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3 years ago

If we push a piece of furniture to move it but we don't manage to move it, are forces acting on

Physics

1 answer:

zubka84 [21]3 years ago

6 0

Answer:

Yes

Explanation:

Forces are pushing such as gravity, and forces from your own body and the couch are pushing against each other.

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A surface receiving sound is moved from its original position to a position three times farther away from the source of the soun

vazorg [7]

Answer:

c. nine times as low.

Explanation:

Sound intensity is defined as the acoustic power transferred by a sound wave per unit of normal area to the direction of propagation:

$I\propto \frac{1}{A}$

Since the sound wave has a spherical wavefront of radius r, then the area is given by:

$A=4\pi r^2$

Here r is the distance from the source of the sound. Thus sound intensity decreases as:

$I\propto \frac{1}{4\pi r^2}\\\\I'\propto \frac{1}{4\pi r'^2}\\\\I'\propto \frac{1}{4\pi (3r)^2}\\\\I'\propto \frac{1}{9} \frac{1}{4\pi r^2}\\\\I'\propto \frac{1}{9} I$

6 0

3 years ago

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen

bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

$f_{a} =(\frac{v+v_{c} }{v} )f_{s}$ (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

$f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s}$ (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

$1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s$

b) Substitute eq. 1 in eq. 2:

$f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz$

7 0

3 years ago

For a particular reaction, the change in enthalpy is 51kJmole and the activation energy is 109kJmole. Which of the following cou

Ronch [10]

Answer

given,

change in enthalpy = 51 kJ/mole

change in activation energy = 109 kJ/mole

when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.

where as activation energy of the product and the reactant decreases.

example:

ΔH = 51 kJ/mole

E_a= 83 kJ/mole

here activation energy decrease whereas change in enthalpy remains same.

5 0

3 years ago

A 2kg watermelon is dropped from a 4m tall roof a) use the appropriate kinematic equations to determine the instantaneous veloci

Ivenika [448]

Answer:

8.85m/s

Explanation:

The potential energy the watermelon held before dropping is Ep=mgh=2*9.8*4=78.4J.

When it strikes the ground, all of its Ep will transfer into Ek, so 1/2*m*v^2=78.4.

We already knew that m=2, so insert that in, we will get the V^2=78.4 m/s, V=8.85 m/s

6 0

3 years ago

An object begins with a speed of 20 meters per second and slows down to a speed of 10 meters per second over a time of 5 seconds

Andreyy89

Answer:

V = 6 m/s

Explanation:

Given that,

Initial speed of an object is 20 m/s

Final speed of an object is 10 m/s

Time, t = 5 s

We need to find the average speed of the object during these 5 seconds. Let it is equal to V. Here, time is same. The average speed is given by :

$V=\dfrac{x+y}{2}\\\\\text{Putting values}\\\\V=\dfrac{20+10}{5}\\\\V=6\ m/s$

So, the average speed of the object is 6 m/s.

3 0

3 years ago