Answer:

The **momentum** of the **car** is **24000 Kg•m/s**

**Momentum** is defined as the **product** of **mass and velocity.** Mathematically, it can be expressed as:

**Momentum = mass × velocity**

With the **above formula,** we can obtain the **momentum** of the **car** as follow:

- Mass = 1200 Kg

- Velocity = 20 m/s

**Momentum =?**

Momentum = mass × velocity

Momentum = 1200 × 20

**Momentum of car = 24000 Kg•m/s**

Learn more about **momentum**:

Answer:

**Answer:**

24000 kg·m/s

**Explanation:**

Momentum is Mass x Velocity, so 1200 kg time 20 m/s = 24000 kg-ms/s

Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was ?s = 1. Around 1962, three companies independently developed racing tires with coefficients of 1.6. This problem shows that tires have improved further since then. The shortest time interval in which a piston-engine car initially at rest has covered a distance of one-quarter mile is about 4.43 s. (A) Assume the car's rear wheels lift the front wheels off the pavement as shown in the figure above. What minimum value of ?s is necessary to achieve the record time?

A circular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing downward, and is released from rest. What is the direction of the induced current in the coil, as viewed from above, as the magnet approaches the coil in free fall?a. clockwiseb. counterclockwisec. There is no induced current in the coil.

Two atoms collide while moving in a dilute gas. The larger atom has a mass M1 = 6 Daltons and a speed v1 = 200 m/s, while the smaller has a mass M2 = 1 Daltons. During the collision both atoms simply bounce off each other. They do not change their speeds, but after the collision they each change their directions, bouncing in the indicated directions. (You may express your results using the mass unit "Daltons". 1 Dalton is approximately equal to the mass of a proton or neutron and is defined as one-twelfth the mass of a single neutral carbon-12 atom in its ground state.)A. What is the magnitude of the change in the momentum, Δp1, of mass M1? B. What is the change in the total momentum of the pair? C. What is the magnitude of the change in the momentum Δp2, of mass M2?

It's a little hard can u help

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50 cm and holds it in position with a force of 150N . If the mass of the arrow is 50g and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?

A circular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing downward, and is released from rest. What is the direction of the induced current in the coil, as viewed from above, as the magnet approaches the coil in free fall?a. clockwiseb. counterclockwisec. There is no induced current in the coil.

Two atoms collide while moving in a dilute gas. The larger atom has a mass M1 = 6 Daltons and a speed v1 = 200 m/s, while the smaller has a mass M2 = 1 Daltons. During the collision both atoms simply bounce off each other. They do not change their speeds, but after the collision they each change their directions, bouncing in the indicated directions. (You may express your results using the mass unit "Daltons". 1 Dalton is approximately equal to the mass of a proton or neutron and is defined as one-twelfth the mass of a single neutral carbon-12 atom in its ground state.)A. What is the magnitude of the change in the momentum, Δp1, of mass M1? B. What is the change in the total momentum of the pair? C. What is the magnitude of the change in the momentum Δp2, of mass M2?

It's a little hard can u help

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50 cm and holds it in position with a force of 150N . If the mass of the arrow is 50g and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?

The **range** of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.

To solve this problem, we need to make use of the concept of** projectile motion** in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.

First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.

Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.

So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.

#SPJ3

**Answer:**

4.875 V

**Explanation:**

N = 1300

diameter = 2.10 cm

radius = half of diameter = 1.05 cm

B1 = 0.130 T

B2 = 0 T

t = 12 ms

According to the law of electromagnetic induction,

Where, Ф be the magnetic flux linked with the coil

e = 4.875 V

Answer:

The gel that is applied before ultrasonic imaging is a **conducting material.** It acts as a medium between transducer and skin. **The ultrasonic waves easily transmit from the probe to the tissues because of gel.** A tight bond is created between the probe and skin layer and the gel acts as a **coupling agent. The density of the gel is similar to the skin layer. This reduces the attenuation of the waves. A thin layer of gel is applied which fills the air gaps and helps in transmission of waves to the tissues. **Hence, the technician apply ultrasound gel to the patient before beginning the examination

The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection.

**Answer:**

≅3.2 nm

**Explanation:**

Using the converter units as know for this case that:

1 ml is 1 cubic centimeter ⇒ 0.1 ml is 0.1 cubic centimeters

32.0 m² so :

32.0 m² *100 *100 cm² ⇒ 0.1 / ( 32.0 * 100 *100 ) = 100,000,000 * 0.1 / (32.0 * 100 * 100 ) nm

v = 100/32.0 nm = 3.125 nm thick.

v ≅3.2 nm

As oil is one molecule thick and the molecules are cubic, length of each oil is 3.2 nm

Part 3) What if the felon then sped up to 30 m/s and all other conditions remained the same?

**1) 621.8 Hz**

**2) 719.3 Hz**

**3) 700 Hz**

**Explanation:**

1)

The Doppler effect occurs when there is a source of a wave in relative motion with respect to an observer.

When this happens, the frequency of the wave appears shifted to the observer, according to the equation:

where

f is the real frequency of the sound

f' is the apparent frequency of the sound

v is the speed of the sound wave

is the velocity of the observer, which is negative if the observer is moving away from the source, positive if the observer is moving towards the source

is the velocity of the source, which is negative if the source is moving towards the observer, positive if the source is moving away

In this problem we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

is the velocity of the car with the siren

is the velocity of the felon (he's moving away from the siren)

So, the frequency heard by the felon is

2)

In this case, the cop does a U-turn and speeds towards the felon at 30 m/s.

This means that now the siren is moving towards the observer (so, becomes positive), while the sign of still remains positive.

So we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

is the velocity of the car with the siren

is the velocity of the felon

So, the frequency heard by the felon is

3)

In this case, the felon speeds up to 30 m/s.

This means that now the felon and the siren are moving with the same relative velocity: so, it's like they are not moving relative to each other, so the frequency will not change.

In fact we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

is the velocity of the car with the siren

is the velocity of the felon

So, the frequency heard by the felon is

So, the frequency will not change.

The last one is the closest

48,800 mi/h2 is the right answer