30.6*X+-2.56 Pls help

Do thermohaline currents flow vertically or horizontally A. Vertically B. Horizontally

Shkiper50 [21]

B

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7 0

3 years ago

Use the SI prefixes in Table 3 of this chapter to convert these hypothetical units of measure into appropriate quantities: a. 10

8090 [49]

A. 10 rations = 1 deca-ration.

b. 2000 mockingbirds = 2 x 10³ = 2 kilo-mockingbirds.

c. 10⁻⁵ phones = 1 micro-phones.

d. 10⁻⁹ goats = 1 nano-goats.

e. 1018 miners = 1.018 x 10³ = 1.018 kilo-miners.

4 0

2 years ago

A spinning tossed baseball veers off course in the direction ofA) reduced air pressure on the ball.B) increased air pressure on

katovenus [111]

Answer:

A) reduced air pressure on the ball.

Explanation:

3 0

2 years ago

If the truck has a mass of 2,000 kilograms , what's its momentum?(v=35 m/s)

katen-ka-za [31]

Answer:

$\boxed {\boxed {\sf 70,000 \ kg*m/s}}$

Explanation:

Momentum is the product of mass and velocity.

$p=m*v$

The mass of the truck is 2,000 kilograms and the velocity is 35 meters per second.

$m= 2000 \ kg \\v= 35 \ m/s$

Substitute the values into the formula and multiply.

$p= 2000 \ kg * 35 \ m/s \\p= 70,000 \ kg*m/s$

The truck's momentum is 70,000 kilograms meters per second.

8 0

3 years ago

A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. W

Liula [17]

Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

Displacement of ball (S) = -1.25 m

Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

$v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s$

Since, the motion is downward, final velocity must be negative. So,

$v_d=-4.95\ m/s$

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity ( $v_d$ ) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

$v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s$

Since, the motion is upward, final velocity must be positive. So,

$v_{up}=3.31\ m/s$

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

$J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns$

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

5 0

2 years ago