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3 years ago

3. A model rocket is launched straight upward at 58.8 m/s.

Physics

1 answer:

SIZIF [17.4K]3 years ago

8 0

Let us assume that rocket only runs in initial energy and not using its own to flying.

Also , let upward direction is +ve and downward direction is -ve .

Initial velocity , u = 58.8 m/s .

Acceleration due to gravity , $g=-9.8\ m/s^2$ .

Final velocity , v - = 0 m/s .

We know , by equation of motion .

$v^2-u^2=2gh\\\\2gh_{max}=0^2-58.8^2\\\\h_{max}=\dfrac{0^2-58.8^2}{-2\times 9.8}\\\\h_{max}= 176.4\ m$

Hence, this is the required solution .

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A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang

koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × $10^{-4}$ Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux is express as

flux Ф = $\int \vec{B}.d\vec{A}$ .................1

and Ф = BA cos∅ ............2

so B = $\frac{\phi }{Acos\theta }$

and we know

A = ab

B = $\frac{\phi }{abcos\theta }$ ..............3

put here value

B = $\frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}$

solve we get

B = 0.692 T

8 0

2 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

7 0

3 years ago

Charge A and charge B are 3.00 m apart, and charge A is +1.44 C and charge B is +3.10 C. Charge C is located between them at a c

Sidana [21]

Answer:

the distance from charge A to C is r₁₃= 1.216 m

Explanation:

following Coulomb's law , the force exerted by 2 point charges between themselves is:

F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant

since C ( denoted as 3) is at equilibrium

F₁₃=F₂₃

k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²

q₁/r₁₃²=q₂/r₂₃²

r₁₃²/q₁=r₂₃²/q₂

r₂₃=r₁₃*√(q₂/q₁)

since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have

r₁₃+r₂₃=d=r₁₂

r₁₃+r₁₃*√(q₂/q₁)=d

r₁₃*(1+√(q₂/q₁))=d

r₁₃=d/(1+√(q₂/q₁))

replacing values

r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m

thus the distance from charge A to C is r₁₃= 1.216 m

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2 years ago

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Answer:

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