In the velocity and acceleration are in the same direction the speed of the object decreases

Assuming a successful launch, what is the speed of the small CO2 rocket given that it travels about 10 meters in about 3 seconds

antiseptic1488 [7]

Probably 30 im not sure

4 0

3 years ago

Which statement is true according to Hooke's law?

Mkey [24]

Hooke's law:

F = -kx

F is the spring force, k is the spring constant, and x is the displacement of the spring. The negative sign is there to indicate that the force acts in the opposite direction of the displacement (e.g. if you stretch a spring towards the right, you will feel a force pulling towards the left).

Looking at Hooke's law, the magnitude of the force is directly proportional to its displacement; that is to say, if the displacement increases, so does the magnitude of the spring force, and if the magnitude of the force increases, this necessitates an increase in the displacement.

1. As the force on a spring increases, the distance it stretches decreases.

An increase in force necessitates an increase in displacement. This statement is false.

2. As the force on a spring increases, the distance it stretches increases.

An increase in force necessitates an increase in displacement. This statement is true.

3. As the force on a spring increases, the spring's length remains unchanged.

An increase in force necessitates an increase in displacement. This statement is false.

4. The distance a spring stretches has nothing to do with how much force is applied to it.

Hooke's law shows that the spring force and its displacement are very much dependent on each other. This statement is false.

Only statement 2 is true. As the force on a spring increases, the distance it stretches increases.

3 0

3 years ago

deon is running at a velocity of 7.9 meters per second. deon has a total mass of 95 kilograms including his helmet, uniform, pad

brilliants [131]

Answer:

712.5 kg m/s

Explanation:

Work out the total momentum before the event (before the collision):

p = m × v

Massof Deon = 95kg

Velocity =7.5m/s

Mass of Chuck = 120kg

Velocity = 0m/s

Momentum of Deon before = 95 × 7.5 = 712.5 kg m/s

Momentum of Chuck before = 120 × 0 = 0 kg m/s

Total momentum before = 712.5+ 0 = 712.5 kg m/s

Working out the total momentum after the event (after the collision):

Because momentum is conserved, total momentum afterwards = 712.5 kg m/s

6 0

4 years ago

When blueshift occurs,the preceived frequency of the wave would be?

LiRa [457]

Answer:

When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.

Explanation:

As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.

Blueshifts happens when the source of the wave and the observer are moving closer towards one another.

Assume that the wave is of frequency $f\; {\rm Hz}$ at the source. In other words, the source of the wave sends out a peak after every $(1/f)\; {\text{seconds}}$ .

Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.

The source of this wave sends out a peak after each period of $(1/f)\; {\text{seconds}}$ . It would appear to the observer that consecutive peaks arrive every $(1/f)\; {\text{seconds}}\!$ . That would correspond to a frequency of $f\; {\rm Hz}$ .

On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of $v \; {\rm m \cdot s^{-1}}$ .

Again, the source of this wave would send out a peak after each period of $(1/f)\; {\text{seconds}}$ . However, by the time the source sends out the second peak, the source would have been $v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}$ closer to the observer then when the source sent out the first peak.

When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than $(1/f)\; {\text{seconds}}$ . The observed frequency of this wave would be larger than the original $f\; {\rm Hz}$ .

6 0

3 years ago

A sealed pressure cooker allows steam to escape only through an opening in the lid. a separate metal object, called a petcock, s

daser333 [38]

The mass of the petcock of the pressure cooker is 0.257 kg.

The total pressure inside the cooker is= atmospheric pressure+Gauge pressure

=100 kPa+101 kPa=201*10^3 Pa

The area of the petcock is= (π/4)d^2=0.785*0.004^2=12.56*10^(-6) m^2

Now the mass of the petcock is calculated as

P=(F/A)=(mg/A)

201*10^3=(m*9.81/(12.56*10^(-6)

m=0.257 kg

Therefore the mass of the petcock is 0.257 kg

5 0

4 years ago