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3 years ago

If the force acting on a cart doubles, what happens to the carts acceleration?

Physics

1 answer:

Effectus [21]3 years ago

8 0

Answer:

If the force on a cart doubles, the acceleration of the cart doubles.

Explanation:

For this problem, we need to consider the following equation:

Force = Mass x Acceleration

We can reasonably assume that the cart will have constant mass in the given force system. With this assumption we can say the following relationship:

Force is directly proportional to Acceleration within the system.

Given that our force on the cart is doubled, then our acceleration of the cart must also be doubled. You can mathematically express this as follows:

F = MA

2F = M * 2A

Hence, if force doubles, the acceleration doubles.

Cheers.

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Chemistry Sem 2

3241004551 [841]

Answer:

Solids - Bricks , wood , Pottery, Bucket

Liquid - Water, soap, Sanitizers.

Gases - Aerosol in Deodorants, Chlorofluorocarbons in Fire extinguishers , Butane in lighters.

4 0

2 years ago

Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa

Lorico [155]

Answer: $r_2=11.81 cm$

Explanation:

Given

$m_1=2.2\times 10^{-8} kg$

$m_2=4.8\times 10^{-8} kg$

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

$qV=\frac{mv^2}{2}$

$v=\sqrt{\frac{2qV}{m}}$

and we know

Force due to magnetic field will Provide centripetal Force

$qvB=\frac{mv^2}{r}$

$B=\frac{\sqrt{\frac{2Vm}{q}}}{r}$

and B is equal for both particles

thus $\frac{m}{r^2}=constant$

$\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}$

$r_2^2=\frac{4.8}{2.2}\times 8^2$

$r_2=11.81 cm$

4 0

2 years ago

How can a soccer ball and a small tennis ball bouncing relates to newtons first law interita?

Anna35 [415]

Answer:

I think it has something to do with both inertia and gravity?

7 0

3 years ago

How fast is a wave traveling if it has a wavelength of 7 meters and a frequency of 11 Hz?

pashok25 [27]

Answer:

$\huge{ \boxed{ \bold{ \sf{77 \: m/s}}}}$

☯ Question :

How fast is a wave travelling if it has a wavelength of 7 meters and a frequency of 11 Hz?

☯ $\underbrace{ \sf{Required \: Answer : }}$

☥ Given :

Wavelength ( λ ) = 7 meters
Frequency ( f ) = 11 Hz

☥ To find :

Speed of sound ( v ) = ?

☄ We know ,

$\boxed{ \sf{v = f \times λ}}$

where ,

v = speed of sound
f = frequency
λ = wavelength

Now, substitute the values and solve for v.

➺ $\sf{v = 11 \times 7}$

➺ $\boxed{ \sf{v = 77 \: m/s}}$

-------------------------------------------------------------------

✑ Additional Info :

Frequency : The number of complete vibrations made by a particle of a body in one second is called it's frequency. It is denoted by the letter f . The SI unit of frequency is hertz ( Hz ).

Wavelength : The distance between two consecutive compressions or rarefactions of a sound wave is called wavelength of that wave. It is denoted by λ ( lambda ) and it's SI unit is m.

Speed of a sound wave : The distance covered by a sound wave in one second is called speed of sound wave. It depends on the product of wavelength and frequency of the wave.

Hope I helped!

Have a wonderful time! ツ

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7 0

2 years ago

Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

ki77a [65]

Answer:

The lowest possible frequency of sound for which this is possible is 1307.69 Hz

Explanation:

From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.

First, we will determine his distance from the second speaker using the Pythagorean theorem

l₂ = √(2.00²+5.00²)

l₂ = √4+25

l₂ = √29

l₂ = 5.39 m

Hence, the path difference is

ΔL = l₂ - l₁

ΔL = 5.39 m - 5.00 m

ΔL = 0.39 m

From the formula for destructive interference

ΔL = (n+1/2)λ

where n is any integer and λ is the wavelength

n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.

Then,

0.39 = (1+ 1/2)λ

0.39 = (3/2)λ

0.39 = 1.5λ

∴ λ = 0.39/1.5

λ = 0.26 m

From

v = fλ

f = v/λ

f = 340 / 0.26

f = 1307.69 Hz

Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.

5 0

3 years ago