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2 years ago

Why does lifting one end of the track lead to constant acceleration?

1 answer:

7 0

We put a motion detector at one end of the track and put a cart on the track. ... Next, we put a motorized fan on the cart and let it push the cart down the track. ... This is what I would expect based on the velocity graph, since acceleration equals the slope of the velocity graph, which remainsconstant in time.

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A force of 10 newtons toward the right is exerted on a

weeeeeb [17]

Answer:

Explanation:

coefficient of kinetic friction of wooden floor μ = .4

force of friction = μ R , R is reaction force of floor

R = mg = weight of body

R = 25 N

force of friction = .4 x 25 = 10 N

Net force on the crate = 10 - 10 = zero .

Net force on the body will be nil.

6 0

1 year ago

A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, $v_{o} = 20.0 m/s$

Angle made by the ramp, $\theta = 22.0^{\circ}$

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH$

where

H = $dsin22^{\circ} = 5sin22^{\circ}$

g = $9.8 m/s^{2}$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ}}$ = 19.06 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m$

(b) now, considering the coefficient of friction bhetween ramp and the ball, $\mu = 0.150$ :

velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH - \mu gd$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5}$ = 18.7 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m$

6 0

1 year ago

An ambulance moving at 42 m/s sounds its siren whose frequency is 450 hz. a car is moving in the same direction as the ambulance

Korvikt [17]

(a) Since the ambulance and the car are moving one relative to each other, we have to use the general formula of the Doppler effect, which gives us the shift of the frequency of the siren as heard by an observer in the car:
$f'=( \frac{v+v_o}{v+v_s} )f$
where
f' is the apparent frequency as heard by the observer in the car
v is the velocity of the wave
$v_o$ is the velocity of the observer (positive if it is moving towards the source, negative if it is moving away)
$v_s$ is the velocity of the source (positive if the source is moving away from the observer, negative if is is moving towards it)
f is the real frequency of the sound

In the first part of the problem:
$v=343 m/s$ (speed of the sound wave)
$v_o =-25 m/s$ (the car is moving away from the ambulance)
$v_s = -42 m/s$ (the ambulance is moving towards the car)
$f=450 Hz$ (original frequency of the sound)

If we plug the numbers into the formula, we find
$f'=( \frac{343 m/s-25 m/s}{343 m/s-42 m/s} )(450 Hz)=475 Hz$

b) This time, the ambulance passes the car, so the ambulance is now moving away from the car; this means that $v_s$ must be positive:
$v_s=+42 m/s$
Moreover, the car is now moving towards the ambulance, so we should reverse also the sign of $v_o$ :
$v_o=+25 m/s$
All the other data do not change, so if we use the same formula as before, we find
$f'=( \frac{343 m/s+25 m/s}{343 m/s+42 m/s} )(450 Hz)=430 Hz$

8 0

1 year ago

You toss a rock of mass m vertically upward. Air resistance can be neglected. The rock reaches a maximum height h above your han

VladimirAG [237]

Answer with Explanation:

We are given that

Mass of rock=m

Maximum height=h

a.At maximum height, velocity,v=0

We know that

$v^2=u^2-2gh$

$0+2gh=u^2$

$u^2=2gh$

Height,h=h/4

Again, $v'^2=u^2-2g\times \frac{h}{4}$

$v'^2=2gh-\frac{gh}{2}=\frac{4gh-gh}{2}=\frac{3gh}{2}$

$v'=\sqrt{\frac{3gh}{2}}=\sqrt{\frac{3\times 9.8 h}{2}}=3.83\sqrt h$

Where $g=9.8 m/s^2$

b.When height,h=3h/4

$v'^2=u^2-2gh$

$v'^2=2gh-2g\times \frac{3h}{4}=2gh-\frac{3gh}{2}=\frac{4gh-3gh}{2}=\frac{gh}{2}$

$v'=\sqrt{\frac{9.8h}{2}}=2.2\sqrt h$

$v'=2.2\sqrt h$

4 0

2 years ago

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Mark and Balthazar are preparing to conduct neutralization reactions in which they add a base to two different solutions, citric

zhuklara [117]

Lab safety equipment prevents damage from accidents and helps keep the people working in the lab safe. The equipment goes hand in hand with the clothing of the person. The first step would be to wear closed shoes and a lab coat.
The equipment that must be worn are goggles to protect the eyes from irritants and latex gloves to protect the skin on the hands.

5 0

2 years ago

Read 2 more answers