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n200080 [17]
2 years ago
7

A 1500kg car traveling along a road is hit by a 0.1kgrock that creates a small crack in the car’s windshield. Which of the follo

wing describes the interaction between the windshield and the rock?a. The car exerts a force on the rock, but the rock does not exert a force on the car. b. The car exerts a force on the rock, and the rock exerts a force on the car. The two forces are not equal in magnitude. c. The car exerts a force on the rock, and the rock exerts a force on the car. The two forces are equal in magnitude.
Physics
2 answers:
Alex17521 [72]2 years ago
3 0

Answer:

The car exerts a force on the rock, and the rock exerts a force on the car. The two forces are equal in magnitude.

Explanation:

It is given that,

Mass of the car, M = 1500 kg

Mass of the rock, m = 0.1 kg

It is mentioned that a car traveling along a road is hit by a rock that creates a small crack in the car’s windshield. We need to find the correct option that describes the interaction between the windshield and the rock. We know that when one object hits another object, the force applied by object 1 on object 2 have equal and opposite reaction as per the Newton's third law of motion.

So, the car exerts a force on the rock, and the rock exerts a force on the car. The two forces are equal in magnitude. Hence, the correct option is (c).

mario62 [17]2 years ago
3 0

Answer:

The interaction between the windshield and the rock is described by

c) The car exerts a force on the rock, and the rock exerts a force on the car. The two forces are equal in magnitude.

Explanation:

Forces can be described as an interaction between two objects and always come in pairs.

We have to remember Newton's third law of motion:

If an object 1 exerts a force on object 2, then object 2 should exert a force of equal magnitude but opposite direction on object 1.

So, despite that the car has 1500 kg and the rock 0.1 kg, the force the car exerts on the rock and the force the rock exerts on the car are equal in magnitude.

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The deepest point of the pacific ocean is 11,033 m, in the mariana trench. what is the gauge pressure in the water at that point
Liula [17]

Given that,

Depth of seawater, h = 11,033 m

Density seawater, p (rho) = 1025 kg/m³

Gauge Pressure , P = ??

Since, we know that:

Pressure, P = pgh

Pressure = 1025 * 9.81 * 11033

Pressure = 1109395723.3 N/m²

or

Pressure = 1.1 x 10∧8 Pascal

4 0
2 years ago
In the metric system, the appropriate unit for weight is the _____. gram newton newton/cm2 gram/cm3
Archy [21]

Answer:

Newton

Explanation:

The earth attracts every body towards its centre. The force with which the earth attracts any body towards its centre, is called its weight.

It is a vector quantity.

It always acts towards the centre of earth.

The SI unit of Newton.

4 0
2 years ago
A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice
Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of \sqrt{3} v_{e}

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , v_{i}=2 v_{e}. Here the radius is R, so find the escape velocity as follows,

            \frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0

            \frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}

            v_{e}^{2}=\frac{2 G M}{R}

            v_{e}=\sqrt{\frac{2 G M}{R}}

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as,  U_{i}=-\frac{G M m}{R}

R tend to infinity when far away from the planet, so v_{f}=0

Then, kinetic energy at initial would be,

                  k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}

Similarly, kinetic energy at final would be,

                k_{f}=\frac{1}{2} m v_{f}^{2}

Here, v_{f}=\text { final velocity }

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

                 U_{i}+k_{i}=k_{f}+v_{f}

                 \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0

                  \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}

'm' and \frac{1}{2} as common on both sides, so gets cancelled, we get as

                   4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}

We know, v_{e}=\sqrt{\frac{2 G M}{R}}, it can be wriiten as \left(v_{e}\right)^{2}=\frac{2 G M}{R}, we get

                4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}

                v_{f}^{2}=3\left(v_{e}\right)^{2}

Taking squares out, we get,

                v_{f}=\sqrt{3} v_{e}

4 0
2 years ago
You should have observed that there are some frequencies where the output is stronger than the input. Discuss how that is even p
nydimaria [60]

Answer:

w = √ 1 / CL

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

Explanation:

This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.

In these circuits the impedance is

             X = √ (R² +  (X_{C} -X_{L})² )

where Xc and XL is the capacitive and inductive impedance, respectively

            X_{C} = 1 / wC

           X_{L} = wL

From this expression we can see that for the resonance frequency

           X_{C} = X_{L}

the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

               V = IR

Since the contribution of the two other components is canceled, this occurs for

                X_{C} = X_{L}

                1 / wC = w L

                w = √ 1 / CL

6 0
2 years ago
Suppose an X-ray binary is found in which the visible star is a 12 M⦿ red giant, the orbital period is 3.65 days, and the semima
nataly862011 [7]

Answer:

If the mass of a star is greater than 3 solar masses, it will create a black hole. If its mass is less, it will create a neutron star.

Explanation:

If a star's gravity is high enough, when it condenses on itself, it will form a black hole. Otherwise, it will create a large amount of highly dense matter, such as a neutron star. It can be said that if the mass of a star is greater than 3 solar masses, it will create a black hole. If its mass is less, it will create a neutron star.

8 0
2 years ago
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