All categories

3 years ago

(answer it or get reported) Please answer it w the steps

Physics

1 answer:

Travka [436]3 years ago

3 0

Answer;

The mass value for the above kinetic energy equation is 400.0000 kg. This is equal to:

■ 400,000.0000 g.

■ 14,109.6000 ounces.

■ 881.8480 pounds.

You might be interested in

3. A model rocket is launched straight upward at 58.8 m/s.

SIZIF [17.4K]

Let us assume that rocket only runs in initial energy and not using its own to flying.

Also , let upward direction is +ve and downward direction is -ve .

Initial velocity , u = 58.8 m/s .

Acceleration due to gravity , $g=-9.8\ m/s^2$ .

Final velocity , v - = 0 m/s .

We know , by equation of motion .

$v^2-u^2=2gh\\\\2gh_{max}=0^2-58.8^2\\\\h_{max}=\dfrac{0^2-58.8^2}{-2\times 9.8}\\\\h_{max}= 176.4\ m$

Hence, this is the required solution .

8 0

3 years ago

A metallic conductor has a resistivity of 35 × 10 −6 Ω⋅m. What is the resistance of a piece that is 20 m long and has a uniform

Nitella [24]

Answer:

Therefore the resistance of the conductor is 175Ω

Explanation:

Resistance:

Resistance of a metallic conductor is directly proportional to its length(l).

Resistance of a metallic conductor is inversely proportional to its cross section area(A).

The notation sign of resistance is R.

The unit of resistance is ohm (Ω).

Therefore,

$R \propto l$

and

$R \propto \frac{1}{A}$

$\therefore R\propto \frac{l}{A}$

$\Rightarrow R=\rho \frac{l}{A}$

ρ is the proportional constant.

It is also known as resistivity of that metal.

Given ρ=35×10⁻⁶Ω-m

l= 20 m

A= 4.0×10⁻⁶m²

$\therefore R=35\times 10^{-6}\times \frac{20}{4.0\times 10^{-6}}$

=175Ω

Therefore the resistance of the conductor is 175Ω

3 0

3 years ago

A 1-kg block is lifted vertically 1 m at constant speed by a boy. The work done by the boy is about:

Igoryamba

Answer:

$9.8\; {\rm J}$ , assuming that the gravitational field strength is $g = 9.8\; {\rm N \cdot kg^{-1}}$ .

Explanation:

Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.

By Newton's Second Law, the net force on this block would be $0$ . External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.

Let $m$ denote the mass of this block. It is given that $m = 1\; {\rm kg}$ . The weight of this block would be:

$\begin{aligned}\text{weight} &= m\, g \\ &= 1\; {\rm kg} \times 9.8\; {\rm N \cdot kg^{-1}} \\ &= 9.8\; {\rm N}\end{aligned}$ .

Hence, the force that the boy applies on this block would be upward with a magnitude of $F = 9.8\; {\rm N}$ .

The mechanical work that a force did is equal to the product of:

the magnitude of the force, and
the displacement of the object in the direction of the force.

The displacement of this block (upward by $s = 1\; {\rm m}$ ) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of: