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vodka [1.7K]
2 years ago
8

The speed of an object undergoing constant acceleration increased from 8.0 meters per second to 16.0 meters per second in 10. Se

conds . How far does the object travel during the 10. Seconds
Physics
1 answer:
saw5 [17]2 years ago
6 0

v₀ = initial speed of the object = 8 meter/second

v = final speed of the object = 16 meter/second

t = time taken to increase the speed = 10 seconds

d = distance traveled by the object in the given time duration = ?

using the kinematics equation

d = (v + v₀) t/2

inserting the above values in the above equation

d = (16 + 8) (10)/2

d = 120 meter


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A 1.0 kg object moving at 4.5 m/s has a wavelength of:
Lisa [10]
By wave particle  duality.

Wavelength , λ = h / mv

where h = Planck's constant = 6.63 * 10⁻³⁴ Js,  m = mass in kg,  v = velocity in m/s.
m = 1kg,  v = 4.5 m/s

λ = h / mv

λ = (6.63 * 10⁻³⁴) /(1*4.5)

λ ≈  1.473 * 10⁻³⁴  m

Option D.
7 0
2 years ago
A puck rests on a horizontal frictionless plane. A string is wound around the puck and pulled on with constant force. What fract
Ivan

Answer:

Explanation:

Let v be the linear velocity ,  ω be the angular velocity  and I be the moment of inertia of the the puck.

Kinetic energy ( linear ) = 1/2 mv²

Rotational kinetic energy = 1/2 I ω²

I = 1/2 m r² ( m and r be the mass and radius of the puck )

Rotational kinetic energy = 1/2 x1/2 m r² ω²

= 1/4 m v² ( v = r ω )

Total energy

= Kinetic energy ( linear ) + Rotational kinetic energy

= 1/2 mv² +  1/4 m v²

= 3/4 mv²

rotational K E / Total K E = 1/4 m v² / 3/4 mv²

= 1 /3

So  1 /3  rd of total energy is rotational K E.

3 0
2 years ago
A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal
Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration (g=9.8 m/s^2) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

v=18.9 m/s

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

v_y^2 - u_y^2 = 2ah (1)

where

u_y = 16.4 m/s is the initial vertical velocity

v_y is the vertical velocity at height h

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, v_y = 0, so we can use the equation to find the maximum height

h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m

So, at half of the maximum height,

h = \frac{13.7}{2}=6.9 m

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s

And so, the speed of the stone at half of the maximum height is

v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

v_1 = 18.9 m/s

while the speed at half of the maximum height (part b) is

v_2 = 22.2 m/s

So the difference is

\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s

And so, in percentage,

\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0
2 years ago
7) A heavy dart and a light dart are launched horizontally on a frictionless table by identical ideal springs. Both springs were
Elenna [48]

Answer:

A, B, and E

Explanation:

The springs are identical, and are compressed the same amount, so they have the same initial elastic potential energy. (E is true)

Energy is conserved, so the darts have the same amount of kinetic energy. (A is true, C is false)

The lighter dart has the same energy as the heavier dart.  Since it has less mass, it must have a greater speed. (B is true, D is false)

6 0
2 years ago
A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which
goldfiish [28.3K]

Answer:

160 Hz  ,  240 Hz  , 400 Hz

Explanation:

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

f_1=\dfrac{320}{4}\ Hz

f₁=80 Hz

Frequency of first harmonic = f₂

f₂=2 f₁

f₂ =2 x 80 = 160 Hz

Frequency of second harmonic = f₃

f₃= 3 f₁=3 x 80 = 240 Hz

Frequency of fifth harmonic = f₅

f₅=  5 f₁= 5 x 80 = 400 Hz

Three frequencies are as follows

160 Hz  ,  240 Hz  , 400 Hz

6 0
2 years ago
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