Which phrase describes an irregular galaxy?

r-ruslan [8.4K]

Momentum is the quantity of moving motion if that helps in anyway :)

8 0

3 years ago

An 1120 kg car traveling at 17 m/s is brought to a stop while skidding 40 m. Calculate the work done on the car by friction forc

sergij07 [2.7K]

Answer:

Approximately $(-1.6 \times 10^{5}\; \rm J)$ (assuming that the car was on level ground.)

Explanation:

When an object of mass $m$ is moving at a speed of $v$ , the kinetic energy of that object would be $(1/2) \, m \cdot v^{2}$ .

Initial kinetic energy of the car:

$\begin{aligned}&\frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2}\times 1120\; \rm kg \times (17\; \rm m \cdot s^{-1})^{2} \\ &\approx 1.6\times 10^{5}\; \rm J \end{aligned}$ .

After the car comes to a stop, the kinetic energy of this car would be $0\; \rm J$ because the car would not be moving.

Change to the kinetic energy of the car: $\text{Final KE} - \text{Initial KE} = -1.6 \times 10^{5}\; \rm J$ .

If the car is traveling on level ground, friction would be the only force that contributed to this energy change. Hence:

$\text{Work of friction} = \text{Energy change} = -1.6\times 10^{5}\; \rm J$ .

The value of the work that friction did is negative. The reason is that at any instant before the car comes to a stop, friction would be exactly opposite to the direction of the movement of the car.

The work of a force on an object is the dot product of that force and the displacement of that object. The dot product of two vectors of opposite directions is negative. Hence, in this question, the work that friction did on the car would be negative because the friction vector would be opposite to the movement of the car.

6 0

3 years ago

In a collision experiment, the ratio of the velocity change between two carts of equal inertia is found to equal 1.

VashaNatasha [74]

Answer:

A) Therefore if I double the masses with are in the two terrine they are simplified and the radii of the speeds remain the same

B) If the masses are maintained and the speeds are doubled, the radius of the two speeds remains the same

Explanation:

A vehicle crash problem must be solved with the equation of the moment,

Initial instant Before crash

p₀ = m v₁ + mv₂

After the crash

$p_{f}$ = m $v_{1f}$ + m $v_{2f}$

p₀ = $p_{f}$

If the speed ratio before and after the crash is one

p₀ / $p_{f}$ = 1

We can assume that initially one of the cars was stopped

m v₁₀ = m $v_{2f}$

v₁₀ = $v_{2f}$

For the two speeds to be equal, the masses of the vehicles must be the same.

A) Therefore if I double the masses with are in the two terrine they are simplified and the radii of the speeds remain the same

B) If the masses are maintained and the speeds are doubled, the radius of the two speeds remains the same

5 0

3 years ago

Defind wave.then explain how vibrations,waves,and energy are related to sound

vagabundo [1.1K]

Waves are either transversal stereotypical pictures of waves or longitudinal it goes in rings waves are just different forms of vibration of any kind of energy

7 0

3 years ago

Read 2 more answers

Twoâ€‹ high-speed ferries leave at the same time from a city to go to the same island. The firstâ€‹ ferry, theâ€‹ Cat, travels a

kumpel [21]

Answer:

1 hour

Explanation:

Since the definition of velocity is $v=x/t$ , we can calculate the position at time t of an object moving at speed v with $x=vt$ .

The position of the first ferry is $x_1=v_1t$

The position of the second ferry is $x_2=v_2t$

We want to know when they will be 6 miles apart. This means, if we call that distance d, that we want to know when the difference between their positions will be d, or $x_1-x_2=d$ (we know that at the beginning the position of the ferry 1 is of higher value than that of ferry 2 since it left before).

We use our previous formulas then:

$d=x_1-x_2=v_1t-v_2t=(v_1-v_2)t$

Since we want the time, we do:

$t=\frac{d}{v_1-v_2}$

And substitute our values:

$t=\frac{6\ miles}{30\ miles/hour-24\ miles/hour}=\frac{6\ miles}{6\ miles/hour} =1\ hour$

7 0

3 years ago