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2 years ago

You wad up a piece of paper and throw it into the wastebasket. How far will

Physics

2 answers:

vitfil [10]2 years ago

6 0

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, $g=9.8 m/s^2$ )

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

$d=\frac{u^2 sin 2\theta}{g}$

where

u is the initial velocity

$\theta$ is the angle of projection

g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

$\theta=65^{\circ}$

Substituting,

$d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

aksik [14]2 years ago

4 0

Answer:

like the other person said, the answer is 1.4 m

Explanation:

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When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$

Solving the above quadratic equation

$\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}$

$\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega$

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0

2 years ago

A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi

Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block $m = 6.60$ kg

Initial speed of block $v _{i} = 1.56$ $\frac{m}{s}$

Final speed of block $v_{f} = 0$ $\frac{m}{s}$

Coefficient of kinetic friction $\mu _{k} = 0.62$

Ramp inclined at angle $\theta =$ 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

$\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U$

Where $\Delta U = mg d\sin 28.4$

$\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4$

$\mu _{k} mgd \cos 28.4 +mgd\sin 28.4 = \frac{1}{2}mv_{i} ^{2}$

$d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}$

$d = 0.122$ m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0

2 years ago

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the jun

AlexFokin [52]

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

$i_1+i_3=i_2$ (1)