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2 years ago

If a 50 N force was applied to a really massive object as well as a tiny object, what can you say

Physics

1 answer:

frozen [14]2 years ago

6 0

Answer:

The tiny object's acceleration will be much greater.

Explanation:

Because of Newton's second law of motion, (which is F=ma or Force= Mass*acceleration), then if the force maintains the same (which in this case it does, because it says 50 N to both obejcts) and one mass was much greater than the other, then the ould be less for the more massive object and much greater for the lighter object.

For example: If 50 N were applied to a 500 kg object and a 50 kg object, then theformulas for each (respectively) are:

50 = 500*acceleration

and

50 = 50*acceleration

(Because of Newtons Second Law of Motion)

Then, solving for the equations, we get for equation 1:

Acceleration = .1 m/s^2

And for equation 2:

Accleration = 1 m/s^2

Thus, you can see that more massive objects (when applied he same amount of force as the smaller object) clearly have less accleration than the smaller objects.

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Jeremy accidently dropped his toy stuffed animal from the balcony of his apartment on the fourth floor. The toy hit the ground a

uysha [10]

F = (mass)(acceleration) = ma

m = 0.25 kg

Vi = 16 m/s

t = 2 s

Vf = 0 m/s (since it was put to stop)

a=(Vf-Vi)/t

a=(0-16)/2

a = 8 m/s^2 (decelerating)

F = ma = (0.25 kg)(8 m/s^2)

F = 2 N

<span>Hope this answer will be a good h<span>elp for you.</span></span>

3 0

3 years ago

Read 2 more answers

A strong lightning bolt transfers an electric charge of about 16 C to Earth (or vice versa). How many electrons are transferred?

zzz [600]

Answer:

Number of electrons, $n=9.98\times 10^{19}$

Explanation:

A strong lightning bolt transfers an electric charge of about 16 C to Earth, q = 16 C

We need to find the number of electrons that transferred. Let there are n electrons transferred. It is given by using quantization of electric charge as :

q = ne

$n=\dfrac{q}{e}$

e is elemental charge

$n=\dfrac{16}{1.602\times 10^{-19}}$

$n=9.98\times 10^{19}$

So, there are $9.98\times 10^{19}$ electrons that gets transferred. Hence, this is the required solution.

3 0

3 years ago

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A

Flura [38]

Answer:

The mass of the wheel is 2159.045 kg

Explanation:

Given:

Radius $r = 0.330$

Force $F = 290$ N

Angular acceleration $\alpha = 0.814 \frac{rad}{s^{2} }$

From the formula of torque,

Γ $= I\alpha$ (1)

Γ $= rF$ (2)

$rF = I \alpha$

Find momentum of inertia $I$ from above equation,

$I = \frac{rF}{\alpha }$

$I = \frac{0.330 \times 290}{0.814}$

$I = 117.56$ $Kg. m^{2}$

Find the momentum inertia of disk,

$I = \frac{1}{2} Mr^{2}$

$M = \frac{2I}{r^{2} }$

$M = \frac{2 \times 117.56}{(0.330)^{2} }$

$M = 2159.045$ Kg

Therefore, the mass of the wheel is 2159.045 kg

8 0

3 years ago

How can you demonstrate that charged objects exert forces, both attractive and repulsive

Tcecarenko [31]

If it attractive it has opposite pole and if it repulsive it has same pole

8 0

3 years ago

Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth r

KIM [24]

The force of gravity between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is $m_1=0.60 kg$ , while the Earth's mass is $m_2=5.97 \cdot 10^{24} kg$ . Their separation is $r=1.3 \cdot 10^7 m$ , therefore the gravitational force exerted on the object is
$F=(6.67 \cdot 10^{-11}m^3 kg^{-1} s^{-2}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=1.4 N$

5 0

2 years ago