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yarga [219]
3 years ago
5

How does gravity effect deceleration

Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
3 0
If you throw an object into the air, higher gravity will decelerate it more quickly than normal.
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Your friend says that for a moving object to continue moving, a force must be continually applied to it. Do you agree with your
Zigmanuir [339]

Answer:

I would disagree with my friend because that according to Newton's first law, an object in motion will continue to be in motion until stopped by another object/force. If the object is already moving, it will stay in motion until something else stops it. There is no need for a force to be <em>continually</em> applied to it while it's already moving.

4 0
2 years ago
A car could move at constant speed on an icy curve which is banked for _______________ (all, one, no) speed(s) of the car.
son4ous [18]
All is the answer I believe.
4 0
3 years ago
Suppose a star the size of our Sun, but of mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 17 days. If
Tanya [424]
Use the conservation of angular momentum; angular momentum at the beginning = angular momentum at the end 
Conservation of angular momentum: 
I1 w1 = I2 w2 
Where I is the moment of inertia. For a sphere, I=2/5 m R^2. Substituting into the equation above we get 
w2 = I1 w1 / I2 = w1 m1 R1^2 / (m2 R2^2) 
w2 = w1 4 * (R1/R2)^2
= 4*(1)*(7E5/7.5)^2
= 3.48E10 revs/(17days)
= 2.04705882 x 10^9 revs/sec
4 0
3 years ago
A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl
zysi [14]

Answer:\alpha =9.302\ rad/s^2

Explanation:

Given

mass of sphere m=250\ gm

diameter of sphere d=4.30\ cm

radius r=\frac{4.30}{2}\ cm

f=0.0200\ N

friction will provide resisting torque so

f\times r=I\times \alpha

where I=\text{moment of Inertia}

f=\text{friction force}

\alpha =\text{angular acceleration}

I=\frac{2}{5}mr^2

0.02\times r=\frac{2}{5}mr^2\times \alpha

\alpha =\frac{5}{2r}\times f

\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02

\alpha =9.302\ rad/s^2

(b)time taken to decrease its rotational speed by 21\ rad/s

t=\dfrac{\Delta \omega }{\alpha }

t=\dfrac{21}{9.302}

t=2.25\ s

6 0
3 years ago
German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) to the uncertainty in its velocity (
Ierofanga [76]

Answer:

The uncertainty in the position of the electron is 5.79x10^{-9}m

Explanation:

The Heisenberg uncertainty principle is defined as:

\Lambda p\Lambda x ≥ \frac{h}{4 \pi}  (1)

Where \Lambda p is the uncertainty in momentum, \Lambda x is the uncertainty in position and h is the Planck's constant.

The momentum is defined as:

p =mv  (2)

Therefore, equation 2 can be replaced in equation 1

\Lambda (mv) \Lambda x ≥ \frac{h}{4 \pi}

Since, the mass of the electron is constant, v will be the one with an associated uncertainty.

m \Lambda v \Lambda x ≥ \frac{h}{4 \pi} (3)

Then, \Lambda x can be isolated from equation 3

\Lambda x ≥ \frac{h}{m \Lambda v 4 \pi}  (4)

\Lambda x = \frac{6.626x10^{-34}J.s}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) 4 \pi}

But 1J = Kg.m^{2}/s^{2}

\Lambda x = \frac{(6.624x10^{-34} Kg.m^{2}/s^{2}.s)}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) (4 \pi)}

\Lambda x = 5.79x10^{-9}m

Hence, the uncertainty in the position of the electron is  5.79x10^{-9}m

7 0
3 years ago
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