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2 years ago

What can cause an increase in density that results in a deep ocean current

2 answers:

6 0

The cause of the increase in density in deep ocean current is the differences in the water, which are typically due to variations in salinity and temperature.

mrs_skeptik [129]2 years ago

5 0

The movement of plates

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What will the resistance be for a lamp that draws 4.6 amps of current from a 120-volt outlet? A. 550 ohms B. 115 ohms C. 26 ohms

Paraphin [41]

R = V/I = 120 / 4.6 ≈ 26 ohms.

Option C.

3 0

2 years ago

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PLEASE HURRY

Sophie [7]

Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)

• The net force in the parallel direction is

∑ F (para) = -mg sin(21°) - f = ma

• The net force in the perpendicular direction is

∑ F (perp) = n - mg cos(21°) = 0

Solving the second equation for n gives

n = mg cos(21°)

n = (0.200 kg) (9.80 m/s²) cos(21°)

n ≈ 1.83 N

Then the magnitude of friction is

f = µn

f = 0.25 (1.83 N)

f ≈ 0.457 N

Solve for the acceleration a :

-mg sin(21°) - f = ma

a = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)

a ≈ -5.80 m/s²

so the block is decelerating with magnitude

a = 5.80 m/s²

down the ramp.

5 0

2 years ago

Two children sit on different sides of a seesaw. The first child of mass 27 kg sits 1.5 m from the center. How far must the seco

erma4kov [3.2K]

Answer:

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Explanation:

jsjszhxhbssbzbbzbdfnndsndnndndnsndnfndndnsmsnsxnbdbsndndndnxbcbwnnbfndbdxbbdbvh go do some gehehhsdhehndnsbxhdbdj

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3 years ago

A purse at radius 2.30 m and a wallet at radius 3.45 m travel in uniform circular motion on the floor of a merry-go-round as the

ivolga24 [154]

Answer:

The acceleration of the wallet is $3\hat{i}+6\hat{j}$

Explanation:

Given that,

Radius of purse r= 2.30 m

Radius of wallet r'= 3.45 m

Acceleration of the purse $a=2\hat{i}+4.00\hat{j}$

We need to calculate the acceleration of the wallet

Using formula of acceleration

$a=r\omega^2$

Both the purse and wallet have same angular velocity

$\omega=\omega'$

$\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}$

$\dfrac{a}{r}=\dfrac{a'}{r'}$

$\dfrac{a'}{a}=\dfrac{r'}{r}$

$\dfrac{a'}{a}=\dfrac{3.45}{2.30}$

$\dfrac{a'}{a}=\dfrac{3}{2}$

$a'=\dfrac{3}{2}\times(2\hat{i}+4.00\hat{j})$

$a'=3\hat{i}+6\hat{j}$

Hence, The acceleration of the wallet is $3\hat{i}+6\hat{j}$

4 0

2 years ago

A 2.1 kg block is dropped from rest from a height of 5.5 m above the top of the spring. When the block is momentarily at rest, t

ella [17]

Answer:

The speed of the block is 8.2 m/s

Explanation:

Given;

mass of block, m = 2.1 kg

height above the top of the spring, h = 5.5 m

First, we determine the spring constant based on the principle of conservation of potential energy

¹/₂Kx² = mg(h +x)

¹/₂K(0.25)² = 2.1 x 9.8(5.5 +0.25)

0.03125K = 118.335

K = 118.335 / 0.03125

K = 3786.72 N/m

Total energy stored in the block at rest is only potential energy given as:

E = U = mgh

U = 2.1 x 9.8 x 5.5 = 113.19 J

Work done in compressing the spring to 15.0 cm:

W = ¹/₂Kx² = ¹/₂ (3786.72)(0.15)² = 42.6 J

This is equal to elastic potential energy stored in the spring,

Then, kinetic energy of the spring is given as:

K.E = E - W

K.E = 113.19 J - 42.6 J

K.E = 70.59 J

To determine the speed of the block due to this energy:

KE = ¹/₂mv²

70.59 = ¹/₂ x 2.1 x v²

70.59 = 1.05v²

v² = 70.59 / 1.05

v² = 67.229

v = √67.229

v = 8.2 m/s

8 0

2 years ago

Read 2 more answers