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3 years ago

A plane moves at 15 km/h for a total distance of 300km. How long did it fly for? *

Physics

1 answer:

strojnjashka [21]3 years ago

5 0

20 hours you do 300 divided by 15 and get 20 hours

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A car with 2 × 10^3 kg moving at a speed of 10 m/s collides and sticks with car B of mass of 3 × 10^3 kg initially at rest. How

stepan [7]

Answer:

$6 \times 10^4 \; \rm J$ .

Explanation:

KE lost = Total KE before Collision - Total KE after Collision.

For each car, the KE before collision can simply be found with the equation:

$\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2$ , where

$m$ is the mass of the car, and
$v$ is the speed of the car.

The $2 \times 10^3\; \rm kg$ car would have an initial KE of:

$\displaystyle \frac{1}{2} \times 2 \times 10^3 \times 10^2 = 10^5\; \rm J$ .

The $3 \times 10^3\; \rm kg$ car was initially not moving. Hence, its speed and kinetic energy would zero before the collision.

To find the velocity of the two cars after the collision, apply the conservation of momentum.

The momentum $p$ of an object is equal to its mass $m$ times its velocity $v$ . In other words, $p = m\cdot v$ .

Let the mass of the two cars be denoted as $m_1$ and $m_2$ , and their initial speeds $v_1$ and $v_2$ . Since the two cars are stuck to each other after the collision, their final speeds would be the same. Let that speed be denotes as $v_3$ .

Initial momentum of the two-car system:

$\begin{aligned}& m_1 \cdot v_1 + m_2 \cdot v_2 \\ &= 2 \times 10^3 \times 10 + 3 \times 10^3 \times 0 \\ &= 2 \times 10^4\; \rm kg \cdot m \cdot s^{-1}\end{aligned}$ .

After the collision, both car would have a velocity of $v_3$ (since they were stuck to each other.) As a result, the final momentum of the two-car system would be:

$m_1\cdot v_3 + m_2 \cdot v_3 = (m_1 + m_2)\, v_3$ .

Since momentum is conserved during the collision, the momentum of the system after the collision would also be $2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}$ . That is: $(m_1 + m_2) \, v_3 = 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}$ .

Solve for $v_3$ :

$\begin{aligned} v_3 &= \frac{(m_1 + m_2)\, v_3}{m_1 + m_2} \\ &= \frac{2 \times 10^4}{2 \times 10^3 + 3 \times 10^3} \\ &= \frac{2 \times 10^4}{5 \times 10^3} \\ &= 4 \; \rm m \cdot s^{-1}\end{aligned}$ .

Hence, the total kinetic energy after the collision would be:

$\begin{aligned} &\frac{1}{2}\, m_1 \, v^2 + \frac{1}{2}\, m_2\, v^2 \\ &= \frac{1}{2}\, (m_1 + m_2)\, v^2 \\ &= \frac{1}{2} \times \left(2 \times 10^3 + 3 \times 10^3\right) \times 4^2 \\ &= 4 \times 10^4\; \rm J\end{aligned}$ .

The amount of kinetic energy lost during the collision would be:

$\begin{aligned}&\text{KE After Collision} - \text{KE Before Collision} \\ &= 10^5 - 4 \times 10^4 \\&= 6\times 10^4\; \rm J \end{aligned}$ .

5 0

3 years ago

Jasper made a list of the properties of electromagnetic waves. Identify the mistake in the list. Electromagnetic Wave Properties

Nata [24]

Answer:

Statement 2 is wrong

Explanation:

To check the statements in this exercise, let's describe the main properties of electromagnetic waves. Let's describe the characteristics

* they are transverse waves

* formed by the oscillations of the electric and magnetic fields

* the speed of the wave is the speed of light

with these concepts let's review the final statements

1) True. Formed by the oscillation of the two fields

2) False. They are transverse waves

3) True. Can travel by vacuum as they are supported by oscillations of the electric and magnetic fields

4) True. They all have the same speed of light

Statement 2 is wrong

6 0

4 years ago

A rifle of mass M is initially at rest, but is free to recoil. It fires a bullet of mass m with a velocity +v relative to the gr

Natali5045456 [20]

Answer:

$V=-\dfrac{mv}{M}$

Explanation:

Given that

Mass of rifle = M

Initial velocity ,u= 0

Mass of bullet = m

velocity of bullet = v

Lets take final speed of the rifle is V

There is no any external force ,that is why linear momentum of the system will be conserve.

Initial linear momentum = Final linear momentum

M x 0 + m x 0 = M x V + m v

0 = M x V + m v

$V=-\dfrac{mv}{M}$

Negative sign indicates that ,the recoil velocity will be opposite to the direction of bullet velocity.

6 0

3 years ago

Given all the positive characteristics of Venus, why has space exploration focused so much on Mars?

Zielflug [23.3K]

The reason space exploration focused more on Mars than Venus with all the advantages, Venus still has a huge disadvantage, which is that the super hot is there and we can actually land on Venus without exploding.

6 0

4 years ago

Read 2 more answers

A semicircular plate with radius 6 m is submerged vertically in water so that the top is 2 m above the surface. Express the hydr

dalvyx [7]

Answer: 313920

Explanation:First, we’re going to assume that the top of the circular plate surface is 2 meters under the water. Next, we will set up the axis system so that the origin of the axis system is at the center of the plate.

Finally, we will again split up the plate into n horizontal strips each of width Δy and we’ll choose a point y∗ from each strip. Attached to this is a sketch of the set up.

The water’s surface is shown at the top of the sketch. Below the water’s surface is the circular plate and a standard xy-axis system is superimposed on the circle with the center of the circle at the origin of the axis system. It is shown that the distance from the water’s surface and the top of the plate is 6 meters and the distance from the water’s surface to the x-axis (and hence the center of the plate) is 8 meters.

The depth below the water surface of each strip is,

di = 8 − yi

and that in turn gives us the pressure on the strip,

Pi =ρgdi = 9810 (8−yi)

The area of each strip is,

Ai = 2√4− (yi) 2Δy

The hydrostatic force on each strip is,

Fi = Pi Ai=9810 (8−yi) (2) √4−(yi)² Δy

The total force on the plate is found on the attached image.

5 0

3 years ago