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stealth61 [152]
3 years ago
13

The arctic tundra can be found in _______. a. Greenland b. Canada c. The United States d. all of the above

Physics
2 answers:
dangina [55]3 years ago
7 0
The artic tundra can be found in D. All of the above.
Juli2301 [7.4K]3 years ago
3 0

Answer: b. Canada

Artic tundra is found across the northern Alaska, Canada and Siberia. These biomes are coolest biomes in the world. The artic tundra exhibit long cool winters and short cool summers. The artic tundra receives low rainfall and dry winds. The climatic conditions are responsible for desert like features of these biomes. Simple vegetations and low biodiversity are features of the artic tundra.

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A particle with a mass of 6.64 × 10–27 kg and a charge of +3.20 × 10–19 C is accelerated from rest through a potential differenc
blondinia [14]

Answer:

Explanation:

Given that,

Mass m = 6.64×10^-27kg

Charge q = 3.2×10^-19C

Potential difference V =2.45×10^6V

Magnetic field B =1.6T

The force in a magnetic field is given as Force = q•(V×B)

Since V and B are perpendicular i.e 90°

Force =q•V•BSin90

F=q•V•B

So we need to find the velocity

Then, K•E is equal to work done by charge I.e K•E=U

K•E =½mV²

K•E =½ ×6.64×10^-27 V²

K•E = 3.32×10^-27 V²

U = q•V

U = 3.2×10^-19 × 2.45×10^6

U =7.84×10^-13

Then, K•E = U

3.32×10^-27V² = 7.84×10^-13

V² = 7.84×10^-13 / 3.32×10^-27

V² = 2.36×10^14

V=√2.36×10^14

V = 1.537×10^7 m/s

So, applying this to force in magnetic field

F=q•V•B

F= 3.2×10^-19 × 1.537×10^7 ×1.6

F = 7.87×10^-12 N

6 0
3 years ago
Read 2 more answers
What are non-contact forces please include example in your answer
Alina [70]

Answer:

Gravitational force.  Magnetic force.  Electrostatics.  Nuclear force.

Explanation:

Apple falling from a tree

raindrops falling from the sky

4 0
3 years ago
Read 2 more answers
Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the
castortr0y [4]

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}.

Now, making the substitutions

d \ = \ 2.00 * 10 ^{-15} \ m,

q_1 = q_2 = 1.60 * 10 ^ {-19} \ C,

\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2},

we can find:

F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}.

F \ = 57.536 N.

Not so big for everyday life, but enormous for subatomic particles.

4 0
3 years ago
Read 2 more answers
Provide an example of when momentum is conserved and explain your answer you can get 10 PTS if answered with a good explaination
dezoksy [38]

Answer:

m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s

Explanation:

<u>Conservation of Momentum </u>

The total momentum of a system of two particles is

p=m_1v_1+m_2v_2

Where m1,m2,v1, and v2 are the respective masses and velocities of the particles at a given time. Then, the two particles collide and change their velocities to v1' and v2'. The final momentum is now

p'=m_1v_1'+m_2v_2'

The momentum is conserved if no external forces are acting on the system, thus

m_1v_1+m_2v_2=m_1v_1'+m_2v_2'

Let's put some numbers in the problem and say

m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s

(8)(12)+(6)(4)=(8)(-6)+(6)(28)

96+24=-48+168

120=120

It means that when the particles collide, the first mass returns at 6 m/s and the second continues in the same direction at 28 m/s

4 0
3 years ago
A 30-mm-diameter copper rod is 1 m long with a yield strength of 70 MPa. Determine the axial force necessary to cause the diamet
ivolga24 [154]

Explanation:

Given data:

d = 30 mm = 0.03 m

L = 1m

S_{y} = 70 Mpa

Δd = -0.0001d

Axial force = ?

validity of elastic deformation assumption.

Solution:

O'₂ = Δd/d = (-0.0001d)/d = -0.0001

For copper,

v = 0.326      E = 119×10³ Mpa

O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶

∵δ = F.L/E.A    and σ = F/A so,

σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa

F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN

S_{y} = 70 MPa > σ = 36.5 MPa

∵ elastic deformation assumption is valid.

so the answer is

F = 25800 K N            and     S_{y} > σ

3 0
3 years ago
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