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alukav5142 [94]
3 years ago
7

Which type of ooze dominates in warm shallow ocean floors?A. CalcareousB. Silieous

Physics
1 answer:
gogolik [260]3 years ago
6 0
Actually it's sediment
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What is cell in electricity?
agasfer [191]

Answer:

An electrical cell is a device used to generate electricity, or to make chemical reactions by applying electricity.

7 0
4 years ago
What is the transfer of energy by collisions between the atoms in the material
Anettt [7]
Conduction is the transfer of energy by collusion between the atoms and molecules in a material.
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3 years ago
Read 2 more answers
Recall that the blocks can only move along the x axis. the x components of their velocities at a certain moment are v1x and v2x.
Contact [7]
The center of mass is given with this formula:
x_c=\frac{\sum_{n=1}^{n=i}m_ix_i}{M}
Velocity is:
v=\frac{dv}{dt}
So, for the velocity of the center of mass we have:
\frac{dx_c}{dt}=\frac{\sum_{n=1}^{n=i}d(m_ix_i)}{Mdt}\\
v_c=\frac{\sum_{n=1}^{n=i}p_i}{M}\\
In our case it is:
v_{xc}=\frac{m_1v_{x1}+m_2v_{x2}}{m_1+m_2}
 
5 0
3 years ago
The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

5 0
3 years ago
A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The ca
sergeinik [125]

Answer:

t = 8.45 sec

car distance d = 132.09  m

bike distance d = 157.08 m

Explanation:

GIVEN :

motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t

for car

by equation of motion

d  = ut + \frac{1}{2}at^2

u = 0 starting from rest

d = \frac{1}{2}at^2

t^2 = \frac{2d}{a}

for bike

d+25 = 0 + \frac{1}{2}*4.40t^2

t^2= \frac{d+25}{2.20}

equating time of both

\frac{2d}{a} = \frac{d+25}{2.20}

solving for d we get

d = 132 m

therefore t is= \sqrt{\frac{2d}{a}}

t =  \sqrt{\frac{2*132}{3.70}}

t = 8.45 sec

each travelled in time 8.45 sec as

for car

d = \frac{1}{2}*3.70 *8.45^2

d = 132.09  m

fro bike

d = \frac{1}{2}*4.40 *8.45^2

d = 157.08 m

7 0
3 years ago
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