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exis [7]
3 years ago
15

If a hockey puck is made to slide along a straight line on a completely frictionless surface, which statement about the motion o

f the hockey puck is true?
The hockey puck will come to a halt immediately.

The hockey puck will continue moving with constant velocity along a straight line.

The hockey puck will accelerate and continue moving in straight line.

The hockey puck will continue moving in random direction.
Physics
2 answers:
77julia77 [94]3 years ago
8 0
C. the hockey puck will accelerate and continue moving in a straight line
Llana [10]3 years ago
7 0

Answer:

The hockey puck will continue moving with constant velocity along a straight line.

Explanation:

As per Newton's first law we know that when there is no net force on an object then the velocity of object will always remains constant

Or we can say that the acceleration of an object will be zero when the force applied on the object is zero.

As per Newton's law we know

F = ma

so we will have

a = 0

so object will continue in straight line with same uniform speed without any change in magnitude as well as any change in direction

so correct answer will be

The hockey puck will continue moving with constant velocity along a straight line.

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arsen [322]
Animal Adaptation 1 Adaptation 2 Arctic Fox It's thick fur and fluffy tail help it survive in it's harsh habitat. Their small, pointy ears can hear their prey moving around in underground tunnels. An Arctic fox's fur changes colors with the seasons of the year. The Arctic Fox has many unique adaptations.
7 0
3 years ago
An object is dropped and is in free fall. Each second, the position of the object is marked. The distance between each mark is m
navik [9.2K]
C. hope this helps :)
8 0
3 years ago
A boat is traveling at an initial velocity of 2.7 meters per second in the positive direction. It accelerates at a rate of 0.15
cupoosta [38]

Answer:

\boxed {\boxed {\sf 4.5 \ m/s \ in \ the  \ positive \ direction}}

Explanation:

We are asked to find the final velocity of the boat.

We are given the initial velocity, acceleration, and time. Therefore, we will use the following kinematic equation.

v_f= v_i + at

The initial velocity is 2.7 meters per second. The acceleration is 0.15 meters per second squared. The time is 12 seconds.

  • v_i= 2.7 m/s
  • a= 0.15 m/s²
  • t= 12 s

Substitute the values into the formula.

v_f = 2.7 \ m/s + (0.15 \ m/s^2)(12 \ s)

Multiply the numbers in parentheses.

v_f= 2.7 \ m/s + (0.15 \ m/s/s * 12 \ s)

v_f = 2.7 \ m/s + (0.15 \ m/s *12)

\v_f=2.7 \ m/s + (1.8 \ m/s)v_f=2.7 \ m/s + (1.8 \ m/s)

Add.

v_f=4.5 \ m/s

The final velocity of the boat is <u>4.5 meters per second in the positive direction.</u>

5 0
3 years ago
A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the pers
jek_recluse [69]

Answer:

0.74 m/s

Explanation:

From the question,

We apply the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Since the skateboard, the person and the brick where stationary, therefore, the total momentum before collision is 0

0 = Total momentum after collision

(m+M)V + m'v = 0

Where m = mass of the  skateboard, M = mass of the person, m' = mass of the brick, V = recoil velocity of the person and the skateboard, v =  velocity of the brick

make V the subject of the equation above

V = -m'v/(m+M)................... Equation 1

Given: m = 4.10 kg, M = 68.0 kg, m' = 2.50 kg, v = 21.0 m/s.

Substitute these values into equation 1

V = -(2.5×21)/(68+2.5)

V = 52.50/70.5

V = 0.74 m/s

4 0
3 years ago
Consider two identical insulated metal spheres, A and B. Sphere A initially has a charge of -6.0 units and sphere B initially ha
Oksi-84 [34.3K]

Answer:

<em>-2 units of charge</em>

Explanation:

charge on A = Qa = -6 units

charge on B = Qb = 2 units

if the spheres are brought in contact with each other, the resultant charge will be evenly distributed on the spheres when they are finally separated.

charge on each sphere will be = \frac{Qa + Qb}{2}

charge on each sphere =  \frac{-6 + 2}{2} = \frac{-4}{2} = <em>-2 units of charge</em>

8 0
3 years ago
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