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iris [78.8K]
3 years ago
11

Which one is it?????

Physics
2 answers:
antoniya [11.8K]3 years ago
5 0

Answer:ummm

Explanation:

masss

EastWind [94]3 years ago
4 0

Answer:

see below

Explanation:

mass divided by the net force of an object

The formula for calculating weight is F = m × 9.8 m/s2, where F is the object's weight in Newtons (N) and m is the object's mass in kilograms.

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A spring stretches 2.6 cm when a 7 g object is hung from it. The object is replaced with a block of mass 28 g which oscillates i
PSYCHO15rus [73]

Answer:

period = 0.65 sec

Explanation:

from the question we are given the following

extension (x) = 2.6 cm = 0.026 m

mass of object (Mo) = 7 g = 0.007 kg

mass of block (Mb) = 28 g = 0.028 g

acceleration due to gravity (g) = 9.8 m/s^{2}

period = 2π x \sqrt{\frac{Mb}{k}}

where k is the spring constant of the spring

and k = \frac{Mo x g}{x}

k =  \frac{0.007 x 9.8}{0.026}

k = 2.64 N/m

now period = 2π x \sqrt{\frac{0.028}{2.64}}

period = 0.65 sec

4 0
3 years ago
What is the frequency of a wave with a wavelength of 12 meters and a velocity of 4 m/s?
trasher [3.6K]

Answer:

0.33 hz

Explanation:

the formula for the frequency in this situation is f=v/wavelength

8 0
2 years ago
A camper stands in a valley between two parallel cliff walls. He claps his hands and notices that the echo from the nearby wall
Stells [14]
5,658 ft is your answer I believe
6 0
2 years ago
Suppose you were bungee jumping from a bridge while blowing a hand-held air horn. How would someone remaining on the bridge hear
Genrish500 [490]

Answer:

The pitch will progressively lower

Explanation:

If i were bungee jumping from  a bridge while blowing a hand-held air horn and someone who remains on the bridge will hear a decreased pitch or frequency as the source is moving away from the stationary listener as per the Dooplers effect. Hence, the pitch will progressively lower as the source is moving away from the observer.

4 0
3 years ago
Read 2 more answers
I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t
dezoksy [38]
Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }
where L is the length of the string, T the tension and m its mass. If  we plug the data of the problem into the equation, we find
f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz

The wavelength of the standing wave is instead twice the length of the string:
\lambda=2 L= 2 \cdot 24 m=48 m

So the speed of the wave is
v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s
7 0
3 years ago
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