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earnstyle [38]
2 years ago
3

A 7- kg object is accelerating to the right at 2 km/s^2. What is the magnitude of the rightward net force acting on it?

Physics
1 answer:
Andreas93 [3]2 years ago
6 0

Answer:

Net force is 14 kN.

Explanation:

Given:

The mass of the object is, m=7\textrm{ kg}

The acceleration of the object is, a=2\textrm{ }km/s^{2}

According to Newton's second law of motion,

Net force is given as the product of mass and acceleration of the object.

Therefore, the net force acting on the object is given as,

F=ma\\F=7\times 2=14\textrm{ kN}

Therefore, the magnitude of the rightward net force acting on it is 14 kN.

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A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice
Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of \sqrt{3} v_{e}

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , v_{i}=2 v_{e}. Here the radius is R, so find the escape velocity as follows,

            \frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0

            \frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}

            v_{e}^{2}=\frac{2 G M}{R}

            v_{e}=\sqrt{\frac{2 G M}{R}}

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as,  U_{i}=-\frac{G M m}{R}

R tend to infinity when far away from the planet, so v_{f}=0

Then, kinetic energy at initial would be,

                  k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}

Similarly, kinetic energy at final would be,

                k_{f}=\frac{1}{2} m v_{f}^{2}

Here, v_{f}=\text { final velocity }

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

                 U_{i}+k_{i}=k_{f}+v_{f}

                 \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0

                  \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}

'm' and \frac{1}{2} as common on both sides, so gets cancelled, we get as

                   4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}

We know, v_{e}=\sqrt{\frac{2 G M}{R}}, it can be wriiten as \left(v_{e}\right)^{2}=\frac{2 G M}{R}, we get

                4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}

                v_{f}^{2}=3\left(v_{e}\right)^{2}

Taking squares out, we get,

                v_{f}=\sqrt{3} v_{e}

4 0
2 years ago
Solve the equation x=3logy2 for y.
melisa1 [442]

X = 3 · log(Y²)

X = 3 · 2·log(Y)

X/6 = log(Y)

10^(X/6) = 10^log(Y)

Y = 10^(X/6)

6 0
2 years ago
Read 2 more answers
You are learning about energy transformations in science class. Mel and Sam's built this set-up to see if light energy could be
Allisa [31]
I think it might be heat energy. light transforms into heat energy
5 0
2 years ago
Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The
Eddi Din [679]

Answer:

(a) Steel rod: 1.1 * 10^{-4}

    Copper rod: 1.88 * 10^{-4}

(b) Steel rod: 8.3 * 10^{-5} m

Copper rod: 1.41 * 10^{-4} m

Explanation:

Length of each rod = 0.75 m

Diameter of each rod = 1.50 cm = 0.015 m

Tensile force exerted = 4000 N

(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as

Strain = \frac{1}{Y} * \frac{F}{A}

where Y = Young modulus

F = Fore applied

A = Cross sectional area

For the steel rod:

Y =  200 000 000 000 N/m^{2}

F = 4000N

A = \pi r^{2}      (r = d/2 = 0.015/2 = 0.0075 m)

=> A = \pi * (0.0075)^{2}

=> A = 0.000177 m^{2}

∴ Strain = \frac{4000}{200000000000 * 0.000177} \\\\Strain = \frac{4000}{35400000}\\ \\Strain = 0.000113 = 1.13 * 10^{-4}

For the copper rod:

Y =  120 000 000 000 N/m²

F = 4000N

A = \pi r^{2}      (r = d/2 = 0.015/2 = 0.0075 m)

=> A = \pi * (0.0075)^{2}

=> A = 0.000177 m^{2}

Strain = \frac{4000}{120 000 000 000 * 0.000177} \\\\Strain = \frac{4000}{21240000}\\ \\Strain =  = 1.88 * 10^{-4}

(b) We can find the elongation by multiplying the Strain by the original length of the rods:

Elongation = Strain * Length

For the steel rod:

Elongation = 1.1 * 10^{-4} * 0.75 = 8.3 * 10^{-5} m

For the copper rod:

Elongation = 1.88 * 10^{-4} * 0.75 = 1.41 * 10^{-4} m

6 0
2 years ago
Fill in the blanks to complete the statements.
Murljashka [212]

Answer:

When an object changes speed (increases/decreases) it results in acceleration/de acceleration, its velocity also changes.

Explanation:

Acceleration is the rate of change in velocity.An object can accelerate when speed increases, decreases or direction changes. All these instances involves a change in velocity.Velocity is a vector quantity thus it has magnitude and the direction.Acceleration due to change in direction is centripetal acceleration.The expression for finding acceleration is;

a=change in velocity/change in time

a=Δv/Δt in m/s²

3 0
2 years ago
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