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Sedaia [141]
3 years ago
14

What is the net force of an object with a mass of 90.0 and accelerating at 3.0 m/s

Physics
1 answer:
Ksivusya [100]3 years ago
3 0

The net force is 270 N

Explanation:

We can solve this problem by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:

F=ma

where

F is the force

m is the mass

a is the acceleration

In this problem, we have

m = 90.0 kg

a=3.0 m/s^2

Substituting, we find the net force on the object:

F=(90.0)(3.0)=270 N

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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Which of the following is represented by the letter A in the diagram below?
Kruka [31]

We have that the letter A in the diagram below given as

Amplitude

Option A

<h3>Amplitude</h3>

Question Parameters:

Amplitude

Crest

Trough

Wavelength

Generally, the amplitude of a wave is the maximum  displacement of the wave in the medium from its initial position.

Amplitude is denoted with the letter A

Therefore,Amplitude

Option A

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6 0
2 years ago
An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s squared. What ar
Iteru [2.4K]
The direction of the electric field would be south. 

qE/m = 115 
<span>       E = 115*m/q </span>
<span>           = 115 * 9.1 * 10^(-31) / 1.67*10^(-19) </span>
<span>           = 762.87 * 10^(-12) </span>
<span>           = 6.27 x 10^-10 N/C
</span>
Hope this answers the question. Have a nice day. Feel free to ask more questions.
6 0
3 years ago
A 5.0 V battery contains 775 C of charge. How much electricity energy can it produce?
KatRina [158]

Answer:

3875J

Explanation:

Energy is defined as the power × time

And it's defined as

Power = IV - I- current and V- voltage

Now quantity of electricity; Q = I × t

Where I is current and t is time

Now Energy = I ×V×t = V× I×t = V× Q;

where Q is quantity of electricity 775C and V is 5.0volt

Hence 775 × (5) =3875J

7 0
3 years ago
Elaborate on the reason why heating water up will increase the rate of dissolving for a solute.
Harrizon [31]
<span> D) <span> <span> <span> The average kinetic energy of the molecules that make up the solution also increases with increasing temperature. This increase in kinetic energy allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions. </span></span></span></span>
5 0
3 years ago
Read 2 more answers
You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.
Neporo4naja [7]

Answer:

The distance is r_2  =  0.24 \  m

Explanation:

From the question we are told that

       The  distance from the conversation is r_1    =  24.0 \ m

       The  intensity of  the sound at your position is  \beta _1 =  40 dB

        The  intensity at the sound at the new position is  \beta_2 =  80.0dB

Generally the intensity in  decibel is  is mathematically represented as

      \beta  =  10dB log_{10}[\frac{d}{d_o} ]

The intensity is  also mathematically represented as

      d =  \frac{P}{A}

So

    \beta  =  10dB *  log_{10}[\frac{P}{A* d_o} ]

=>   \frac{\beta}{10}  =  log_{10} [\frac{P}{A (l_o)} ]

From the logarithm definition

=>    \frac{P}{A  *  d_o}  =  10^{\frac{\beta}{10} }

=>      P =  A (d_o ) [10^{\frac{\beta }{ 10} } ]

Here P is the power of the sound wave

 and  A is the cross-sectional area of the sound wave  which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

               P_1 =  A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]

Now the power of the sound wave at the second  position is mathematically represented as

               P_2 =  A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]

Generally  power of the wave is constant at both positions  so  

    A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]  = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]

      4 \pi r_1 ^2   [10^{\frac{\beta_1 }{ 10} } ]  = 4 \pi r_2 ^2   [10^{\frac{\beta_2 }{ 10} } ]

        r_2 =  \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}

       substituting value

        r_2 =   \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}

        r_2  =  0.24 \  m

     

7 0
3 years ago
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