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Artyom0805 [142]
3 years ago
12

A boy is running to the east and jumps into a stationary boat in a lake. Once the boy lands in the boat, the boat moves to the e

ast. Which best compares the initial momentum of the boy with the final momentum of the boy/boat system?
Physics
1 answer:
Hunter-Best [27]3 years ago
4 0
<span>The initial momentum of the boy is equal to the boy/boat because the final velocity of the boat is less than the initial velocity of the boy.</span>
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If two arm wrestlers exert a force on each other’s hands, and the hands don’t move, the forces must be *
jeyben [28]
Balanced. They’re equally as strong so as their arm wrestling, neither of the men’s hands go down. Because they’re equally/balanced as strong.
3 0
3 years ago
A saddle weighs 250 newtons. The mass of the saddle is __________ kilograms. Use g = 9.8 N/kg for gravity.
Irina18 [472]

Answer:

saddle weighs 250 newtons. The mass of the saddle is ____250/9.8 kg______ kilograms. Use g = 9.8 N/kg for gravity.

7 0
2 years ago
Physics laws of motion
expeople1 [14]

The acceleration is 0.095 m/s^2

Explanation:

We can solve the problem by applying Newton's second law of motion: in fact, the net force acting on an object is equal to the product between the mass of the object and its acceleration. Therefore we can write:

\sum F = ma

where:

\sum F is the resultant force acting on the object

m is its mass

a is its acceleration

In this problem, we have the following forces acting on the system:

F_1 = 2500 N (forward)

F_2 = -2400 N (backward)

So, Newton's second law can be rewritten as:

F_1 -F_2 = ma

where:

m = 1050 kg is the mass of all the students

Solving the formula for a, we find the acceleration of the system:

a=\frac{F_1-F_2}{m}=\frac{2500+(-2400)}{1050}=0.095 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

8 0
3 years ago
8. Two forces of 10 N and 30 N are applied to a 10 kg box. Find (1) the box’s acceleration when both forces point due east and (
love history [14]

(1) acceleration, a = 4 m/s^{2}  (2) acceleration of 10 N, a_{1} = 1 m/s^{2} and acceleration of 30 N, a_{2} = 3 m/s^{2}

Explanation:

  • Here, the acceleration of the object could be found using the equation derived in the second law of motion. The equation is given as, F = ma where m is the acceleration of the object, m is the mass of the object and F is the applied on the object.
  • Let a_{1} be the acceleration for force 10 N, to find acceleration rearrange the equation to a = \frac{F}{m}. When we substitute 10 N force and 10 kg mass of the box in the equation. We will get a_{1} = 1 m/s^{2}
  • Let a_{2}  be the acceleration for force 30 N, to find acceleration rearrange the equation to F = \frac{F}{m}. When we substitute 30 N force and 10 kg mass of the box in the equation. We will get a_{2} = 3 m/s^{2}
  • To find the combined, just add the force and substitute in the above equation. Hence, a = 4 m/s^{2}
3 0
3 years ago
A pendulum of length L = [02]____________________ cm and mass m = 169 g is released from rest when the cord makes an angle of 65
arsen [322]

Complete question:

A pendulum of length L = 48.5 cm and mass m = 169 g is released from rest when the cord makes an angle of 65.4° with the vertical. What is the speed of the mass (m/s) upon reaching its lowest point?

Answer:

The speed of the mass upon reaching its lowest point is 2.36m/s

Explanation:

To obtain the speed of the mass upon reaching its lowest point, we apply the principle of conservation of mechanical energy. At the lowest point, the kinetic energy of the pendulum is maximum and at the highest point, the vertical displacement is maximum, thus potential energy is maximum.

Kinetic energy at the lowest point  = Potential energy at the highest point

mgh = \frac{1}{2}mv^2\\\\gh = \frac{1}{2}v^2\\\\v^2 = 2gh\\\\v =\sqrt{2gh}

From my explanation above, h is the vertical displacement, when potential energy of the pendulum is maximum. Considering a right angled triangle, this vertical displacement, h is the adjacent of the triangle, and it is equal to

L - Lcosθ.

h = 48.5 - 48.5cos(65.4) = 28.31 cm = 0.2831 m

v =\sqrt{2gh} = v =\sqrt{2*9.8*0.2831} =2.36 \frac{m}{s}

Therefore, the speed of the mass upon reaching its lowest point is 2.36m/s

7 0
3 years ago
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