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3 years ago

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An object of mass m has these three forces acting on it (there is no normal force, "no surface"). F1 = 1 N, F2 = 9 N, and F3 = 5

N. When answering the questions below, assume the x-direction is to the right, and the y-direction is straight upwards.

1 answer:

hammer [34]3 years ago

7 0

Answer:

solved

Explanation:

a) F_net = (F2 - F3)i - F1 j

b) |Fnet| = sqrt( (F2 - F3)^2 + F1^2)

= sqrt( (9- 5)^2 + 1^2)

= 4.123 N

c) θ = tan^-1( (Fnet_y/Fnet_x)

= tan^-1( -1/(9-5) )

= -14.036°

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Two long, parallel wires are separated by a distance of 2.50 cm. The force per unit length that each wire exerts on the other is

Kisachek [45]

Answer:

The current in the second wire is 8.33 A and the two currents are flowing in the opposite directions

Explanation:

Given that,

The separation between two long parallel wires, d = 2.5 cm

The force per unit length that each wire exerts on the other is, $\dfrac{F}{l}=4\times 10^{-5}\ N/m$

The current in one wire, $I_1=0.6\ A$

(a) The force per unit length of the wire is given by :

$\dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}$

On putting all the values we get :

$4\times 10^{-5}=\dfrac{4\pi \times 10^{-7}\times 0.6I_2}{2\pi \times (2.5\times 10^{-2})}$

$I_2=8.33\ A$

So, the current in the second wire is 8.33 A.

(b) It is given that, both the wires repel each other, so the current in other wire is flowing in the opposite direction of the current in the first wire.

Hence, this is the required solution.

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3 years ago

The toxic effects of water pollution are generally _______.

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3 years ago

A paintball’s mass is 0.0032kg. A typical paintball strikes a target moving at 85.3 m/s.

vekshin1

Answer:

A) If the paintball stops completely the magnitude of the change in the paintball’s momentum is $p=0.273kg*m/s$

B) If the paintball bounces off its target and afterward moves in the opposite direction with the same speed, the change in the paintball’s momentum is $p=0.546kg*m/s$

C) A paintball bouncing off your skin in the opposite direction with the same speed hurts more than a paintball exploding upon your skin because of the strength exerted is twice than if it explodes.

Explanation:

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A) We use the formula of momentum $p=mv$ , so we have $p=0.0032kg*85.3m/s=0.273kg*m/s$

B) We use the same formula above, then due we have a change of direction at the same speed, therefore the change in the momentum is the double so

$p=2*0.0032kg*85.3m/s=0.546kg*m/s$ .

C) The average strength of the force an object exerts during impact is determined by the amount the object’s momentum changes. therefore

$F=\frac{\Delta p}{\Delta t}$ , as we don't have any data about the impact time but we know momentum is twice, time does no matter and strength is twice too.

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3 years ago

Starting fom rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far doe

JulsSmile [24]

Answer:

(A) $a=2.0.37m/sec^2$

(B) s = 146.664 m

Explanation:

We have given car starts from the rest so initial velocity u = 0 m /sec

Final velocity v = 88 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So $88km/hr=88\times \frac{1000}{3600}=24.444m/sec$

Time is given t = 12 sec

(A) From first equation of motion v = u+at

So $24.444=0+a\times 12$

$a=2.0.37m/sec^2$

So acceleration of the car will be $a=2.0.37m/sec^2$

(b) From third equation of motion $v^2=u^2+2as$

So $24.444^2=0^2+2\times 2.037\times s$

s = 146.664 m

Distance traveled by the car in this interval will be 146.664 m

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3 years ago

Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed o

Mrac [35]

Answer:

a)α= 53.13°

b)The velocity at the highest point = 15 m/s

The acceleration at the highest point = 9.8 $m/s^2$

c)h=15 m

V=18.02 m/s

Explanation:

Speed of water ,u= 25 m/s

So the horizontal component of speed u = u cos α

Given that horizontal distance cover by water in 3 s is 45 m.

So We know that in projectile motion horizontal acceleration is zero.

In horizontal direction

Distance = Velocity x time

45 = u cos α x 3

u cos α = 45

45 = 25 cos α x 3

cos α = 45/75

α= 53.13°

So the velocity at the highest point = u cos α

The velocity at the highest point = 15 m/s

The acceleration at the highest point = 9.8 $m/s^2$

Now the velocity along vertical direction(Vo) = u sin α

Vo= 25 sin 53.13°

Vo =20 m/s

$h=V_o.t-\dfrac{1}{2}gt^2$

$h=20\times 3-\dfrac{1}{2}\times 10\times 3^2$

h=15 m

So at 15 m above the ground water will strike .

The y-component of velocity after 3 sec

Vy= Vo - g t

Vy = 20 - 10 x 3

Vy= -10 m/s

The horizontal component of velocity will remain 15 m/s.

The resultant velocity

$V=\sqrt{10^2+15^2}\ m/s$

V=18.02 m/s

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4 years ago