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Korolek [52]
3 years ago
14

according to newton's second law of motion of the net force acting on the object increases while the mass of the object remains

constant, what happens to the acceleration?
Physics
1 answer:
Licemer1 [7]3 years ago
4 0

Answer:

The Acceleration will increase

Explanation:

Newton's Second Law of motion: It states that the rate of change of momentum is directly proportional to the applied force and takes places along the direction of the force.

It can be expressed mathematically as,

F ∝ m(v-u)/t

Where (v-u)/t = a

F  = kma.

F = force, m = mass of the body, a = acceleration, k = constant of proportionality which tend to unity for a unit force, a unit mass, and a unit acceleration.

Therefore,

F = ma.

From the equation above,

If the net force acting on a body increase, while the mass of the body remains constant, the acceleration will also increase.

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A softball player moving 3.89 m/s
Ahat [919]

Answer:

0.119 s

Explanation:

Given that

U=3.89\ m/s\\a=-1.44\ m/s^2\\S=4.8\ m

Lets take final speed of the softball after covering 4.8 m = V

We know that

V^2=U^2+2aS\\V^2=3.89^2-2\times 1.44\times 4.8\\V=3.7\ m/s

Also We know that

V=U+at\\

Putting the value of V ,U and a in the previous equation  We get

3.7=3.89-1.44\times t\\t=0.119\ s

Therefore slide time will be 0.119 s

3 0
3 years ago
A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t
kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV

hope this helps!!

5 0
3 years ago
This problem has been solved!
lana66690 [7]

Answer:

Charge on each metal sphere will be 8\times 10^{8}C

Explanation:

We have given number of electron added to metal sphere A n=10^{12}electron

As both the spheres are connected by rod so half -half electron will be distributed on both the spheres.

So electron on both the spheres =\frac{10^{12}}{2}=5\times 10^{11}electron

We know that charge on each electron e=1.6\times 10^{-19}C

So charge on both the spheres will be equal to q=1.6\times 10^{-19}\times 5\times 10^{11}=8\times 10^{8}C

So charge on each metal sphere will be equal to 8\times 10^{8}C

6 0
3 years ago
How are electric motors and generators similar? 
kotykmax [81]

Answer:

they both produce energy

Explanation:

7 0
3 years ago
What is the correct explanation of Newton's Third Law of Motion?
viva [34]

Answer:

a

Explanation:

it explains the most, and it is the correct theorem

8 0
3 years ago
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