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Katena32 [7]
3 years ago
7

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

Physics
2 answers:
astraxan [27]3 years ago
6 0

Answer:

Since the paper is wadded up tight, and if there's any

air resistance left we assume there isn't any, it might

just as well be a stone that's tossed.  This is just a

stripped down projectile situation.

You said "an angle of 36 degrees", but you didn't say relative

to what.  I'll assume that it's 36 degrees above horizontal, and

now I'll proceed to answer the question with the information that

I just gave myself.

-- The vertical component of the velocity is  1.4 sin(36)

                                                                        = 0.823 m/s up.

-- The projectile rises for (0.823/9.8) second, runs out of gas,

and then falls for another (0.823/9.8) second to its original height.

So it's in the air for

                                  2 (0.823/9.8) = 0.168 second

                                                            (not very long at all)

-- The horizontal component of the velocity is  1.4 cos(36)

                                                                           = 1.133 m/s  

                                                             and it doesn't change.

-- During the 0.168 second that it's in the air,

the wad travels horizontally

                                              (0.168 s) x (1.133 m/s)

                                          =            0.19 meter

                                              (19 cm, ~ 7.5 inches)

If you find my mistake on this one, please please tell me.  

As of now, it looks like with that velocity at that angle, your

paper wad only makes it 7.5 inches from your hand into the can.

Explanation:

Talja [164]3 years ago
4 0
I used this kinematics equation:
H = Vt + 0.5At^2
H is height, t is time, V is initial vertical velocity, and A is acceleration.
Given values:
V = 3.1sin(32) = 1.643m/s
A = -9.81m/s^2
H = 1.643t - 4.905t^2
Solve for t when H = 0:
1.643t - 4.905t^2 = 0
t = 0.3349, 0
Reject t = 0
t = 0.3349s

Use this to calculate horizontal displacement:
X = Vt
X is horizontal displacement, V is horizontal velocity, and t is time.
Given values:
V = 3.1xcos(32) = 2.629m/s
t = 0.3349s
X = 2.629x0.3349
X = 0.88m

Choice C.
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A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu
Alekssandra [29.7K]

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

F = BIl \ sin(\theta)

(a) When the angle, θ = 60 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N

(b) When the angle, θ = 90 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N

(c) When the angle, θ = 120 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N

6 0
3 years ago
General relativity, rather than special relativity, must be used for which kind of frame of reference
gregori [183]
The answer is B. A frame of reference that is accelerating.
6 0
3 years ago
Read 2 more answers
Two cylindrical containers are placed on a turntable. canister A is empty; canister B contains lead shots. each canister is the
devlian [24]

Answer:

c) both containers slide off the turntable at the same turntable speed.

Explanation:

The greater the mass of the cylinder, the greater the centrifugal force acting on it. That makes up for the greater force needed to move the heavier cylinder.

5 0
3 years ago
During an earthquake, where is the greatest motion felt on the surface?
dalvyx [7]
That would be the epicenter. 
5 0
3 years ago
Energy from the Sun arrives at the top of the Earth's atmosphere with an intensity of 1.36 kW/m 2 . How long does it take for 3.
alina1380 [7]

Answer:

The time taken by 3.55\times 10^9\ J to arrive on an area of 4.25\ m^2 is 6.14\times 10^5\ seconds.

Explanation:

Given the intensity of the energy is 1.36\ kW/m^2

And the arriving energy is 3.55\times 10^9\ J

Also, the area in which energy is being arriving is 4.25\ m^2

Now, we will use relation between energy (E), intensity of energy (p), area (A) and time (T).

Where energy is in Joule, intensity is in kW/m^2, area is in m^2 and time is in seconds.

The equation is

E=pAT\\\\T=\frac{E}{pA}\\\\T=\frac{3.55\times 10^9}{1.36\times 10^3\times 4.25}\\\\T=0.614\times 10^6\ s\\T=6.14\times 10^5\ seconds

So, the time taken by 3.55\times 10^9\ J to arrive on an area of 4.25\ m^2 is 6.14\times 10^5\ seconds.

4 0
3 years ago
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