A 70 watt light bulb is run by a 120 v circuit. how many amps does it use

The two forces in each pair can either both act on the same body or they can act on different bodies. The two forces in each pai

kotykmax [81]

Answer: The given statement is false.

Explanation:

According to Newton's third law of motion, every action has an equal and opposite reaction. So, when we apply force in one direction on an object then the object also applies a force in the opposite direction.

Hence, it is true that two forces in each pair of forces act in opposite directions.

For example, when we push a wooden box of 20 kg in the forward direction then the box will also apply a force in the opposite direction.

But the statement two forces in each pair can either both act on the same body or they can act on different bodies is false.

5 0

2 years ago

Describe what happens to the moving boat when the oars are out of the water and the forward thrust is zero

Tatiana [17]

Answer:

The boat won't be able to move if the oars were out and there was no thruster. If there was a flow of the water then yes there would be a moving boat.

6 0

2 years ago

Podrian ayudarme? Llevo dias intentando pero no se como 1. el profesor no explico

Paraphin [41]

You have to upload this in the area of mathematicians..!

8 0

3 years ago

A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug

natali 33 [55]

Answer:

Approximately $1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}$ , assuming friction between the vehicle and the ground is negligible.

Explanation:

Let $m$ denote the mass of the vehicle. Let $v$ denote the initial velocity of the vehicle. Let $k$ denote the spring constant (needs to be found.) Let $x$ denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

$\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}$ .

Initial kinetic energy ( ${\rm KE}$ ) of the vehicle:

$\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}$ .

When the vehicle is brought to a rest, the elastic potential energy ( $\text{EPE}$ ) stored in the spring would be:

$\displaystyle \frac{1}{2}\, k\, x^{2}$ .

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial $\text{KE}$ of the vehicle should be equal to the ${\rm EPE}$ of the vehicle. In other words:

$\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}$ .

Rearrange this equation to find an expression for $k$ , the spring constant:

$\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}$ .

Substitute in the given values $m = 1508\; {\rm kg}$ , $v \approx 1.908\; {\rm m\cdot s^{-1}}$ , and $x = 6.87\; {\rm m}$ :

$\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}$

8 0

1 year ago

A fireworks shell is accelerated from rest to a velocity of 68.0 m/s over a distance of 0.230 m.

Ugo [173]

Answer:

(A) 10052.2 m/s²

(B) 0.00678 seconds

Explanation:

From the question,

(A) Applying

V² = U²+2as..................... Equation 1

Where V = Final velocity, U = Initial velocity, a = acceleration due to gravity, s = distance.

make a the subject of the equation

a = (V²-U²)/2s........................ Equation 2

Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m

Substitute these values into equation 2

a = (68²-0²)/(2×0.230)

a = 10052.2 m/s²

(B) Using,

a = (V-U)/t......................... Equation 3

Where t= time.

make t the subject of the equation

t = (V-U)/a......................... Equation 4

Given: V = 68 m/s, U = 0 m/s, a = 10052.2

Substitute into equation 4

t = (68-0)/10052.2

t = 0.00678 seconds

5 0

3 years ago