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3 years ago

What is the momentum of a 0.16-kg apple that moves with a speed of 2.7m/s

Physics

2 answers:

Ksju [112]3 years ago

8 0

Answer: 0.432kgm/s

Explanation:

Momentum of a body is defined as the product of the mass (m) of the body and its velocity (v).

Momentum = Mass × Velocity

Given mass = 0.16kg and velocity = 2.7m/s

Substituting the datas into the formula,

Momentum = 0.16×2.7

Momentum = 0.432kgm/s

kirill115 [55]3 years ago

6 0

Answer:

Explanation:

The concept of momentum is defined as,

Momentum = mass × velocity

So for the apple,

Momentum of the apple = mass × velocity

= 0.16×2.7 = 0.432 Ns

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An object moving from a large positive velocity to a smaller negative velocity value

olga55 [171]

Answer:

Observe that the object below moves in the positive direction with a changing velocity. An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration). The dot diagram shows that each consecutive dot is not the same distance apart (i.e., a changing velocity). The position-time graph shows that the slope is changing (meaning a changing velocity) and positive (meaning a positive velocity). The velocity-time graph shows a line with a negative (downward) slope (meaning that there is a negative acceleration); the line is located in the positive region of the graph (corresponding to a positive velocity). The acceleration-time graph shows a horizontal line in the negative region of the graph (meaning a negative acceleration).

Animn of a Positive Velocity and Negative Accel'n

3 0

4 years ago

Particles q1 = -53.0 uc, q2 = +105 uc, and

Nimfa-mama [501]

Answer:

$-180.38\ \text{N}$

Explanation:

$q_1=-53\ \mu\text{C}$

$q_2=105\ \mu\text{C}$

$q_3=-88\ \mu\text{C}$

r = Distance between the charges

$r_{12}=0.5\ \text{m}$

$r_{23}=0.95\ \text{m}$

$r_{13}=1.45\ \text{m}$

k = Coulomb constant = $9\times 10^9\ \text{Nm}^2/\text{C}^2$

Net force is given by

$F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}$

The force on the particle $q_1$ is $-180.38\ \text{N}$ .

8 0

3 years ago

Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. A trained, fit person we

jeka57 [31]

(a) $4.14 rad/s^2$

The relationship beween centripetal acceleration and angular speed is

$a=\omega^2 r$

where

$\omega$ is the angular speed

r is the radius of the circular path

Here we gave

$a = 9g = 88.2 m/s^2$ is the centripetal acceleration

r = 5.15 m is the radius

Solving for $\omega$ , we find:

$\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{88.2 m/s^2}{5.15 m}}=4.14 rad/s^2$

(b) 21.3 m/s

The relationship between the linear speed and the angular speed is

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular speed

r is the radius of the circular path

In this problem we have

$\omega=4.14 rad/s$

r = 5.15 m

Solving the equation for v, we find

$v=(4.14 rad/s)(5.15 m)=21.3 m/s$

7 0

3 years ago

Read 2 more answers

How do I find the work done by something if I am only given force, mass, and acceleration?

Anna [14]

Yeah oppp because this is yes

8 0

3 years ago

How long does it take a person to skate the width of a hockey rink (85 feet) at a constant speed of 15 feet per second?

Zepler [3.9K]

S(travel distance)=85 ft
v (velocity)=15 ft/s
-----------------------------------
t (time)=?

Calculate the time with the formula for the velocity:
v=S/t
t=S/v
t=85 ft/(15 ft/s)
t=5.666s

3 0

3 years ago