Ask question

Ask question

All categories

3 years ago

6

Which of the following is right?

1 answer:

barxatty [35]3 years ago

7 0

Scientific theory because it's a theory it's already an answer but it might change depending on the condition.

You might be interested in

A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of

Sunny_sXe [5.5K]

Answer:

$5.8\; {\rm kg\cdot m \cdot s^{-1}}$ .

Explanation:

If the mass of an object is $m$ and the velocity of that object is $v$ , the linear momentum of that object would be $m\, v$ .

Assume that the initial velocity of the mass is positive ( $6.0\; {\rm m\cdot s^{-1}}$ .) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be $(-4.0\; {\rm m\cdot s^{-1}})$ .

Thus, the change in the velocity of the mass would be:

$\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}$ .

The change in the linear momentum of the mass would be:

$\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\ =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}$ .

Thus, the magnitude of the change of the linear momentum would be $5.8\; {\rm kg \cdot m \cdot s^{-1}}$ .

7 0

2 years ago

Heyyy! Okay riddle of the day

vekshin1

Answer:

Time to get a new fence...?

C'mon that's pretty cringe lol

Please give brainliest

have a nice day :D

3 0

3 years ago

Jeremy walked at an average rate of speed of 5 km/h. how far had Jeremy walked in 15 minutes?

astraxan [27]

Answer:

1 half kilometer is right answer

5 0

3 years ago

Read 2 more answers

Some enterprising physics students working on a catapult decide to have a water balloon fight in the school hallway. The ceiling

sergejj [24]

Answer:

$\alpha =54.7$ º

Explanation:

From the exercise we have our initial information

$y=3.4m\\v_{o}=10m/s\\g=-9.8m/s^2$

When the balloon gets to the ceiling its velocity at that moment is 0 m/s. Being said that we can calculate velocity at the vertical direction

$v_{y}^2=v_{oy}^2+ag(y-y_{o})$

Since $v_{y}=0$ and $y_{o}=0$

$0=v_{oy}^2-2(9.8m/s^2)(3.4m)$

$v_{oy}=\sqrt{2(9.8m/s^2)(3.4m)}=8.16m/s$

Knowing that

$v_{oy}=v_{o}sin\alpha$

$sin\alpha =\frac{v_{oy} }{v_{o} }$

$\alpha =sin^{-1}(\frac{v_{oy}}{v_{o}})=sin^{-1}(\frac{8.16m/s}{10m/s})=54.7$ º

8 0

3 years ago

A string fixed at both ends is 8.40 m long and has a mass of 0.120 kg. It is subjected to a tension of 96.0 N and set oscillatin

Luden [163]

Answer:

81.9756 m/s

16.8 m

4.8795 Hz

Explanation:

m = Mass of string = 0.12 kg

L = Length of string = 8.4 m

T = Tension on string = 96 N

Linear density is given by

$\mu=\dfrac{m}{L}\\\Rightarrow \mu=\dfrac{0.12}{8.4}$

Spee of the wave is given by

$v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow v=\sqrt{\dfrac{96}{\dfrac{0.12}{8.4}}}\\\Rightarrow v=81.9756\ m/s$

The speed of the waves on the string is 81.9756 m/s

Wavelength is given by

$\lambda=2L\\\Rightarrow \lambda=2\times 8.4\\\Rightarrow \lambda=16.8\ m$

The longest possible wavelength is 16.8 m

Frequency is given by

$f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{81.9756}{16.8}\\\Rightarrow f=4.8795\ Hz$

The frequency of the wave is 4.8795 Hz

3 0

3 years ago