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Simora [160]
3 years ago
5

Imagine that a loudspeaker is producing a quiet tone with a low pitch. How will its vibrations change:

Physics
1 answer:
iVinArrow [24]3 years ago
5 0
I believe it is A :) hope this helped
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3 0
3 years ago
An object of mass 25 kg acted upon by a net force of 10 N will experience an acceleration of?
Zolol [24]

Answer:

0.4 m/s2

Explanation:

mass: 25kg

net force: 10 N

acceleration: ?

net force ÷ by mass= acceleration

10 N ÷ 25 Kg = 0.4 m/s2

6 0
3 years ago
you push with an 18-N horizontal force on a 5-kg box of coffee resting on a on a horizontal surface. the force of friction on th
KatRina [158]

The acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.

Answer:

Explanation:

As per Newton's second law of motion, acceleration of any object is directly proportional to the net external unbalanced force acting on that object and inversely proportional to the mass of the object.

Since there are two forces acting on the box in opposite direction, the net force will be the difference of horizontal and frictional force acting on the object.

Net force = Horizontal force - Frictional force = 18 N - 8 N = 10 N.

Now, from second law of motion, Acceleration = \frac{Net force}{Mass}

So, acceleration = 10 N /5 kg = 2 m/s².

Since, acceleration exerted by the box is found to be 2 m/s², we can determine the final velocity of the object after 10 seconds using the first equation of motion.

v = u + at, Here v is the final velocity and u is the initial velocity which is zero for the present case. Other parameters like a is found to be 2 m/s² and time is 10 seconds.

So Final velocity v = 0+(2×10)=20 m/s.

And the final position can be determined using the second equation of motion.

s = ut+1/2at²

Final position = (0×10)+(0.5×2×10×10)= 100 m.

So the final position is 100 m.

Thus, the acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.

3 0
3 years ago
2. Choose one food chain in the web. Name all the organisms in that chain. Start with the producer and end with the top-level co
Neko [114]

Answer:

sun alfalfa rabbit coyote

Explanation:

8 0
2 years ago
A 200 kg wood crate sits in the back of a truck. The coefficients of friction between the crate and the truck are ????s = 0.9 an
Xelga [282]

Answer:

Maximum acceleration of the truck is 5.25 m/s^2

Explanation:

To find the maximum acceleration the truck first we need to calculate friction between truck and wood. Because of wood is not moving, coefficient of c_{s} need to be used for it. Then the formula will be:

F_{s}=c_{s}*200*9.8*cos(20)\\F_{s}=0.9*200*9.8*0.94\\F_{s}=1657.62

1657.62 Newton is the friction.

So force against friction need to be at most 1657.62 N. Then equation will be:

F_{s}=F_{a}+F_{m}\\1657.62=200*9.8*sin(20)+200*a*cos(20)\\1657.62=670.36+187.94*a\\987.26=187.94*a\\a=5.25

Maximum acceleration of the truck need to be 5.25 m/s^2

4 0
3 years ago
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