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Masja [62]
3 years ago
5

A+20 N force acts on a car and at the same time, a -30 N force acts on the

Physics
1 answer:
katovenus [111]3 years ago
5 0

Answer:

-10 N, not balanced

Explanation:

Force is a vector quantity, so in order to find the net force on an object, we must use vector addition rule.

This means that if two forces acting on an object along the same line, we have to choose one direction as positive, and write the two forces with the correct sign.

In this problem, we have two forces acting along a line on the car:

- A first force of

F_1 = +20 N

- A second force of

F_2=-30 N

Therefore, the net force on the car is:

\sum F=F_1+F_2=+20 +(-30)=-10 N

Moreover, the net force on an object is said to be "balanced" if it is zero: in this case, it is not zero, so it is not balanced.

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Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?
eduard

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = \frac{1}{2}mv^{2}

Plug in the numbers = \frac{1}{2}(4.0)(8^{2})

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

7 0
3 years ago
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4 years ago
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What is the magnitude of the momentum change of two gallons of water (inertia about 7.3 kg ) as it comes to a stop in a bathtub
aliya0001 [1]

We know that the change in momentum is equals to the product of force and time that is impulse (  F \times t). Therefore, we need to determine the value of that the water is in air by using the second equation of motion,

s=ut+\frac{1}{2} gt^2

Here, u is initial velocity which is zero.

s= \frac{1}{2} gt^2 \\\\ t = \sqrt{\frac{2s}{g} }.

Thus, impulse

= F \times \sqrt{\frac{2s}{g} }

From Newton`s second law,

F =mg

Therefore, impulse

= mg \times \sqrt{\frac{2s}{g} } = m \sqrt{2gs}

Given,  m = 7.3 kg and s = 2.0 m

Substituting these values, we get

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