A water wave has a frequency of 2 hertz and a wavelength fo 5 meters. Calculate is speed

Find the impulse a 1500 kg car experiences slowing down from 40m/s to 10m/s?

Kitty [74]

Answer: - 45000 N.s

Explanation: Impulse is equal to the change in momentum

J = Δp

To solve for impulse we calculate the change in momentum

Δp = m ( Δv)

= 1500 kg ( 10 m/s - 40 m/s)

= - 45000 N.s

6 0

3 years ago

Is space expanding within clusters of galaxies? why or why not?

GarryVolchara [31]

No, since their gravity is powerful enough to keep them together even while the universe expands as a whole. Space is not expanding within clusters of galaxies.

<h3>What is a galaxy?</h3>

A galaxy is a massive clump of gas, dust, and billions of stars and their solar systems bound together by gravity.

No, since their gravity is powerful enough to keep them together even while the universe expands as an entire.

Hence,space is not expanding within clusters of galaxies.

To learn more about the galaxy, refer to the link;

brainly.com/question/2905713

#SPJ1

6 0

2 years ago

The fastest time ever run at Pikes Peak is 7:57. That’s way too long. If we removed all the curves, making a straight line from

anygoal [31]

Answer: 271.4 s

Explanation:

We are told the top speed (maximum speed) $V_{max}$ the car has is:

$V_{max}=203 mph=90.74 m/s$ taking into account $1 mile=1609.34 m$

And the car's average acceleration $a_{ave}$ is:

$a_{ave}=0.091 g=2.93 ft/s^{2}=0.89 m/s^{2}$

Since:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}$ (1)

Where:

$V_{f}=V_{max}=90.74 m/s$ is the car's final speed (top speed)

$V_{o}=0 m/s$ because it starts from rest

$\Delta t$ is the time it takes to reach the top speed

Finding this time:

$\Delta t=\frac{V_{f}-V_{o}}{a_{ave}}$ (2)

$\Delta t=\frac{90.74 m/s - 0 m/s}{0.89 m/s^{2}}$ (3)

$\Delta t=t_{1}=101.95 s$ (4)

Now we have to find the distance $d$ the car traveled at this maximum speed with the following equation:

$V_{f}^{2}=V_{o}^{2} + 2a_{ave} d$ (5)

Isolating $d$ :

$d=\frac{V_{f}^{2}}{2a_{ave}}$ (6)

$d=\frac{(90.74 m/s)^{2}}{2(0.89 m/s^{2})}$ (7)

$d=4625.70 m$ (8)

On the other hand, we know the total distance $D$ traveled by the car is:

$D=12.42 miles = 19988.052 m$

Hence the remaining distance is:

$d_{remain}=D-d=19988.052 m - 4625.70 m$ (9)

$d_{remain}=15362.35 m$ (10)

So, we can calculate the time $t_{2}$ it took to this car to travel this remaining distance $d_{remain}$ at its top speed $V_{max}$ , with the following equation:

$V_{max}=\frac{d_{remain}}{t_{2}}$ (11)

Isolating $t_{2}$ :

$t_{2}=\frac{d_{remain}}{V_{max}}$ (12)

$t_{2}=\frac{15362.35 m}{90.74 m/s}$ (13)

$t_{2}=169.45 s$ (14)

With this time $t_{2}$ and the value of $t_{1}$ calculated in (4) we can finally calculate the total time $t_{TOTAL}$ :

$t_{TOTAL}=t_{1}+ t_{2}$ (15)

$t_{TOTAL}=101.95 s + 169.45 s$ (16)

$t_{TOTAL}=271.4 s s$

5 0

4 years ago

You've always wondered about the acceleration of the elevators in the 101-story-tall Empire State Building. One day, while visit

kondor19780726 [428]

Answer:

(i) The maximum acceleration upward is 2.02 m/s².

(ii) The maximum acceleration downward is 1.39 m/s².

Explanation:

Let a be the maximum acceleration of the elevator.

The mass of the person at ground is 150 lb. We have to convert the mass into kg,

1 lb = 0.453592 kg

150 lb = 68 kg

170 lb = 77 kg

120 lb = 54 kg

(i) The person experience a force due to Earth's gravity in the downwards direction. The magnitude of this force is:

F₁ = mg = 68 x 9.8 = 666.4 N

The weight of the person decreases as the elevator is moving upwards. So, the force experienced by the person in this case due to gravity is:

F₂ = 54 x 9.8 = 529.2 N

Applying Newton's force equation;

(F₁ - F₂) = ma

(666.4 - 529.2) = 68 x a

a = 2.02 m/s²

(ii) The weight of the person increases as the elevator is moving downwards. So, the force experienced by the person in this case due to gravity is:

F₂ = 77 x 9.8 = 754.6 N

Applying Newton's force equation;

(F₁ - F₂) = ma

(666.4 - 754.6) = 68 x a

a = -1.30 m/s²

8 0

4 years ago

9. Kokio dydžo Archimedo jėga veikia 0,5 m² tūrio medinį rastą vandenyje?

ICE Princess25 [194]

Answer:

i dont knowwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

6 0

3 years ago