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vampirchik [111]
4 years ago
10

6. A car is approaching a hill at 30 m/s when its engine is stopped at the bottom of the hill. The car moves up the hill with an

acceleration of - 2 m/s/s. Determine the maximum distance the car travels up the hill.
Physics
1 answer:
irinina [24]4 years ago
7 0

Answer:

225 m.

Explanation:

Applying the equation of motion.

v² = u²+2as..................... Equation 1

Where v = final velocity of the car, u = initial velocity of the car, a = acceleration of the car, s = distance traveled by the car.

make s the subject if the equation.

s = (v²-u²)/2a....................  Equation 2

Given: v = 0 m/s, u = 30 m/s, a = -2 m/s²

Substitute into equation 2

s = (0²-30²)/(-2×2)

s = -900/-4

s = 225 m.

Hence the maximum distance the car travels up the hill =  225 m.

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Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters
Vesnalui [34]

Answer:

C. \frac{3F}{8}

Explanation:

Let initial charges on both spheres be,q

F=\frac{Kq^2}{d^2}   \ \ \  \ \ \  \ \ \  \ \_i

When the sphere C is touched by A, the final charges on both will be,\frac{q}{2}

#Now, when C is touched by B, the final charges on both of them will be:

q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\

Now the force between A and B is calculated as:

F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}

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3 years ago
What type of wave is infrared light
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Answer: Heat waves? I’m not %100 sure

4 0
3 years ago
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A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total
Harlamova29_29 [7]

Answer:

a)0.48 m/s

b) 0.583 m/s

Explanation:

As the wagon rolls,

momentum'p'= m x v => 95.8 x 0.530 = 50.774 Kgm/s

(a)Rock is thrown forward,

momentum of rock = 0.325 x 15.1 =  4.9075 Kgm/s

Conservation of momentum says momentum of wagon is given by

50.774 - 4.9075 = 45.8665

Therefore, Speed of wagon = 45.8665 / (95.8-0.325) = 0.48 m/s

(b) Rock is thrown backward,

momentum of wagon = 50.774  + 4.9075 = 55.68  Kgm/s

Therefore, speed of wagon = 55.68 / (95.8-0.325) = 0.583 m/s

4 0
4 years ago
A proton traveling at 27.1° with respect to the direction of a magnetic field of strength 2.33 mt experiences a magnetic force o
Elan Coil [88]
(a) The magnetic force experienced by a charged particle is:
F=qvB \sin \theta
where
q is the charge
v is the velocity of the particle
B is the magnitude of the magnetic field
\theta is the angle between the directions of v and B

The proton in our problem has a charge of q=1.6 \cdot 10^{-19} C, and it travels through a magnetic field with strength 
B=2.33 mT=2.33 \cdot 10^{-3} T
The direction between its velocity and B is \theta=27.1 ^{\circ}, and the force exerted on the proton is F=6.54 \cdot 10^{-17}N. Re-arranging the previous equation and using these data, we can find the value of v:
v= \frac{F}{qB \sin \theta} = \frac{6.54 \cdot 10^{-17}N}{(1.6 \cdot 10^{-19} C)(2.33 \cdot 10^{-3}T)(\sin 27.1^{\circ})}=3.85 \cdot 10^5 m/s

(b) Using the mass of the proton, m=1.67 \cdot 10^{-27}kg, we find its kinetic energy:
K= \frac{1}{2} mv^2= \frac{1}{2}(1.67 \cdot 10^{-27} kg)(3.85 \cdot 10^5 m/s)^2=1.24 \cdot 10^{-16} J

And keeping in mind that 1 eV = 1.6 \cdot 10^{-19}J, we can convert this value into electronvolts:
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6 0
3 years ago
 circuit in Mike’s home is almost at maximum current. If it is a 120 V circuit with about 1/3 of an amp available to safely use,
NikAS [45]

Answer:

40 W

Explanation:

The maximum wattage (power) that Mike can use is given by:

P=VI

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V is the voltage

I is the maximum current

In this problem, we have:

V=120 V is the voltage

I=\frac{1}{3}A=0.33 A is the maximum current available

Therefore, the maximum power that could be used is

P=VI=(120 V)(0.33 A)=40 W

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