Which of the following sketches represents a possible configuration for this problem?

Suppose you push a hockey puck of mass m across frictionless ice for a time 1.0 s, starting from rest, giving the puck speed v a

EleoNora [17]

Newton's second law ...Force = momentum change/time.momentum change = Forcextme.also, F=ma -> a=F/m - the more familiar form of Newton's second law
using one of the kinematic equations for m ... V=u+at; u=0; a=F/m -> V=(F/m)xt.-> t=mV/F using one of the kinematic equations for 2m ... V=u+at; u=0; a=F/2m -> V=(F/2m)xt. -> t=2mV/F (twice as long, maybe ?)
I think I've made a mistake somewhere below, but I think that the principle is right ...using one of the kinematic equations for m ... s=ut + (1/2)at^2); s=d;u=0;a=F/m; t=1; -> d=(1/2)(F/m)=F/2musing one of the kinematic equations for 2m ... s=ut + (1/2)at^2); s=d;u=0;a=F/2m; t=1; -> d=(1/2)(F/2m)=F/4m (half as far ????? WHAT ???)

3 0

2 years ago

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Nicki rides her bike at a constant speed for 6 km. That part of her ride takes her 1 h. She then rides her bike at a constant sp

Savatey [412]

km x h = km/h

First trial: 6 x 1 = 6km/h

Second trial: 9 x 2 = 18km/h

6 + 18 = <u>24km/h</u> (Total)

Or

6 + 9 = 15 km

2 + 1 = 3h

15 + 3 = 18

15 x 2 = 30

3 x 2 = 6

30 - 6 = <u>24km/h</u>

8 0

2 years ago

U-238 has protons and146 neutrons. A particular isotope of plutonium has 94 protons, neutrons, and a mass number of 241. Thorium

enyata [817]

#1

$^{238}U$

so mass number = 238

mass number = protons + neutrons

given that

neutrons = 146

238 = protons + 146

protons = 92

#2

$^{241}Pu$

so mass number = 241

mass number = protons + neutrons

given that

Protons = 94

241 = 94 + neutrons

neutrons = 147

#3

$^ATh$

A = mass number

Protons = 90

Neutrons = 137

A = protons + Neutrons

A = 90 + 137 = 227

5 0

2 years ago

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You are piloting a small airplane in which you want to reach a destination that is 750 km due north of your starting location. O

alexira [117]

Answer:

v_wind = 101.46 km / h , θ = 61.8

Explanation:

This is a velocity composition exercise.

Let's do the problem in parts. Let's start by knowing the speed of the plane without air.

v = d / t

v = 750 / 3.14

v = 238.85 km / h

This is the speed of the plane relative to the Earth and it does not change.

In the second part, when there is wind, the travel time is greater than when there is no wind, therefore the wind delays the plane. To be more general, suppose that the wind has two components vₓ and $v_{y}$

Let's use trigonometry to find the components of the plane's speed

cos θ = v_N / v

sin θ = v_W / v

v_N = v cos θ

v_W = v sin θ

let's calculate

V _N = 238.85 cos 22 = 221.46 km / h

v_W = -238.85 sin 22 = -89.47

the negative sign is because the plane is going west and the positive sign is the east direction.

As it indicates that the destination of the avine is towards the north, the x component of the wind must be

vₓ - v_W = 0

vₓ = v-w

vₓ = 89.47 km / h

in the direction to the East.

Now let's analyze the component of the wind in the Nort-South direction,

Indicate the travel time, let's calculate the speed that the component must have the speed of the plane

v_total = d / t

v_total = 750 / 4.32

v_total = 173.61 km / h

This is the final speed of the plane, which can be written

v_total = v_n - vy

vy = v_n - v_total

vy = 221.46 - 173.61

vy = 47.85 km

this component is directed towards the south

Let's use the Pythagorean Theorem, to find the magnitude

v_wind² = vₓ² + vy²

v_wind = √ (89.47² + 47.85²)

v_wind = 101.46 km / h

the address will then be found using trigonometry

θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹1 (47.85 / 89.47)

θ = 28.14

Therefore, the magnitude of the wind speed is 101.5 km / h and its direction is 28º south of the East, to give this value

90- θtea = 90- 28.2

θ = 61.8

East of South

5 0

2 years ago

for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha

VLD [36.1K]

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

$R=\dfrac{u^2\ \sin2\theta}{g}$

Here,

u is the speed of launch of projectile

Here, $\theta=30^{\circ}$

We need to find the other launch angle when the projectile have the same range, such that,

$\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}$

$\sin(60)=\sin2\alpha$

$\dfrac{\sqrt3}{2}=\sin2\alpha$

$\alpha =30^{\circ}$

So, the other angle is 30 degrees. Hence, this is the required solution.

3 0

2 years ago