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katrin [286]
3 years ago
10

What is the most abundant type of sediment found on the ocean floor?

Physics
1 answer:
makvit [3.9K]3 years ago
4 0
Its D that  so easy look at it
You might be interested in
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
4 years ago
On Planet Y, which has no air, a dropped object falls 9 m in 3 seconds. What is g, the acceleration due to gravity, on that plan
slavikrds [6]

Answer:

a = 2 m/s^2

which agrees with the third answer option provided.

Explanation:

Recall the kinematic formula for displacement under the action of a constant acceleration "a":

yf - yi = 1/2 a  t^2

using the information provided this equation becomes:

9 = 1/2 a (3)^2

solve for a:

9 * 2 / 9 = a

then a = 2 m/s^2

which agrees with the third answer option provided.

7 0
3 years ago
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
4 years ago
A meter stick is supported by a pivot at its center of mass. Assume that the meter stick is uniform and that the center of mass
iragen [17]

Answer:

64.5 cm

Explanation:

30 * 80 + x * 110 = 50* 190 => x = 64.5

5 0
3 years ago
A 3.00 x 10^2-W electric immersion heater is
andre [41]

Answer

t = 367.77 s = 6.13 min

Explanation:

According to the law of conservation of energy:

Heat\ Supplied\ By \ Heater = Heat\ Absorbed\ by\ Glass + Heat\ Absorbed\ by\ Water\\Pt = m_gC_g\Delta T_g + m_wC_w\Delta T_w\\

where,

P = Electric Power of Heater =  300 W

t = time required = ?

m_g = mass of glass = 300 g = 0.3 kg

m_w = mass of water = 250 g = 0.25 kg

C_g = speicific heat of glass = 840 J/kg.°C

C_w =  specific heatof water = 4184 J/kg.°C

ΔT_g = ΔT_w = Change in Temperature of Glass and water = 100°C - 15°C

ΔT_g = ΔT_w = 85°C

Therefore,

(300\ W)(t) = (0.3\ kg)(840\ J/kg.^oC)(85^oC)+(0.25\ kg)(4184\ J/kg.^oC)(85^oC)\\

<u>t = 367.77 s = 6.13 min</u>

8 0
3 years ago
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