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Scilla [17]
3 years ago
5

Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running

start and jumps with an initial horizontal velocity of 25 m/s. Neither person experiences any significant air resistance. Just as they reach the lake below.A) Alice reaches the surface the lake first. B) Tom of reaches the surface of the first. C) Alice and Tom will reach the surface of the lake at the same time. D) Information is not enough
Physics
1 answer:
crimeas [40]3 years ago
3 0

Answer:c

Explanation:

Given

Alice launches with horizontal velocity u=25\ m/s

Tom simply drops straight down from the edge

Time taken by both the person is same as they have same initial vertical velocity i.e. zero so the time taken to reach the ground is zero.

Although Alice will travel more horizontal distance compared to Tom.

Thus option c is correct

You might be interested in
A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be
lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × 10^{-3} m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36  

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ)       ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v  /  2     ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × 10^{-3})²×ω²× 34.64 )   /  2

0.050 ×(6.43 × 10^{-3})²×ω²× 34.64 = 108

ω² = 108 / 7.160  × 10^{-5}

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

7 0
3 years ago
Two astronauts (each with mass 100 kg) are drifting together through space. They are connected to each other by a rope 5 m in le
Nana76 [90]

Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

600 J

Explanation:

m = Mass of astronauts = 100 kg

d = Diameter

r = Radius = \frac{d}{2}

v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s

The angular momentum of the system is 1000 kgm²/s

Rotational energy is given by

K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J

The rotational energy of the system is 400 J

There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s

Energy

E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J

The new energy will be 1000 J

Work done will be the change in the kinetic energy

W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J

The work done is 600 J

5 0
3 years ago
Billiard balls moving across a pool table is an example of _____ friction.
frozen [14]
It is an example of rolling friction because balls roll.

Answer is ROLLING
8 0
3 years ago
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
3 years ago
Find the slit separation of a double-slit arrangement that will produce interference fringes 0.018 rad apart on a distant screen
inessss [21]

Answer:

1.64 * 10^(-5) m

Explanation:

Parameters given:

Angular separation, θ = 0.018 rad

Wavelength, λ = 589 nm = 5.89 * 10^(-7) m

The angular separation when there are 2 slots is given as

θ = λ/2d

where d = separation between slits

d = λ/2θ

d = (589 * 10^(-9))/(2 * 0.018)

d = 1.64 * 10^(-5) m

5 0
2 years ago
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