARRIOR [948]

Answer:

(a) Magnitude of force is 262.51 N

(b) Angle with East direction is -14.75^{o}

Explanation:

Force by Jack in vector form

\overrightarrow F _1} = 64.7{\rm{ N}}\left( {\hat i} \right)  

Force by Jill in Vector form is given by

\begin{array}{c}\\{\overrightarrow F _2} = 86.5{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 86.5{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\widehat j} \right)\\\\ = 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right)\\\end{array}

Force by Jane is

\begin{array}{c}\\{\overrightarrow F _3} = 181{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 181{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( { - \widehat j} \right)\\\\ = 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\end{array}

Net force is:

\overrightarrow F = {\overrightarrow F _1} + {\overrightarrow F _2} + {\overrightarrow F _3}

Hence

\begin{array}{c}\\\overrightarrow F = 64.7{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right) + 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\\ = 253.86{\rm{ N}}\left( {\hat i} \right) - 66.84{\rm{ N}}\left( {\widehat j} \right)\\\end{array}

The net force will be given by

F = \sqrt {{{\left( {{F_x}} \right)}^2} + {{\left( {{F_y}} \right)}^2}

Since F_{x}=253.86N and F_{y}=-66.84N

\begin{array}{c}\\F = \sqrt {{{\left( {253.86{\rm{ N}}} \right)}^2} + {{\left( { - 66.84{\rm{ N}}} \right)}^2}} \\\\ = {\bf{262.51 N}}\\\end{array}

The direction of net force is:

\theta = {\tan ^{ - 1}}\left {\frac{{{F_y}}}{{{F_x}}}}

Since F_{x}=253.86N and F_{y}=-66.84N  

\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{ - 66.84{\rm{ N}}}}{{253.86{\rm{ N}}}}} \right)\\\\ = {\tan ^{ - 1}}\left( { - 0.2633} \right)\\\\ = {\bf{ - 14}}{\bf{.7}}{{\bf{5}}^{\bf{o}}}\\\end{array}

The angle with East direction is -14.75^{o}

Net force exerted on the donkey is in the south-east direction. So, the angle of net force from the east direction is -14.75^{o} and it is 14.75^{o} from the south.

5 0
3 years ago
: Consider a transformer used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 4 turns in
Aneli [31]

Answer:

a) 096V  b) 0.0288A  c) 0.3456W

Explanation:

a) Vp/Vs= Np/Ns

 120/Vs= 500/4

Vs= 096V

b) Np/Ns= Is/Ip

500/4= 3.6/Ip

Ip= 0.0288A

c) P= VI

P=(120)(0.0288)

P= 0.3456W

8 0
3 years ago
A car with a mass of 833 kg rounds an unbanked curve in the road at a speed of 28.0 m/s. If the
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I had the same question

6 0
3 years ago
I shared a picture of the problem. It’s a basic Physics question and an Algebra question.
julia-pushkina [17]

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

Given the expressions;

T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}

Equating both expressions we will have;

2 \pi \sqrt{\frac{m}{k} }  = \frac{2 \pi}{\omega}

Divide both equations by 2π

\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\

Square both sides

(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}

Take the square root of both sides

\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

3 0
3 years ago
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