Ask question

Ask question

All categories

3 years ago

12

Can instantaneous velocity ever be negative?

2 answers:

kirill115 [55]3 years ago

6 0

Answer:

yes, explained

Explanation:

The instantaneous velocity of an object can be negative. Even though the average velocity of the object in an entire trip is positive, the instantaneous velocity can be negative if the object is moving in negative direction. This direction depends upon the observer. If he considers direction away from him in the right as positive then the direction towards him in the left is negative.

SSSSS [86.1K]3 years ago

4 0

Instantaneous velocity in definition is the average velocity in given particular moment time. Hence, it underlies the context of each velocity at each occupied time and space at that particular moment where you are. Moreover, talking about velocity is vector quantity, which means it both has magnitude and direction. Furthermore, this positive and negative sign attributed to their number means only the specific direction in which the object is going. For example in analogy to average velocity, the initial position of the object at the moment is 2 m/s to the right then the object suddenly changes his position to 3 m/s to the left as his final position. Which is, the exact opposite of the former direction then indicates that the final position is negative, contrary to the initial direction. Just remember that the negative values connoted on the numbers can indicate the opposite direction of either the initial position or an object is downward.

You might be interested in

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

Mashutka [201]

Answer:

$a = 0.53 m/s^2$

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi ( \frac{20}{60})$

$\omega = 2.10 rad/s$

so final tangential speed is given as

$v = r\omega$

$v = 1.71 (2.10) = 3.58 m/s$

now average acceleration of the girl is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{3.58 - 0}{6.73}$

$a = 0.53 m/s^2$

8 0

3 years ago

A cupboard is placed in front of a heater. Air can move through a gap under the cupboard.

expeople1 [14]

Answer:

B

Explanation:

7 0

3 years ago

A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit

djverab [1.8K]

Answer:

The current is $I = 6.68 \ A$

Explanation:

From the question we are told that

The radius of the loop is $r = 6 \ cm = 0.06 \ m$

The earth's magnetic field is $B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T$

The number of turns is $N =1$

Generally the magnetic field generated by the current in the loop is mathematically represented as

$B = \frac{\mu_o * N * I}{2 r }$

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

$B = B_e$

=> $B_e = \frac{\mu_o * N * I }{ 2 * r}$

Where $\mu$ is the permeability of free space with value $\mu _o = 4\pi * 10^{-7} N/A^2$

$0.7 *10^{-4}= \frac{ 4\pi * 10^{-7} * 1 * I}{2 * 0.06}$

=> $I = \frac{2 * 0.06 * 0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}$

$I = 6.68 \ A$

3 0

3 years ago

While hovering motionless 3.0 meters an asteroid, a small

AnnZ [28]

I have no idea what you are trying to ask sorry

4 0

3 years ago

A person is standing on a level floor. His head, upper

BabaBlast [244]

Answer:

$y_{cg} = 1.03 m$

Explanation:

Given data:

weigh (head+arms + head) w_1 = 438 N

centre of gravity y_1= 1.28 m

weigh (upper leg) w_2 = 144 N

Center of gravity y_2 = 0.760 m

weigh ( lower leg + feet) = 87 N

centre of gravity = y_3 = 0.250 m

location of center of gravity $= \frac{w_1 y_1 + w_2 y_2 + w_3 y_3}{w_1 +W_2 +w_3}$

$y_{cg} = \frac{438 \times 1.28 + 144\times 0.760 + 87 \times 0.250}{438+144+87}$

$y_{cg} = 1.03 m$

8 0

3 years ago