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Flura [38]
3 years ago
4

If someone is attempting to focus on multiple tasks at one time, it could be said that they have divided attention.

Physics
2 answers:
umka21 [38]3 years ago
7 0

Answer:

the correct answer is true

Explanation:

edge 2020

pantera1 [17]3 years ago
4 0
It is true. There you go!
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A mass of 2.50 kg is in a gravitational field of 14.0 N/kg. What force acts on the mass?
Olenka [21]
To determine the force that acts on the mass, just multiply the mass by the gravitational field. Using the given data,
                             F = (2.50 kg)(14 N/kg) = 35 N
Therefore, the force that acts on the mass is equal to 35 N. 
7 0
4 years ago
Read 2 more answers
What is the current in a 120V circuit if the resistance 10?
densk [106]

Explanation:

V=IR; I=V/R

V=120V, R=10

I= 120/10=12

The current in the circuit is 12A

4 0
4 years ago
A billiard ball moving at 5 m/s strikes a stationary ball of the same mass. After the collision, the original ball moves at a ve
SIZIF [17.4K]

90 degrees - 30 = 60 degrees

Velocity = (5m/s - 4.35m/s x cos(30)) / cos(60)

Velocity = 2.47 m/s

The answer is D) 2.47 m/s at 61.9 degrees

8 0
3 years ago
Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football
faltersainse [42]

Answer:

a=159.32\ m/s^2

Explanation:

It is given that,

Angular speed of the football spiral, \omega=6.9\ rev/s=43.35\ rad/s

Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

v=r\omega

v=0.085\times 43.35

v = 3.68 m/s

The centripetal acceleration is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(3.68)^2}{0.085}

a=159.32\ m/s^2

So, the centripetal acceleration of the laces on the football is 159.32\ m/s^2. Hence, this is the required solution.

6 0
3 years ago
Tom is throwing an baseball at an aluminum can,
pishuonlain [190]

Answer:

The question relates to the conservation of energy principle, the conservation of the linear momentum, and Newton's Laws of motion

Part A

1) Tom throwing a baseball at a can

The initial velocity of the baseball = v₂

The initial kinetic energy of the baseball, K.E.₂ = (1/2)·m₂·v₂²

∴ The final kinetic energy of the baseball, K.E.₂' = (1/2)·m₂·v₂'² < (1/2)·m₂·v₂²

Therefore, the energy of the ball before the collision is lesser than the energy of the ball after the collision

2) The evidence that would likely support the claim is that the baseball's height above the ground reduces rapidly immediately after the collision which is due to the reduced velocity, and therefore, the reduced (kinetic) energy

The final velocity of the baseball v₂' < v₂

Part B

1) The argument

The initial velocity of the can = v₁ = 0 (The can is initially  at rest)

The initial kinetic energy of the can, K.E.₁ = (1/2)·m₁·v₁² = 0

The final velocity of the can v₁' > v₁ = 0

∴ The final kinetic energy of the can, K.E.₁ = (1/2)·m₁·v₁² > 0

Given that the velocity of the can increases from zero to a positive value after collision with the baseball, the kinetic energy of the can is increased from zero before the collision to a positive value after the collision

2) An evidence in support of the argument is the motion of the can which was initially at rest which is an indication of increase in energy podded by the can

Explanation:

8 0
3 years ago
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