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myrzilka [38]
3 years ago
6

PLEASE HELP AND HURRY

Physics
2 answers:
Ahat [919]3 years ago
6 0

Answer:

B

Explanation:

Vikentia [17]3 years ago
6 0

Answer:

D. m/s^2

Explanation:

meters / second ^2

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What force is necessary to accelerate a 5.0 kg mass from rest to a final velocity of 10.0 m/s in 5.0 s?
vesna_86 [32]

Answer:

10 N

Explanation:

F = ma = m(Δv/t) = 5.0(10.0 - 0)/5.0 = 10 N

4 0
3 years ago
An alpha particle (a helium nucleus, consisting of two protons and two neutrons) has a radius of approximately 1.6 × 10-15 m. A
Snezhnost [94]

Answer:

9.96x10^-20 kg-m/s

Explanation:

Momentum p is the product of mass and velocity, i.e

P = mv

Alpha particles, like helium nuclei, have a net spin of zero. Due to the mechanism of their production in standard alpha radioactive decay, alpha particles generally have a kinetic energy of about 5 MeV, and a velocity in the vicinity of 5% the speed of light.

From this we calculate the speed as

v = 5% 0f 3x10^8 m/s (speed of light)

v = 1.5x10^7 m/s

The mass of an alpha particle is approximately 6.64×10−27 kg

Therefore,

P = 1.5x10^7 x 6.64×10^−27

P = 9.96x10^-20 kg-m/s

8 0
3 years ago
What happens during the process of deposition
frez [133]
Deposition is the process in which sediments, soil and rocks are added to a landform or landmass. When previous weathers surface material , is deposited to a building layer of sediment .
8 0
3 years ago
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Veseljchak [2.6K]

Answer:

The speed of q₂ is 4\sqrt{10}\ m/s

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

E_{i}=E_{f}

\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2

\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})

Put the value into the formula

\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})

0.00075(400-v_{f}^2)=0.18


400-v_{f}^2=\dfrac{0.18}{0.00075}

-v_{f}^2=240-400

v_{f}^2=160

v_{f}=4\sqrt{10}\ m/s

Hence, The speed of q₂ is 4\sqrt{10}\ m/s

7 0
3 years ago
As we move above up from one trophic level to another in an energy pyramid, what happens to the energy?
Shkiper50 [21]

As we move above up from one trophic level to another in an energy pyramid, what happens to the energy?

a. It decreases from one trophic level to another.
b. It remains the same for each trophic level.
c. It increases from one trophic level to another.

As we move above up from one trophic level to another in an energy pyramid, the energy level decreases from one trophic level to another. The answer is letter A.

3 0
3 years ago
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