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Tpy6a [65]
2 years ago
6

. A girl runs and jumps horizontally off a platform 10m above a pool with a speed of 4.0m/s. As soon as she leaves the platform,

she starts flipping, spinning with a constant angular acceleration of 15.0rad/s2. How many revolutions does she make before hitting the water? (Note that maintaining constant angular acceleration in the air is not realistic, but let’s do it here anyway.)\
Physics
1 answer:
faust18 [17]2 years ago
3 0

Answer:

2.39 revolutions

Explanation:

As she jumps off the platform horizontally at a speed of 10m/s, the gravity is the only thing that affects her motion vertically. Let g = 10m/s2, the time it takes for her to fall 10m to water is

h = gt^2/2

10 = 10t^2/2

t^2 = 2

t = \sqrt{2} = 1.414 s

Knowing the time it takes to fall to the pool, we calculate the angular distance that she would make at a constant acceleration of 15 rad/s2:

\theta = \alpha t^2/2

\theta = 15 * 2/2 = 15 rad

As each revolution is 2π, the total number of revolution that she could make is: 15 / 2π = 2.39 rev

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NeX [460]

Answer:

Force applied to smaller cross section is

= 82.63 N

Explanation:

As we know

F_2 A_1 = F_1 A_2

where F1, F2 signifies the weight of the two chair in a hydraulic-lift system

And A_1, A_2 signifies the area of the two respective chairs in a hydraulic-lift system

Given -

F2=1400 N

A1 =1220 Square centimeter

A_2 = 72 Square centimeter

Substituting the given values in above equation, we get -

1400 * 72 = F1 * 1220\\F2 = 82.63

Force applied to smaller cross section is

= 82.63 N

8 0
2 years ago
A fisherman has caught a very large, 5.0kg fish from a dock that is 2.0m above the water. He is using lightweightfishing line th
agasfer [191]

Answer:

t = 2 s

Explanation:

As we know that fish is pulled upwards with uniform maximum acceleration

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T - mg = ma

here we know that maximum possible acceleration of so that string will not break is given as

T = 54 N

now we have

54 - (5 \times 9.8) = 5 a

a = 1 m/s^2

now for such acceleration we can use kinematics

d = \frac{1}{2}at^2

2 = \frac{1}{2}(1) t^2

t = 2 s

7 0
2 years ago
Wave A has an amplitude of 2 and wave B has an amplitude of 2 as shown below. What will happen when the crest of wave A meets th
puteri [66]
Since the two waves have equal amplitudes, if the crest of one wave
meets the trough of the other one, they'll add to produce a level of zero
at that location.
6 0
2 years ago
Read 2 more answers
A particle with a charge of -1.24 x 10"° C is moving with instantaneous velocity (4.19 X 104 m/s)î + (-3.85 X 104 m/s)j. What is
astra-53 [7]

Answer:

(a) F= 6.68*10¹¹⁴ N (-k)

(b) F =( 6.68*10¹¹⁴ i  + 7.27*10¹¹⁴ j  ) N

Explanation

To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

F=q* v X B Formula (1 )

q: charge (C)

v: velocity (m/s)

B: magnetic field (T)

vXB : cross product between the velocity vector and the magnetic field vector

Data

q= -1.24 * 10¹¹⁰ C

v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j

B  =(1.40 T)i  

B  =(1.40 T)k

Problem development

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =

            = - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)

1T= 1 N/ C*m/s

We apply the formula (1)

  F= 1.24 * 10¹¹⁰ C*  5.39* 10⁴ m/s* N/ C*m/s (-k)

   F= 6.68*10¹¹⁴ N (-k)

a)  vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =

             =( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C*  (  5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s

F =( 6.68*10¹¹⁴  i  + 7.27*10¹¹⁴  j  ) N

8 0
2 years ago
If R is the total resistance of three resistors, connected in parallel, with resistances R1, R2, R3, then 1 R = 1 R1 + 1 R2 + 1
ANTONII [103]

Answer:

\Delta R_{max} = 0.46 ohm

Explanation:

Resistance is given of different values

now we have

R_1 = 64 ohm

R_2 = 20 ohm

R_3 = 8 ohm

possible error in all resistance is 0.5 %

now we know that

\Delta R_1 = 0.005\times 64 = 0.32 ohm

\Delta R_2 = 0.005 \times 20 = 0.1 ohm

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Now the maximum possible error when all resistance is connected in series

\Delta R_{max} = \Delta R_1 + \Delta R_2 + \Delta R_3

\Delta R_{max} = 0.32 + 0.1 + 0.04

\Delta R_{max} = 0.46 ohm

7 0
2 years ago
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