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Natalija [7]
3 years ago
9

Please help me!

Physics
2 answers:
mihalych1998 [28]3 years ago
7 0
The answer would be mass movement 
Kay [80]3 years ago
5 0
Runoff because the mud is a liquid and moves on an amount of water that is in it like with quicksand
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Need thy help once again
Thepotemich [5.8K]
B cause I took the quiz
8 0
3 years ago
Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist
andreyandreev [35.5K]

Answer:

a) It is moving at 43.15\frac{m}{s^{2}} when reaches the ground.

b) It is moving at 101.44\frac{m}{s^{2}} when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

W=K_f-K_i (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

K=\frac{mv^2}{2} (2)

with m the mass and v the velocity.

Using (2) on (1):

W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2} (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

W=Fd=wd=mgd (4)

Using (4) on (3):

mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2} (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}

v_f=43.15\frac{m}{s^{2}}

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

-mgd=-\frac{mv_i^2}{2}

Solving for initial velocity (when the boulder left the volcano):

v_i=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*525}

v_i=101.44 \frac{m}{s^{2}}

3 0
3 years ago
A pitcher delivers a fast ball with a velocity of 43 m/s to the south. the batter hits the ball and gives it a velocity of 51 m/
hoa [83]
Assuming north as positive direction, the initial and final velocities of the ball are:
v_i=-43 m/s (with negative sign since it is due south)
v_f=+51 m/s
the time taken is t=1.0 ms=0.001 s, so the average acceleration of the ball is given by
a= \frac{v_f-v_i}{t}= \frac{51 m/s-(-43 m/s)}{0.001 s}=9.4 \cdot 10^4 m/s^2
And the positive sign tells us the direction of the acceleration is north.
4 0
3 years ago
I NEED HELP ASAP WILL MARK BRAINLIEST:
Evgen [1.6K]

Answer:

It is b for sure.

Explanation:

because they are examining urine now that's pure science.

7 0
3 years ago
Sound level B in decibels is defined as
Lelechka [254]

Answer:

The approximate combined sound  intensity is I_{T}=1.1\times10^{-4}W/m^{2}

Explanation:

The decibel  scale intensity for busy traffic is 80 dB. so intensity will be

10log(\frac{I_{1}}{I_{0}} )=80, therefore I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}

In the same way for the loud conversation having a decibel intensity of 70 dB.

10log(\frac{I_{2}}{I_{0}} )=70, therefore I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}

Finally we add both of them I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}, is the approximate combined sound  intensity.

3 0
3 years ago
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