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kogti [31]
4 years ago
11

What’s the answer to this one?

Physics
1 answer:
Pie4 years ago
6 0

V = d/t

The speed of light is 3.0 x 10^8 m/s

The distance traveled is 149 x 10^9 m

Filling in the equation

3.0 x 10^8 = 149 x 10^9 / t

Multiply by t

3.0 x 10^8 x t = 149 x 10^9

Divide by the speed

t = 149 x 10^9 / 3.0 x 10^8

t = 496.67 s or roughly 500 s

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A change in position of an object relating to time
MaRussiya [10]
The answer is: Motion!

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3 years ago
Read 2 more answers
A teacher pushed a 98 newton desk across a floor for a distance of 5 meter he exerted a horizontal force of 20 newton for four s
Rom4ik [11]
Missing question: "how much work is done by the teacher?" (found on internet)

Solution:
The work done to move the desk across the floor is equal to
W=Fd
where F is the horizontal force applied to move the desk, and d is the distance covered by the desk. If we use F=20 N and d=5 m, we find the work done:
W=Fd=(20 N)(5 m)=100 J
3 0
3 years ago
A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.
Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

    k x = m g

k = \dfrac{mg}{x}

k = \dfrac{6.4 \times 9.8}{0.28}

      k = 224 N/m

b) f = \dfrac{\omega}{2\pi}

    \omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s

   f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

   f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}

  f =0.94\ Hz

c)  v_b = -v cos \omega t

    v_b = -5.1 \times cos (5.92 \times 0.42)

    v_b = 4.04\ m/s

d)  a_{max} = v \omega

    a_{max} = 4.04 \times 5.92

    a_{max} =23.94\ m/s^2

e)  Y =- A sin (\omega t)

    A = \dfrac{v}{\omega}

    A = \dfrac{4.04}{5.92}

        A = 0.682 m

    Y =- 0.682 \times sin (5.92 \times 0.42)

    Y =- 0.42

Force =m \omega^2 |Y|

          =6.4 \times 5.92^2\times 0.42

F = 94.20 N

4 0
3 years ago
A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp
Anna71 [15]

\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}

Explanation:

       Natural length of a spring is 0.5\text{ }m. The spring is streched by 0.02\text{ }m. The resultant energy of the spring is 0.5\text{ }J.

       The potential energy of an ideal spring with spring constant k and elongation x is given by \dfrac{1}{2}kx^{2}.

       So, in the current problem, the natural length of the spring is not required to find the spring constant k.

       \text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}

∴ The spring constant of the spring = 2500\text{ }\frac{kg}{s^{2}}

4 0
4 years ago
A crane carries a 1600 kg car at 1.5 m/s^2 with a chain that has a negligible mass. If the coefficient of kinetic friction betwe
padilas [110]

Answer:

option (d) 7.1 kN

Explanation:

Given:

Mass of the car, m = 1600 kg

Acceleration of the car, a = 1.5 m/s²

Coefficient of kinetic friction = 0.3

let the tension be 'T'

Now,

ma = T - f .................(1)

where f is the frictional force

also,

f = 0.3 × mg

where g is the acceleration due to the gravity

thus,

f = 0.3 × 1600 × 9.81 =

therefore,

equation 1 becomes

1600 × 1.5 = T - 4708.8

or

T = 2400 + 4708.8

or

T = 7108.8 N

or

T = 7.108 kN

Hence,

The correct answer is option (d) 7.1 kN

6 0
3 years ago
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