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2 years ago

Can velocity of an object be negative when it’s acceleration is positive?

Physics

2 answers:

Bad White [126]2 years ago

6 0

Answer:

Yes

Explanation:

When a bike is coming to a stop the velocity is negative while the acceleration is positive.

Margaret [11]2 years ago

5 0

Answer
Yes
Example
When a car is coming to rest by braking the acceleration will be negative and the velocity will be positive.

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When all group iia elements lose their valence electrons, the remaining electron configurations are the same as for what family

Orlov [11]

Noble Gases

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8 0

3 years ago

Read 2 more answers

A cheetah can go from the state of rest to running at 20m/s in just two seconds. What is the Cheetahs average acceleration

babymother [125]

acceleration = change in velocity/change in time

so...

a = 20 m/s / 2 seconds

a = 10

hope that helps :)

P.S. found this from Brainly User, sometimes all you have to do is search to find the answer.

7 0

2 years ago

A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal

suter [353]

Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

$m_c$ = The center of mass of the stick = $\frac{L}{2}-0.22=0.5-0.22=0.28\ m$

$\omega$ = Angular velocity

Moment of inertia of the system is given by

$I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)$

As the energy in the system is conserved

$mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s$

The maximum angular velocity is 5.82812 rad/s

4 0

2 years ago

Complete the table. can I please get help will give points to whoever

MA_775_DIABLO [31]

Answer: proton mass 1 and neutron has no mass number

Explanation: proton because of positive charge neutron because of negative charge

4 0

1 year ago

Four point charges of magnitudes +3q, -q, +2q, and -4q are arranged in the corners of a square of side length L. The charge -q c

mafiozo [28]

Answer:

d) 0 V

Explanation:

It can be showed that the potential due to a point charge q, to a distance d from the charge, can be expressed as follows:

$V = \frac{k*q}{r}$

where k = $\frac{1}{4*\pi*\epsilon0} = 9e9 N*m2/C2$

As the potential is an scalar, and is linear with the charge, we can apply the superposition principle, which means that we can find the potential due to one of the charges, as if the other were not present.

By symmetry, all four charges are at the same distance from the center, so we can write the total potential, as follows:

$V = \frac{k}{d} ( q1 + q2 + q3 + q4) (1)$

where d, is the semi-diagonal of the square, that we can find applying Pythagorean theorem, as follows:

$d = \sqrt{\frac{L^{2}}{4} + \frac{L^{2}}{4} } = L*\frac{\sqrt{2}}{2}$

Replacing by the values in (1) we have:

$V = \frac{9e9N*m2/C2}{\frac{L}{2}*\sqrt{2} }* ( +3q -q + 2q + -4q) = 0 V$

which is equal to the option d).

6 0

3 years ago