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bulgar [2K]
3 years ago
6

If net force is zero, the what will be constant.

Physics
2 answers:
vodomira [7]3 years ago
8 0

Answer:

The ans is velocity

Explanation:

Hope it helps you

PtichkaEL [24]3 years ago
3 0

momentum remain constant.

when net force=0=acceleration=0-velocity doesn't change....

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A 10 Kg dog is running with a speed of 5.0 m/s what is the minimum work required to stop the dog
ra1l [238]

Answer:

25J

Explanation:

Given parameters:

Mass of the dog  = 10kg

Speed of the dog  = 5m/s

Unknown:

The minimum energy required to stop the dog  = ?

Solution:

The dog is moving with a kinetic energy and to stop the dog, an equal amount of kinetic energy generated must be applied to the dog.

 To find the kinetic energy;

        K.E  = \frac{1}{2} m v²

m is the mass

v is the velocity

Now insert the parameters and solve;

      K.E  =  \frac{1}{2}  x 10 x 5  = 25J

8 0
3 years ago
A common magnifying glass is an example of.​
Misha Larkins [42]

Answer:

planoconvex lens

Explanation:

8 0
3 years ago
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Choose the +x-direction to point to the right. • Object 1 has a mass 1.66 kg and is moving to the right at 11.2 m/s. • Object 2
egoroff_w [7]

Answer:

M = 49.4kgm/s (towards the left)

Explanation:

Momentum is the product of mass and velocity of an object

Momentum = mass * velocity

Momentum of Object 1 with mass 1.66 kg moving to the right at 11.2 m/s, is expressed as:

M1 = 1.66 * 11.2

M1 = 18.592kgm/s

Momentum of Object 2 with mass 6.59 kg moving to the left at 10.4 m/s is expressed as:

M2 = 6.59 * -10.4

M2 = -68.536kgm/s (negative since it is moving towards the left)

The total momentum will be the sum of momentum along the x-component as shown:

M = M1+M2

M = 18.592kgm/s--68.536kgm/s

M = -49.944kgm/s

M = 49.4kgm/s (towards the left)

5 0
3 years ago
An 80 g, 40 cm long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 15 g ball of clay trav
fiasKO [112]

Answer:

θ  = 12.95º

Explanation:

For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height

Let's start by finding the speed of the bar plus clay ball system, using amount of momentum

The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)

Initial before the crash

      p₀ = m v₀

Final after the crash before starting the movement

     p_{f} = (m + M) v

     p₀ = p_{f}

     m v₀ = (m + M) v

     v = v₀ m / (m + M)

     v = 2.0 0.015 / (0.015 +0.080)

     v = 0.316 m / s

With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy

Lower

    Em₀ = K = ½ (m + M) v²

Higher

    E_{mf} = U = (m + M) g y

   Em₀ = E_{mf}

   ½ (m + M) v² = (m + M) g y

   y = ½ v² / g

   y = ½ 0.316² / 9.8

   y = 0.00509 m

Let's look for the angle the height from the pivot point is

    L = 0.40 / 2 = 0.20 cm

The distance that went up is

     y = L - L cos θ

     cos θ  = (L-y) / L

     θ  = cos⁻¹ (L-y) / L

     θ  = cos⁻¹-1 ((0.20 - 0.00509) /0.20)

      θ  = 12.95º

4 0
3 years ago
A toy cart is pulled a distance of 6.00 m in a straight line across the floor. The force pulling the cart has a magnitude of 20.
aivan3 [116]

Answer:

W = 95.8J

Explanation:

Given

S = 6.00m

F =20.0N

Theta = 37°

W = FSCos(theta)

W = 20×6.0×Cos(37)

W = 95.8J

8 0
3 years ago
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